## Class 10 study materials part 7

Watch the videos and answer the questions

**1.Multiple choice questions.**

**(i) Based on the information given below classify each of the situations as ‘suffering from water scarcity’ or ‘not suffering from water scarcity’.**

**(a) A region with high annual rainfall.**

**(b) A region having high annual rainfall and large population.**

**(c) A region having high annual rainfall but water is highly polluted.**

**(d) A region having low rainfall and low population.**

**Answer:**

(a) A region with high annual rainfall – Not suffering from water scarcity

(b) A region having high annual rainfall and large population – Suffering from water scarcity

(c) A region having high annual rainfall but water is highly polluted – Suffering from water scarcity

(d) A region having low rainfall and low population- Not suffering from water scarcity

**(ii) Which one of the following statements is not an argument in favour of multipurpose river projects?**

**(a) Multi-purpose projects bring water to those areas which suffer from water scarcity.**

**(b) Multi-purpose projects by regulating water flow help to control floods.**

**(c) Multi-purpose projects lead to large scale displacements and loss of livelihood.**

**(d) Multi-purpose projects generate electricity for our industries and our homes.**

**Answer:**

(c) Multi-purpose projects lead to large scale displacements and loss of livelihood

**(iii) Here are some false statements. Identify the mistakes and rewrite them correctly.**

**(a) Multiplying urban centres with large and dense populations and urban lifestyles have helped in proper utilisation of water resources.**

**(b) Regulating and damming of rivers does not affect the river’s natural flow and its sediment flow.**

**(c) In Gujarat, the Sabarmati basin farmers were not agitated when higher priority was given to water supply in urban areas, particularly during droughts.**

**(d) Today in Rajasthan, the practice of rooftop rainwater water harvesting has gained popularity despite high water availability due to the Indira Gandhi Canal.**

**Answer:**

(a) Multiplying urban centres with large and dense populations and urban lifestyles have resulted in improper utilisation of water resources.

(b) Regulating and damming of rivers does affect the river’s natural flow and its sediment flow.

(c) In Gujarat, the Sabarmati basin farmers were agitated when higher priority was given to water supply in urban areas, particularly during droughts.

(d) Today in Rajasthan, the practise of rooftop rainwater water harvesting popularity has declined due to high water availability from Indira Gandhi Canal.

**2. Answer the following questions in about 30 words.**

**(i) Explain how water becomes a renewable resource.**

**Answer:**

Water can be considered as a renewable resource since there will be rains and surface water and groundwater will get recharged continuously due to the 3 process involved in the hydrological cycle.

The 3 processes of the hydrological cycle are

- Evaporations
- Condensation
- Precipitation

**(ii) What is water scarcity and what are its main causes?**

**Answer:**

Many of our cities are such examples. Thus, water scarcity may be an outcome of large and growing population and consequent greater demands for water, and unequal access to it. A large population requires more water not only for domestic use but also to produce more food. Hence, to facilitate higher food-grain production, water resources are being over-exploited to expand irrigated areas for dry-season agriculture. Irrigated agriculture is the largest consumer of water. Most farmers have their own wells and tube-wells in their farms for irrigation to increase their productivity. This has adversely affected water availability and food security of the people.

**(iii) Compare the advantages and disadvantages of multi-purpose river projects.**

**Answer:**

Advantages:

- Irrigation
- Electricity generation
- Flood control
- Water supply for industrial and domestic purposes.
- Tourist attraction
- Inland navigation

Disadvantages:

- The natural flow of water is affected
- Aquatic life gets affected
- Submergence of land in the surrounding areas
- Ecological consequences
- Large scale displacement of local people.

**3. Answer the following questions in about 120 words.**

**(i) Discuss how rainwater harvesting in semi-arid regions of Rajasthan is carried out.**

**Answer:**

In the semi-arid and arid regions of Rajasthan, particularly in Bikaner, Phalodi and Barmer, almost all the houses traditionally had underground tanks or tankas for storing drinking water. The tanks could be as large as a big room; one household in Phalodi had a tank that was 6.1 metres deep, 4.27 metres long and 2.44 metres wide. The tankas were part of the well-developed rooftop rainwater harvesting system and were built inside the main house or the courtyard. They were connected to the sloping roofs of the houses through a pipe. Rain falling on the rooftops would travel down the pipe and be stored in these underground ‘tankas’. The first spell of rain was usually not collected as this would clean the roofs and the pipes. The rainwater from the subsequent showers was then collected. The rainwater can be stored in the tankas till the next rainfall making it an extremely reliable source of drinking water when all other sources are dried up, particularly in the summers. Rainwater, or ‘palar pani’, as commonly referred to in these parts, is considered the purest form of natural water.

**(ii) Describe how modern adaptations of traditional rainwater harvesting methods are being carried out to conserve and store water.**

**Answer:**

Water harvesting system is a viable alternative, both socio-economically and environmentally. In ancient India, along with the sophisticated hydraulic structures, there existed an extraordinary tradition of water-harvesting system. People had in-depth knowledge of rainfall regimes and soil types and developed wide-ranging techniques to harvest rainwater, groundwater, river water and floodwater in keeping with the local ecological conditions and their water needs. In hill and mountainous regions, people built diversion channels like the ‘guls’ or ‘kuls’ of the Western Himalayas for agriculture. ‘Rooftop rainwater harvesting’ was commonly practised to store drinking water, particularly in Rajasthan. In the flood plains of Bengal, people developed inundation channels to irrigate their fields. In arid and semi-arid regions, agricultural fields were converted into rain-fed storage structures that allowed the water to stand and moisten the soil like the ‘khadins’ in Jaisalmer and ‘Johads’ in other parts of Rajasthan. In Gendathur, a remote backward village in Mysuru, Karnataka, villagers have installed, in their household’s rooftop, rainwater harvesting system to meet their water needs. Nearly 200 households have installed this system and the village has earned the rare distinction of being rich in rainwater. Rooftop rainwater harvesting is the most common practice in Shillong, Meghalaya. It is interesting because Cherapunjee and Mawsynram situated at a distance of 55 km. from Shillong receive the highest rainfall in the world, yet the state capital Shillong faces an acute shortage of water. Nearly every household in the city has a rooftop rainwater harvesting structure. Nearly 15-25 per cent of the total water requirement of the household comes from rooftop water harvesting. Tamil Nadu is the first state in India which has made rooftop rainwater harvesting structure compulsory to all the houses across the state.

**1. In which of the following situations, does the list of numbers involved make as arithmetic progression and why?**

**(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.**

**(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.**

**Solution:**

We can write the given condition as;

Taxi fare for 1 km = 15

Taxi fare for first 2 kms = 15+8 = 23

Taxi fare for first 3 kms = 23+8 = 31

Taxi fare for first 4 kms = 31+8 = 39

And so on……

Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.

**(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.**

**Solution:**

Let the volume of air in a cylinder, initially, be *V* litres.

In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time. Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.

Therefore, volumes will be *V*, 3*V*/4 , (3*V*/4)^{2} , (3*V*/4)^{3}…and so on

Clearly, we can see here, the adjacent terms of this series do not have the common difference between them. Therefore, this series is not an A.P.

**(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.**

**Solution:**

We can write the given condition as;

Cost of digging a well for first metre = Rs.150

Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200

Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250

Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300

And so on..

Clearly, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.

**(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.**

**Solution:**

We know that if Rs. P is deposited at *r*% compound interest per annum for n years, the amount of money will be:

P(1+r/100)^{n}

Therefore, after each year, the amount of money will be;

10000(1+8/100), 10000(1+8/100)^{2}, 10000(1+8/100)^{3}……

Clearly, the terms of this series do not have the common difference between them. Therefore, this is not an A.P.

**2. Write first four terms of the A.P. when the first term a and the common difference are given as follows**:

**(i) a = 10, d = 10**

(ii) a = -2, d = 0

(iii) a = 4, d = – 3

(iv) a = -1 d = 1/2

(v) a = – 1.25, d = – 0.25

**Solutions:**

(i) *a* = 10, *d* = 10

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = 10

*a*_{2} = *a*_{1}+*d* = 10+10 = 20

*a*_{3} = *a*_{2}+*d* = 20+10 = 30

*a*_{4} = *a*_{3}+*d* = 30+10 = 40

*a*_{5} = *a*_{4}+*d* = 40+10 = 50

And so on…

Therefore, the A.P. series will be 10, 20, 30, 40, 50 …

And First four terms of this A.P. will be 10, 20, 30, and 40.

(ii) *a* = – 2, *d* = 0

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = -2

*a*_{2} = *a*_{1}+*d* = – 2+0 = – 2

*a*_{3} = *a*_{2}+d = – 2+0 = – 2

*a*_{4} = *a*_{3}+*d* = – 2+0 = – 2

Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …

And, First four terms of this A.P. will be – 2, – 2, – 2 and – 2.

(iii) *a* = 4, *d* = – 3

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = 4

*a*_{2} = *a*_{1}+*d* = 4-3 = 1

*a*_{3} = *a*_{2}+*d* = 1-3 = – 2

*a*_{4} = *a*_{3}+*d* = -2-3 = – 5

Therefore, the A.P. series will be 4, 1, – 2 – 5 …

And, First four terms of this A.P. will be 4, 1, – 2 and – 5.

(iv) *a* = – 1, *d* = 1/2

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{2} = *a*_{1}+*d* = -1+1/2 = -1/2

*a*_{3} = *a*_{2}+*d* = -1/2+1/2 = 0

*a*_{4} = *a*_{3}+*d* = 0+1/2 = 1/2

Thus, the A.P. series will be-1, -1/2, 0, 1/2

And First four terms of this A.P. will be -1, -1/2, 0 and 1/2.

(v) *a* = – 1.25, *d* = – 0.25

Let us consider, the Arithmetic Progression series be *a _{1}, a_{2}, a_{3}, a_{4}, a_{5} …*

*a*_{1} = *a* = – 1.25

*a*_{2} = *a*_{1} + *d* = – 1.25-0.25 = – 1.50

*a*_{3} = *a*_{2} + *d* = – 1.50-0.25 = – 1.75

*a*_{4} = *a*_{3} + *d* = – 1.75-0.25 = – 2.00

Therefore, the series will be 1.25, – 1.50, – 1.75, – 2.00 ……..

And first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.

**3. For the following A.P.s, write the first term and the common difference.**

(i) 3, 1, – 1, – 3 …

(ii) -5, – 1, 3, 7 …

(iii) 1/3, 5/3, 9/3, 13/3 ….

(iv) 0.6, 1.7, 2.8, 3.9 …

**Solutions**

**(i) Given series,**

**3, 1, – 1, – 3 …**

First term, *a* = 3

Common difference, *d* = Second term – First term

⇒ 1 – 3 = -2

⇒ d = -2

**(ii) Given series, – 5, – 1, 3, 7 …**

First term, *a* = -5

Common difference, *d* = Second term – First term

⇒ ( – 1)-( – 5) = – 1+5 = 4

**(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….**

First term, *a* = 1/3

Common difference, *d* = Second term – First term

⇒ 5/3 – 1/3 = 4/3

**(iv) Given series, 0.6, 1.7, 2.8, 3.9 …**

First term, *a* = 0.6

Common difference, *d* = Second term – First term

⇒ 1.7 – 0.6

⇒ 1.1

**4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.**

**(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) **

*a*, 2

*a*, 3

*a*, 4

*a*… (xi)

*a*,

*a*

^{2},

*a*

^{3},

*a*

^{4}… (xii) √2, √8, √18, √32 … (xiii) √3, √6, √9, √12 … (xiv) 1

^{2}, 3

^{2}, 5

^{2}, 7

^{2}… (xv) 1

^{2}, 5

^{2}, 7

^{2}, 7

^{3}…

**Solution**

(i) Given to us,

**2, 4, 8, 16 …
**

Here, the common difference is;

*a*_{2} – *a*_{1} = 4-2 = 2

*a*_{3} – *a*_{2} = 8-4 = 4

*a _{4}* –

*a*

_{3}= 16-8 = 8

Since, *a _{n}*

_{+1}–

*a*or the common difference is not the same every time.

_{n }Therefore, the given series are not forming an A.P.

**(ii) Given, 2, 5/2, 3, 7/2 ….**

Here,

*a*_{2} – *a*_{1} = 5/2-2 = 1/2

*a*_{3} – *a*_{2} = 3-5/2 = 1/2

*a*_{4} – *a*_{3} = 7/2-3 = 1/2

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = 1/2 and the given series are in A.P.

The next three terms are;

*a*_{5} = 7/2+1/2 = 4

*a*_{6} = 4 +1/2 = 9/2

*a*_{7} = 9/2 +1/2 = 5

**(iii) Given, -1.2, – 3.2, -5.2, -7.2 …
**

Here,

*a*_{2} – *a*_{1} = (-3.2)-(-1.2) = -2

*a*_{3} – *a*_{2} = (-5.2)-(-3.2) = -2

*a*_{4} – *a*_{3} = (-7.2)-(-5.2) = -2

Since, *a _{n}*

_{+1}–

*a*or common difference is same every time.

_{n}Therefore, *d* = -2 and the given series are in A.P.

Hence, next three terms are;

*a*_{5} = – 7.2-2 = -9.2

*a*_{6} = – 9.2-2 = – 11.2

*a*_{7} = – 11.2-2 = – 13.2

**(iv) Given, -10, – 6, – 2, 2 …
**

Here, the terms and their difference are;

*a*_{2} – *a*_{1} = (-6)-(-10) = 4

*a*_{3} – *a*_{2} = (-2)-(-6) = 4

*a*_{4} – *a*_{3} = (2 -(-2) = 4

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = 4 and the given numbers are in A.P.

Hence, next three terms are;

*a*_{5} = 2+4 = 6

*a*_{6} = 6+4 = 10

*a*_{7} = 10+4 = 14

**(v) Given, 3, 3+√2, 3+2√2, 3+3√2**

Here,

*a*_{2} – *a*_{1} = 3+√2-3 = √2

*a*_{3} – *a*_{2} = (3+2√2)-(3+√2) = √2

*a*_{4} – *a*_{3} = (3+3√2) – (3+2√2) = √2

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = √2 and the given series forms a A.P.

Hence, next three terms are;

*a*_{5} = (3+√2) +√2 = 3+4√2

*a*_{6} = (3+4√2)+√2 = 3+5√2

*a*_{7} = (3+5√2)+√2 = 3+6√2

**(vi) 0.2, 0.22, 0.222, 0.2222 ….
**

Here,

*a*_{2} – *a*_{1} = 0.22-0.2 = 0.02

*a*_{3} – *a*_{2} = 0.222-0.22 = 0.002

*a*_{4} – *a*_{3} = 0.2222-0.222 = 0.0002

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, and the given series doesn’t forms a A.P.

**(vii) 0, -4, -8, -12 …
**

Here,

*a*_{2} – *a*_{1} = (-4)-0 = -4

*a*_{3} – *a*_{2} = (-8)-(-4) = -4

*a*_{4} – *a*_{3} = (-12)-(-8) = -4

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = -4 and and the given series forms a A.P.

Hence, next three terms are;

*a*_{5} = -12-4 = -16

*a*_{6} = -16-4 = -20

*a*_{7} = -20-4 = -24

**(viii) -1/2, -1/2, -1/2, -1/2 ….
**

Here,

*a*_{2} – *a*_{1} = (-1/2) – (-1/2) = 0

*a*_{3} – *a*_{2} = (-1/2) – (-1/2) = 0

*a*_{4} – *a*_{3} = (-1/2) – (-1/2) = 0

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = 0 and the given series forms a A.P.

Hence, next three terms are;

*a*_{5} = (-1/2)-0 = -1/2

*a*_{6} = (-1/2)-0 = -1/2

*a*_{7} = (-1/2)-0 = -1/2

**(ix) 1, 3, 9, 27 …
**

Here,

*a*_{2} – *a*_{1} = 3-1 = 2

*a*_{3} – *a*_{2} = 9-3 = 6

*a*_{4} – *a*_{3} = 27-9 = 18

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, and the given series doesn’t forms a A.P.

**(x) a, 2a, 3a, 4a …**

Here,

*a*_{2} – *a*_{1} = 2*a*–*a *= *a*

*a*_{3} – *a*_{2} = 3*a*-2*a* = *a*

*a*_{4} – *a*_{3} = 4*a*-3*a* = *a*

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = *a* and the given series forms a A.P.

Hence, next three terms are;

*a*_{5} = 4*a*+*a* = 5*a*

*a*_{6} = 5*a*+*a* = 6*a*

*a*_{7} = 6*a*+*a* = 7*a*

**(xi) a, a^{2}, a^{3}, a^{4} …**

Here,

*a*_{2} – *a*_{1} = *a*^{2}–*a* = a(*a*-1)

*a*_{3} – *a*_{2} = *a*^{3 }–^{ }*a*^{2 }= *a*^{2}(*a*-1)

*a*_{4} – *a*_{3} = *a*^{4} – *a*^{3 }= *a*^{3}(*a*-1)

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, the given series doesn’t forms a A.P.

**(xii) √2, √8, √18, √32 …
**

Here,

*a*_{2} – *a*_{1} = √8-√2 = 2√2-√2 = √2

*a*_{3} – *a*_{2} = √18-√8 = 3√2-2√2 = √2

*a*_{4} – *a*_{3} = 4√2-3√2 = √2

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = √2 and the given series forms a A.P.

Hence, next three terms are;

*a*_{5} = √32+√2 = 4√2+√2 = 5√2 = √50

*a*_{6} = 5√2+√2 = 6√2 = √72

*a*_{7} = 6√2+√2 = 7√2 = √98

**(xiii) √3, √6, √9, √12 …**

Here,

*a*_{2} – *a*_{1} = √6-√3 = √3×√2-√3 = √3(√2-1)

*a*_{3} – *a*_{2} = √9-√6 = 3-√6 = √3(√3-√2)

*a*_{4} – *a*_{3} = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, the given series doesn’t forms a A.P.

**(xiv) 1 ^{2}, 3^{2}, 5^{2}, 7^{2} …**

Or, 1, 9, 25, 49 …..

Here,

*a*_{2} − *a*_{1} = 9−1 = 8

*a*_{3} − *a*_{2 }= 25−9 = 16

*a*_{4} − *a*_{3} = 49−25 = 24

Since, *a _{n}*

_{+1}–

*a*or the common difference is not same every time.

_{n}Therefore, the given series doesn’t forms a A.P.

**(xv) 1 ^{2}, 5^{2}, 7^{2}, 73 …**

Or 1, 25, 49, 73 …

Here,

*a*_{2} − *a*_{1} = 25−1 = 24

*a*_{3} − *a*_{2 }= 49−25 = 24

*a*_{4} − *a*_{3} = 73−49 = 24

Since, *a _{n}*

_{+1}–

*a*or the common difference is same every time.

_{n}Therefore, *d* = 24 and the given series forms a A.P.

Hence, next three terms are;

*a*_{5} = 73+24 = 97

*a*_{6} = 97+24 = 121

*a*_{7 }= 121+24 = 145

#### Exercise 5.2 Page: 105

**1. Fill in the blanks in the following table, given that a is the first term, d the common difference and a_{n} the n^{th} term of the A.P.**

**Solutions:**

**(i) **Given,

First term, *a* = 7

Common difference, *d* = 3

Number of terms, *n* = 8,

We have to find the nth term, *a _{n}* = ?

As we know, for an A.P.,

*a _{n}* =

*a*+(

*n*−1)

*d*

Putting the values,

=> 7+(8 −1) 3

=> 7+(7) 3

=> 7+21 = 28

Hence,* a _{n}* = 28

**(ii)** Given,

First term, *a* = -18

Common difference, *d* = ?

Number of terms, *n* = 10

Nth term, *a _{n}* = 0

As we know, for an A.P.,

*a _{n}* =

*a*+(

*n*−1)

*d*

Putting the values,

0 = − 18 +(10−1)*d*

18 = 9*d*

*d* = 18/9 = 2

Hence, common difference, *d *= 2

**(iii)** Given,

First term, *a* = ?

Common difference, *d* = -3

Number of terms, *n* = 18

Nth term, *a _{n}* = -5

As we know, for an A.P.,

*a _{n}* =

*a*+(

*n*−1)

*d*

Putting the values,

−5 = *a*+(18−1) (−3)

−5 = *a*+(17) (−3)

−5 = *a*−51

*a* = 51−5 = 46

Hence, *a* = 46

**(iv)** Given,

First term, *a* = -18.9

Common difference, *d* = 2.5

Number of terms, *n* = ?

Nth term, *a _{n}* = 3.6

As we know, for an A.P.,

*a _{n}* =

*a*+(

*n*−1)

*d*

Putting the values,

3.6 = − 18.9+(*n* −1)2.5

3.6 + 18.9 = (*n*−1)2.5

22.5 = (*n*−1)2.5

(*n* – 1) = 22.5/2.5

*n* – 1 = 9

*n* = 10

Hence, *n* = 10

**(v)** Given,

First term, *a* = 3.5

Common difference, *d* = 0

Number of terms, *n* = 105

Nth term, *a _{n}* = ?

As we know, for an A.P.,

*a _{n}* =

*a*+(

*n*−1)

*d*

Putting the values,

*a _{n}* = 3.5+(105−1) 0

*a _{n}* = 3.5+104×0

*a _{n}* = 3.5

Hence, *a _{n}* = 3.5

**3. In an AP
(i) Given a = 5, d = 3, a_{n} = 50, find n and S_{n}.
(ii) Given a = 7, a_{13} = 35, find d and S_{13}.
(iii) Given a_{12} = 37, d = 3, find a and S_{12}.
(iv) Given a_{3} = 15, S_{10} = 125, find d and a_{10}.
(v) Given d = 5, S_{9} = 75, find a and a_{9}.
(vi) Given a = 2, d = 8, S_{n} = 90, find n and a_{n}.
(vii) Given a = 8, a_{n} = 62, S_{n} = 210, find n and d.
(viii) Given a_{n} = 4, d = 2, S_{n} = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a.**

**Solutions:**

**(i)** Given that, *a* = 5, *d* = 3, *a _{n}* = 50

As we know, from the formula of the nth term in an AP,

*a*=

_{n}*a*+(

*n*−1)

*d*,

Therefore, putting the given values, we get,

⇒ 50 = 5+(

*n*-1)×3

⇒ 3(

*n*-1) = 45

⇒

*n*-1 = 15

⇒

*n*= 16

Now, sum of n terms,

*S _{n}* =

*n*/2 (

*a*+

*a*)

_{n}*S*= 16/2 (5 + 50) = 440

_{n}**(ii)** Given that, *a* = 7, *a*_{13} = 35

As we know, from the formula of the nth term in an AP,

*a _{n}* =

*a*+(

*n*−1)

*d*,

Therefore, putting the given values, we get,

⇒ 35 = 7+(13-1)

*d*

⇒ 12

*d*= 28

⇒

*d*= 28/12 = 2.33

Now,

*S*=

_{n}*n*/2 (

*a*+

*a*)

_{n}*S*= 13/2 (7+35) = 273

_{13}**(iii)**Given that, *a*_{12} = 37, *d* = 3

As we know, from the formula of the n^{th} term in an AP,

*a _{n}* =

*a*+(

*n*−1)

*d*,

Therefore, putting the given values, we get,

⇒

*a*

_{12}=

*a*+(12−1)3

⇒ 37 =

*a*+33

⇒

*a*= 4

Now, sum of nth term,

*S*=

_{n}*n*/2 (

*a*+

*a*)

_{n}*S*=

_{n}*12*/2 (4+37)

= 246

**(iv)** Given that, *a*_{3} = 15, *S*_{10} = 125

As we know, from the formula of the nth term in an AP,

*a _{n}* =

*a*+(

*n*−1)

*d*,

Therefore, putting the given values, we get,

*a*

_{3}=

*a*+(3−1)

*d*

15 =

*a*+2

*d*…………………………..

**(i)**

Sum of the nth term,

*S _{n}* =

*n*/2 [2

*a*+(

*n*-1)

*d*]

*S*= 10/2 [2

_{10}*a*+(10-1)

*d*]

125 = 5(2

*a*+9

*d*)

25 = 2

*a*+9

*d*………………………..

**(ii)**

On multiplying equation **(i)** by **(ii)**, we will get;

30 = 2*a*+4*d* ………………………………. **(iii)**

By subtracting equation **(iii)** from **(ii)**, we get,

−5 = 5*d*

*d* = −1

From equation **(i)**,

15 = *a*+2(−1)

15 = *a*−2

*a* = 17 = First term

*a*_{10} = *a*+(10−1)*d*

*a*_{10} = 17+(9)(−1)

*a*_{10} = 17−9 = 8

**(v)** Given that, *d* = 5, *S*_{9} = 75

As, sum of n terms in AP is,

*S _{n}* =

*n*/2 [2

*a*+(

*n*-1)

*d*]

Therefore, the sum of first nine terms are;

*S*= 9/2 [2

_{9}*a*+(9-1)

*5*]

25 = 3(

*a*+20)

25 = 3

*a*+60

3

*a*= 25−60

*a*= -35/3

As we know, the n

^{th}term can be written as;

*a*=

_{n}*a*+(

*n*−1)

*d*

*a*

_{9}=

*a*+(9−1)(5)

= -35/3+8(5)

= -35/3+40

= (35+120/3) = 85/3

**(vi)** Given that, *a* = 2, *d* = 8, *S _{n}* = 90

As, sum of n terms in an AP is,

*S _{n}* =

*n*/2 [2

*a*+(

*n*-1)

*d*]

90 =

*n*/2 [2

*a*+(

*n*-1)

*d*]

⇒ 180 =

*n*(4+8

*n*-8) =

*n*(8

*n*-4) = 8

*n*

^{2}-4

*n*

⇒ 8

*n*

^{2}-4

*n –*180 = 0

⇒ 2

*n*

^{2}–

*n*-45 = 0

⇒ 2

*n*

^{2}-10

*n*+9

*n*-45 = 0

⇒ 2

*n*(

*n*-5)+9(

*n*-5) = 0

⇒ (2

*n*-9)(2

*n*+9) = 0

So,

*n*= 5 (as it is positive integer)

∴

*a*

_{5}_{ }= 8+5×4 = 34

**(vii)** Given that, *a* = 8, *a _{n}* = 62,

*S*= 210

_{n}As, sum of n terms in an AP is,

*S*=

_{n}*n*/2 (

*a*+

*a*)

_{n}210 =

*n*/2 (8 +62)

⇒ 35

*n*= 210

⇒

*n*= 210/35 = 6

Now, 62 = 8+5*d*

⇒ 5*d* = 62-8 = 54

⇒ *d* = 54/5 = 10.8

**(viii)** Given that, n^{th} term, *a _{n}* = 4, common difference,

*d*= 2, sum of n terms,

*S*= −14.

_{n}As we know, from the formula of the n

^{th}term in an AP,

*a*=

_{n}*a*+(

*n*−1)

*d*,

Therefore, putting the given values, we get,

4 =

*a*+(

*n*−1)2

4 =

*a*+2

*n*−2

*a*+2

*n*= 6

*a*= 6 − 2

*n*………………………………………….

**(i)**

As we know, the sum of n terms is;

*S _{n}* =

*n*/2 (

*a*+

*a*)

_{n}-14 =

*n*/2 (

*a*+

*4*)

−28 =

*n*(

*a*+4)

−28 =

*n*(6 −2

*n*+4) {From equation

**(i)**}

−28 =

*n*(− 2

*n*+10)

−28 = − 2

*n*

^{2}+10

*n*

2

*n*

^{2}−10

*n*− 28 = 0

*n*

^{2}−5

*n*−14 = 0

*n*

^{2}−7

*n+*2

*n*−14 = 0

*n*(

*n*−7)+2(

*n*−7) = 0

(

*n*−7)(

*n*+2) = 0

Either *n* − 7 = 0 or *n* + 2 = 0

*n* = 7 or *n* = −2

However, *n* can neither be negative nor fractional.

Therefore, *n* = 7

From equation **(i)**, we get

*a* = 6−2*n*

*a* = 6−2(7)

= 6−14

= −8

(ix) Given that, first term, *a* = 3,

Number of terms, *n* = 8

And sum of n terms, *S* = 192

As we know,

*S _{n}* =

*n*/2 [2

*a*+(

*n*-1)

*d*]

192 = 8/2 [2×3+(8 -1)

*d*]

192 = 4[6 +7

*d*]

48 = 6+7

*d*

42 = 7

*d*

*d =*6

(x) Given that, *l* = 28,*S* = 144 and there are total of 9 terms.

*Sum of n terms formula,*

*S _{n}* =

*n*/2 (

*a*+

*l*)

144 = 9/2(

*a*+28)

(16)×(2) =

*a*+28

32 =

*a*+28

*a*= 4

**4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?**

**Solutions:**

Let there be *n* terms of the AP. 9, 17, 25 …

For this A.P.,

First term, *a* = 9

*Common difference, d* = *a*_{2}−*a*_{1} = 17−9 = 8

As, the sum of n terms, is;

*S _{n}* =

*n*/2 [2

*a*+(

*n*-1)

*d*]

636 =

*n*/2 [2×

*a*+(8-1)×8]

636 =

*n*/2 [18+(

*n*-1)×8]

636 =

*n*[9 +4

*n*−4]

636 =

*n*(4

*n*+5)

4

*n*

^{2}+5

*n*−636 = 0

4

*n*

^{2}+53

*n*−48

*n*−636 = 0

*n*(4

*n*+ 53)−12 (4

*n*+ 53) = 0

(4

*n*+53)(

*n*−12) = 0

Either 4*n*+53 = 0 or *n*−12 = 0

*n* = (-53/4) or *n* = 12

*n *cannot be negative or fraction, therefore, *n* = 12 only.

**5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.**

**Solution:**Given that,

*first term, a* = 5

*last term, l* = 45

*Sum of the AP, S _{n}* = 400

As we know, the sum of AP formula is;

*S*=

_{n}*n*/2 (

*a*+

*l*)

400 =

*n*/2(5+45)

400 =

*n*/2(50)

*Number of terms, n*=16

As we know, the last term of AP series can be written as;

*l = a+*(

*n*−1)

*d*

45 = 5 +(16 −1)

*d*

40 = 15

*d*

*Common difference, d*= 40/15 = 8/3

**6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?**

**Solution: **Given that,

*First term, a* = 17

*Last term, l* = 350

*Common difference, d* = 9

Let there be *n* terms in the A.P., thus the formula for last term can be written as;

*l = a+*(*n* −1)*d*

350 = 17+(*n* −1)9

333 = (*n*−1)9

(*n*−1) = 37

*n* = 38

*S _{n}* =

*n*/2 (

*a*+

*l*)

*S*= 13/2 (17+350)

_{38}= 19×367

= 6973

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

**7. Find the sum of first 22 terms of an AP in which d = 7 and 22^{nd} term is 149.**

Solution:Given,

Common difference, d = 7

22^{nd} term, a_{22} = 149

Sum of first 22 term, S_{22} = ?

By the formula of nth term,

*a _{n}* =

*a*+(

*n*−1)

*d*

*a*

_{22}=

*a*+(22−1)

*d*

149 =

*a*+21×7

149 =

*a*+147

*a*= 2 = First term

Sum of n terms,

*S*=

_{n}*n*/2(

*a*+

*a*)

_{n}= 22/2 (2+149)

= 11×151

= 1661

**8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
**

**Solution:**Given that,

*Second term, a*

_{2}= 14

*Third term, a*

_{3}= 18

*Common difference, d*=

*a*

_{3}−

*a*

_{2}= 18−14 = 4

*a*

_{2}=

*a*+

*d*

14 =

*a*+4

*a*= 10 = First term

Sum of n terms;

*S*=

_{n}*n*/2 [2

*a*+ (

*n*– 1)

*d*]

*S*= 51/2 [2×10 (51-1) 4]

_{51}= 51/2[2+(20)×4]

= 51×220/2

= 51×110

= 5610

**9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.**

**Solution: **Given that,

*S*_{7} = 49

S_{17} = 289

We know, Sum of nth term;

*S _{n}* =

*n*/2 [2

*a*+ (

*n*– 1)

*d*]

Therefore,

*S*=

_{7}*7*/2 [2

*a*+(

*n*-1)

*d*]

*S*= 7/2 [2

_{7}*a*+ (7 -1)

*d*]

49 = 7/2 [2

*a*+16

*d*]

7 = (

*a*+3

*d*)

*a*+ 3

*d*= 7 ………………………………….

**(i)**

In the same way,

*S _{17}* = 17/2 [2

*a*+(17-1)

*d*]

289 = 17/2 (2

*a*+16

*d*)

17 = (

*a*+8

*d*)

*a*+8

*d*= 17 ……………………………….

**(ii)**

Subtracting equation **(i)** from equation **(ii)**,

5*d* = 10

*d* = 2

From equation **(i)**, we can write it as;

*a*+3(2) = 7

*a+ *6 = 7

*a = *1

Hence,

*S _{n}* =

*n*/2[2

*a*+(

*n*-1)

*d*]

=

*n*/2[2(1)+(

*n*– 1)×2]

=

*n*/2(2+2

*n*-2)

=

*n*/2(2

*n*)

=

*n*

^{2}

^{ }

**10. Show that**

*a*_{1},*a*_{2 }… ,*a*, … form an AP where_{n}*a*is defined as below_{n}**(i)**

(ii)

Also find the sum of the first 15 terms in each case.

*a*= 3+4_{n}*n*(ii)

*a*= 9−5_{n}*n*Also find the sum of the first 15 terms in each case.

**Solutions:**

(i) *a _{n}* = 3+4

*n*

*a*

_{1}= 3+4(1) = 7

*a*

_{2}= 3+4(2) = 3+8 = 11

*a*

_{3}= 3+4(3) = 3+12 = 15

*a*

_{4}= 3+4(4) = 3+16 = 19

We can see here, the common difference between the terms are;

*a*_{2} − *a*_{1} = 11−7 = 4

*a*_{3} − *a*_{2} = 15−11 = 4

*a*_{4} − *a*_{3} = 19−15 = 4

Hence, *a _{k}*

_{ + 1}−

*a*is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.

_{k}Now, we know, the sum of nth term is;

*S*=

_{n}*n*/2[2

*a*+(

*n*-1)

*d*]

*S*= 15/2[2(7)+(15-1)×4]

_{15 }= 15/2[(14)+56]

= 15/2(70)

= 15×35

= 525

(ii) *a _{n}* = 9−5

*n*

*a*

_{1}= 9−5×1 = 9−5 = 4

*a*

_{2}= 9−5×2 = 9−10 = −1

*a*

_{3}= 9−5×3 = 9−15 = −6

*a*

_{4}= 9−5×4 = 9−20 = −11

We can see here, the common difference between the terms are;

*a*_{2} − *a*_{1} = −1−4 = −5

*a*_{3} − *a*_{2} = −6−(−1) = −5

*a*_{4} − *a*_{3} = −11−(−6) = −5

Hence, *a _{k}*

_{ + 1}−

*a*is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

_{k}Now, we know, the sum of nth term is;

*S*=

_{n}*n*/2 [2

*a*+(

*n*-1)

*d*]

*S*= 15/2[2(4) +(15 -1)(-5)]

_{15 }= 15/2[8 +14(-5)]

= 15/2(8-70)

= 15/2(-62)

= 15(-31)

= -465

**11. If the sum of the first n terms of an AP is 4n − n^{2}, what is the first term (that is S_{1})? What is the sum of first two terms? What is the second term? Similarly find the 3^{rd}, the10^{th} and the n^{th} terms.**

**Solution: **Given that,

*S _{n}* = 4

*n*−

*n*

^{2}

First term,

*a*=

*S*

_{1}= 4(1) − (1)

^{2}= 4−1 = 3

Sum of first two terms =

*S*

_{2}= 4(2)−(2)

^{2}= 8−4 = 4

Second term,

*a*

_{2}=

*S*

_{2}−

*S*

_{1}= 4−3 = 1

*Common difference, d*=

*a*

_{2}−

*a*= 1−3 = −2

N

^{th}term,

*a*=

_{n}*a*+(

*n*−1)

*d*

= 3+(

*n*−1)(−2)

= 3−2

*n*+2

= 5−2

*n*

Therefore, *a*_{3} = 5−2(3) = 5-6 = −1

*a*_{10} = 5−2(10) = 5−20 = −15

Hence, the sum of first two terms is 4. The second term is 1.

The 3^{rd}, the 10^{th}, and the *n*^{th} terms are −1, −15, and 5 − 2*n* respectively.

**12. Find the sum of first 40 positive integers divisible by 6.**

**Solution: **The positive integers that are divisible by 6 are 6, 12, 18, 24 ….

We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.

*a* = 6

*d* = 6

*S*_{40}* = *?

By the formula of sum of n terms, we know,

*S _{n}* =

*n*/2 [2

*a*+(

*n*– 1)

*d*]

Therefore, putting n = 40, we get,

*S*= 40/2 [2(6)+(40-1)6]

_{40 }= 20[12+(39)(6)]

= 20(12+234)

= 20×246

= 4920

**13. Find the sum of first 15 multiples of 8.**

**Solution:**The multiples of 8 are 8, 16, 24, 32…

The series is in the form of AP, having first term as 8 and common difference as 8.

Therefore, *a* = 8

*d* = 8

*S*_{15} = ?

By the formula of sum of nth term, we know,

*S _{n}* =

*n*/2 [2

*a*+(

*n*-1)

*d*]

*S*= 15/2 [2(8) + (15-1)8]

_{15}= 15/2[6 +(14)(8)]

= 15/2[16 +112]

= 15(128)/2

= 15 × 64

= 960

**14. Find the sum of the odd numbers between 0 and 50.**

**Solution: **The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.

Therefore, we can see that these odd numbers are in the form of A.P.

Hence,

First term, a = 1

Common difference, d = 2

Last term, *l* = 49

By the formula of last term, we know,

*l* = *a*+(*n*−1) *d*

49 = 1+(*n*−1)2

48 = 2(*n* − 1)

*n* − 1 = 24

*n* = 25 = Number of terms

By the formula of sum of nth term, we know,

*S _{n}* =

*n*/2(

*a*+

*l*)

*S*= 25/2 (1+49)

_{25}= 25(50)/2

=(25)(25)

= 625

**15. A contract on construction job specifies a penalty for delay of completion beyond a certain dateas follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.**

**Solution:**

We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.

*Therefore, a* = 200 and *d* = 50

Penalty that has to be paid if contractor has delayed the work by 30 days = *S*_{30}

By the formula of sum of nth term, we know,

*S _{n}* =

*n*/2[2

*a*+(

*n*-1)

*d*]

Therefore,

*S*

_{30}= 30/2[2(200)+(30 – 1)50]

= 15[400+1450]

= 15(1850)

= 27750

Therefore, the contractor has to pay Rs 27750 as penalty.

**1. Did Döbereiner’s triads also exist in the columns of Newlands’ Octaves? Compare and find out.**

**Solution:**

Döbereiner’s triads did exist in the columns of Newlands’ Octaves; For example, the elements

Lithium (Li), Potassium (K) and Sodium (Na) constitute a Dobereiner’s Triad but are also found in the second column of Newland’s Octaves.

**2. What were the limitations of Döbereiner’s classification?**

**Solution:**

(i) They were not applicable for very low mass or very high mass elements.

(ii) All the elements couldn’t fit into Dobereiner’s triads.

(iii) As the methods to calculate atomic mass improved, Dobereiner’s triads validity began to decrease. For example, in the triad of F, Cl and Br, the arithmetic mean of atomic masses of F and Br are not equal to the atomic mass of CI.

**3. What were the limitations of Newlands’ Law of Octaves?**

**Solution:**

- Limitations of Newlands’ Law of Octaves are as follows
- Newlands’ Law of Octaves applicable to elements up to Calcium
- Newland assumed there are 56 elements in the nature and no more elements would be discovered in the future.
- To fit elements into table Newland put two elements into one slot. Newland introduced unlike elements with different properties into one column.
- Iron (Fe) was placed away from elements that resembles in properties. Ex: Nickel and cobalt

**Questions Page number** **85**

**1. Use Mendeleev’s Periodic Table to predict the formulae for the oxides of the following elements: K, C, AI, Si, Ba.**

**Solution:**

K- K_{2}O

C-C_{2}O_{4} or CO_{2}

Al- Al_{2}O_{3}

Si-Si_{2}O_{4} or SiO_{2}

Ba_{2}O_{2} or BaO

Oxygen is a member of group VI A in Mendeleev’s periodic table. Its valency is 2. Similarly. The valencies of all the elements listed can be predicted from their respective groups. This will help in writing the formulae of their oxides.

(i) Potassium (K) is a member of group IA. Its valency is 1. Therefore, the formula of it is K_{2}O.

(ii) Carbon (C) is a member of group IV A. Its valency is 4. Therefore, the formula of it is C_{2}O_{4} or CO_{2}.

(iii) Aluminium (Al) belongs to groups III A and its valency is 3. The formula of its oxide is Al_{2}O_{3}.

(iv) Silicon (Si) is present in group IV A after carbon. Its valency is also 4. The formula oxide is Si_{2}O_{4} or SiO_{2}.

(v) Barium (Ba) belongs to group II A and the valency of the element is 2. The formula of oxide of the element is Ba_{2}O_{2} or BaO.

**2. Besides gallium, which other elements have since been discovered that were left by Mendeleev in his Periodic Table? (Any two)**

**Solution:**

Germanium and Scandium are the element that are left by Mendeleev in his Periodic Table since its discovery.

**3. What were the criteria used by Mendeleev in creating his Periodic Table?**

**Solution:**

Mendeleev concentrated on various compounds formed by the elements with Hydrogen and

Oxygen. Among physical properties, he observed the relationship between the atomic masses

of various elements while creating his periodic table.

**4. Why do you think the noble gases are placed in a separate group?**

**Solution:**

Noble gases are placed in a separate group because of their inert nature and low concentration in our atmosphere. They are kept in a separate group called Zero group so that they don’t disturb the existing order.

**Questions Page number** **90**

**1. How could the Modern Periodic Table remove various anomalies of Mendeleev’s Periodic Table?**

**Solution:**

(a) In the Modern Periodic table elements are arranged in the increasing order of their atomic number. This removes the anomaly regarding certain pairs of elements in Mendeleev’s periodic table.

(b) Atomic number of cobalt is 27 and nickel is 28. Hence cobalt will come before nickel even though its atomic mass is greater.

c) All isotopes of the same elements have different atomic masses, but same atomic number; therefore they are placed in the same position in the modern periodic table.

**2. Name two elements you would expect to show chemical reactions similar to magnesium. What is the basis for your choice?**

**Solution:**

Calcium and Beryllium are similar to Magnesium because all the three elements belong to the same group and have 2 valence electrons in their outer shell.

**3. Name**

**(a) Three elements that have a single electron in their outermost shells.**

**(b) Two elements that have two electrons in their outermost shells.**

**(c) Three elements with filled outermost shells**

**Solution:**

- Lithium (Li), Sodium (Na) and potassium (k) have a single electron in their outermost shells.
- Magnesium (Mg) and Calcium (Ca) have two electrons in their outermost shells
- Neon (Ne), Argon (Ar and Xenon (Xe) filled outermost shells

**4. a) Lithium, sodium, potassium are all metals that react with water to liberate hydrogen gas. Is there any similarity in the atoms of these elements?**

**(b) Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common?**

**Solution:**

They’ve one valence electron in their outermost shells and as a result of this, they are very unstable. So, they readily react with water to liberate hydrogen. They are also called alkali metals.

Their outermost shells are full leading to high stability. They react only in extreme

circumstances and hence are called noble gases.

**5. In the Modern Periodic Table, which are the metals among the first ten elements?**

**Solution:**

Lithium and Beryllium are the metals among the first ten elements in Modern Periodic Table.

**6. By considering their position in the Periodic Table, which one of the following elements would you expect to have maximum metallic characteristic? Ga Ge As Se Be**

**Solution:**

Among the elements listed in the question. Be and Ga are expected to be most metallic. Out of Be and Ga, Ga is bigger in size and hence has greater tendency to lose electrons than Be. Therefore, Ga is more metallic than Be.

**Exercise questions Page number** **91-92**

**1. Which of the following statements is not a correct statement about the trends when going from left to right across the periods of periodic Table.**

**(a) The elements become less metallic in nature.**

**(b) The number of valence electrons increases.**

**(c) The atoms lose their electrons more easily.**

**(d) The oxides become more acidic**

**Solution:**

Correct answer is c .The atoms lose their electrons more easily.

The atoms lose their electrons more easily is a wrong statement because as we move from left to right across the periods of the periodic table, the non-metallic character increases.

Therefore tendency to lose an electron decreases.

**2. Element X forms a chloride with the formula XCl _{2}, which is a solid with a high melting point. X would most likely be in the same group of the Periodic Table as (a) Na (b) Mg (c) AI (d) Si**

**Solution:**

Answer is Magnesium because Mg has the valency 2 which is same as the group (a) Na (b) Mg (c) AI (d) Si

Also Mg when combines chloride forms MgCl_{2}.

**3. Which element has?**

**(a) Two shells, both of which are completely filled with electrons?**

**(b) The electronic configuration 2, 8, 2?**

**(c) A total of three shells, with four electrons in its valence shell?**

**(d) A total of two shells, with three electrons in its valence shell?**

**(e) twice as many electrons in its second shell as in its first shell?**

**Solution:**

a) Neon has two shells which are completely filled.

b) Silicon has the electronic configuration 2, 8, 2

c) Carbon has a total of three shells, with four electrons in its valence shell

d) Boron a total of two shells, with three electrons in its valence shell

e) Magnesium has twice as many electrons in its second shell as in its first shell

**4. (a) What property do all elements in the same column of the Periodic Table as boron have in common?**

**(b) What property do all elements in the same column of the Periodic Table as fluorine have in common?**

**Solution:**

(a)All the elements which lie in me same column as that of boron belong to group 13. Therefore, they have three electrons in their respective valence shells. Except, boron which is a non-metal, all other elements (i.e., aluminum, gallium, indium and thallium) in this group are metals.

(b) All elements in the same column of the Periodic Table as fluorine have in common three electrons in their valence shell and they all are belong to group thirteen.

**5. An atom has electronic configuration 2, 8, 7.**

**(a) What is the atomic number of this element?**

**(b) To which of the following elements would it be chemically similar? (Atomic numbers are given in parentheses.) N(7), F(9), P(15), Ar(18)**

**Solution:**

(a)The element with electronic configuration (2+8+7) 17 is chlorine.

The no. of atomic number = no. of electrons

Therefore, atomic number is 17.

(b) An atom with electronic configuration 2, 8, 7 would be chemically similar to F (9)

**6. The position of three elements A, B and C in the Periodic Table are shown below–**

**Group 16 Group 17**

**– –**

**– A**

**– –**

**B C**

**(a) State whether A is a metal or non-metal.**

**(b) State whether C is more reactive or less reactive than A.**

**(c) Will C be larger or smaller in size than B?**

**(d) Which type of ion, cation or anion, will be formed by element A?**

**Solution:**

- Element A is a non-metal
- Element C is less reactive than Element A
- C is smaller in size than B
- A will form anion

**7. Nitrogen (atomic number 7) and phosphorus (atomic number 15) belong to group 15 of the Periodic Table. Write the electronic configuration of these two elements. Which of these will be more electronegative? Why?**

**Solution:**

Atomic number of Nitrogen is 7 hence Electronic configuration of Nitrogen is 1s^{2} 2s^{2} 2p^{3}

Atomic number of Nitrogen is 15 hence Electronic configuration of Phosphorous is 1s^{2} 2s^{2} 2p^{6} 3s^{2} 3p^{3}

On moving down a group in the periodic table, the number of shell increases. Because of which valence electrons move away from the electrons and the effective nuclear charge decreases. This causes the decrease in the tendency to attract electron and hence electro negativity decreases. Because of all these reasons Nitrogen is more electronegative than phosphorus.

**8. How does the electronic configuration of an atom relate to its position in the Modern Periodic Table?**

**Solution:**

The number of valence electrons decides an atom’s position in the periodic table while the electronic configuration decides the number of valence electrons.

**9. In the Modern Periodic Table, calcium (atomic number 20) is surrounded by elements with atomic numbers 12, 19, 21 and 38. Which of these have physical and chemical properties resembling calcium?**

**Solution:**

Calcium has an atomic number of 20, and thus has an electronic configuration of 2, 8, 8, 2. Thus, calcium has 2 valence electrons. The electronic configuration of the element having atomic number 12 is 2, 8.2. Thus, this element with 2 valence electrons resemble calcium the most.

**10. Compare and contrast the arrangement of elements in Mendeleev’s Periodic Table and the Modern Periodic Table.**

**Solution:**

Mendeleev’s Periodic Table |
Modern Periodic Table |

Elements are arranged in the increasing order of their atomic masses. | Elements are arranged in the increasing order of their atomic numbers. |

There are 8 groups | There are 18 groups |

Each groups are subdivided into sub group ‘a’ and ‘b’ | Groups are not subdivided into sub-groups. |

Groups for Noble gas was not present as noble gases were not discovered by that time | A separate group is meant for noble gases. |

There was no place for isotopes. | This problem has been rectified as slots are determined according to atomic number. |

Also Access |

NCERT Exemplar Solutions for class 10 Science Chapter 5 |

CBSE Notes for Class 10 Science Chapter 5 |

## NCERT Solutions for class 10 Science Chapter 5- Periodic Classification Of Elements

NCERT Class 10 Science Chapter 5- Periodic Classification of Elements is categorized under Unit Chemical Substances – Nature and behavior. This unit, on the whole, comprises of 25 marks in the board exam, most compared to any chapter in the syllabus, hence all chapters in Unit Chemical Substances needs to be studied thoroughly. Periodic classification of elements chapter alone holds up 5 marks in the board exam as expected and from a careful analysis of previous question papers.

**List of subtopics covered in Chapter 5 – Periodic Classification Of Elements:**

5.1 Making Order Out of Chaos – Early Attempts At the Classification of Elements

We have been learning how various things or living beings can be classified on the basis of their properties. Even in other situations, we come across instances of organisations based on some properties. The topic discusses concepts of Dobereiner’s Triads and Newlands’ Law of Octaves. Both concepts are explained with suitable examples.

5.2 Making Order Out of Chaos – Mendeleev’s Periodic Table

The topic explains Mendeleev’s Periodic table, achievements of Mendeleev’s Periodic Table and Limitations of Mendeleev’s Classification. The concepts are presented along with the periodic table and Dmitri Ivanovich Mendeleev’s creation of periodic table.

5.3 Making Order Out of Chaos – The Modern Periodic Table

The Modern Periodic Table stated as Properties of elements are a periodic function of their atomic number. The topic further discusses Position of Elements in the Modern Periodic Table, trends in the Modern Periodic table such as valency, Atomic Size, Metallic and Non – metallic Properties.

**List of Exercise with questions**

Number – 5.1 – Making order out of chaos 3 Question ( 2 long, 1 short)

-Early attempts at the classification of elements

Number – 5.2 – Making order out of chaos

– Mendeleev’s Periodic Table 4 Question ( 1 long, 3 short)

Number – 5.3 – Making order out of chaos

-The Modern Periodic table 6 Question ( 1 long, 5 short)

Exercise Solutions – 10 Question ( 2 long, 8 short)

## NCERT Solutions for class 10 Science Chapter 5 – Periodic Classification of Elements

Periodic Classification of elements gives a brief idea about how elements are classified based on the similarities in the properties shared by different elements. It briefly discusses how Dobereiner classified various elements into triads and how Newland proposed the Law of Octaves. The chapter briefly discusses arrangement if elements in increasing order of their atomic numbers and chemical properties as suggested by Mendeleev, as he even predicted the existence and presence of some more elements that are to be discovered yet which he arrived at due to gaps in the Periodic table.

The chapter also talks about the limitations of Mendeleev’s classification and hence the modern periodic table. It gives an idea of the positioning of elements in the modern periodic elements and also the vertical and horizontal arrangement known as groups and periods respectively.

### Key Features of NCERT Solutions for class 10 Science Chapter 5 – Periodic Classification Of Elements

- NCERT class 10 science solutions are answered considering chapterwise marks distribution and question type aligning with the CBSE blueprint
- Solutions have been elaborated enabling students to better understand
- Numerical have been solved giving a step-by-step explanation
- Tabular columns have been used wherever necessary which enables students to learn at a faster pace.
- The language used in the chapter is simplified and logical.

## Frequently Asked Questions on Periodic Classification Of Elements

### Did Döbereiner’s triads also exist in the columns of Newlands’ Octaves ?

Döbereiner’s triads did exist in the columns of Newlands’ Octaves; For example, the elements

Lithium (Li), Potassium (K) and Sodium (Na) constitute a Dobereiner’s Triad but are also found in the second column of Newland’s Octaves.

### Besides gallium, which other elements have since been discovered that were left by Mendeléev in his Periodic Table ?

Germanium and Scandium are the element that are left by Mendeleev in his Periodic Table since its discovery.

### What were the criteria used by Mendeléev in creating his Periodic Table?

Mendeléev concentrated on various compounds formed by the elements with Hydrogen and

Oxygen. Among physical properties, he observed the relationship between the atomic masses

of various elements while creating his periodic table.

### Why do you think the noble gases are placed in a separate group ?

Noble gases are placed in a separate group because of their inert nature and low concentration in our atmosphere. They are kept in a separate group called Zero group so that they don’t disturb the existing order.

### In the Modern Periodic Table, which are the metals among the first ten elements?

Lithium and Beryllium are the metals among the first ten elements in Modern Periodic Table.

### How does the electronic configuration of an atom relate to its position in the Modern Periodic Table ?

The number of valence electrons decides an atom’s position in the periodic table while the electronic configuration decides the number of valence electrons.