Arithmetic Progressions ⇒⇒ Exercise 5.2

Question 1

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

	a	d	n	an
(i)	7	3	8	...
(ii)	-18	...	10	0
(iii)	..	-3	18	-5
(iv)	-18.9	2.5	...	3.6
(v)	3.5	0	105	...

Question 1 (i)

(i) Given, a = 7,d = 3,n = 8, $ a_n = ? $

Solution :

According to the question

Given, First term, a = 7

Common difference, d = 3

Number of terms, n = 8

We have to find the nth term $a_n = ?$

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get :,

$$ a_8 = { 7 + ( 8 - 1) × 3 } $$

$$ ⇒ a_8 = { 7 + 21 } $$

$$ ⇒ a_8 = { 28 } $$

Therefore, nth term of an A.P = 28

Question 1 (ii)

(ii) Given, a = -18 , d = ? , n = 10 , $ a_n = 0$

Solution :

According to the question

Given, First term, a = -18

Common difference, d = ?

Number of terms, n = 10

Nth term, $a_n = 0?$

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get :,

$$ 0 = { -18 + ( 10 - 1) × d } $$

$$ ⇒ 0 = { -18 } + 9d $$

$$ ⇒ 9d = { 18 } $$

$$ ⇒ d = {2 } $$

Therefore, Common difference = d = 2

Question 1 (iii)

(iii) Given, a = ? , d = -3 , n = 18 , $ a_n = -5 $

Solution :

According to the question

Given, First term, a = ?

Common difference, d = -3

Number of terms, n = 18

Nth term, $a_n = -5 $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ -5 = { a + ( 18 - 1)×(-3) } $$

$$ -5 = { a + 17 ×(-3) } $$

$$ ⇒ -5 = a - 51 $$

$$ ⇒ a = -5 + 51 $$

$$ ⇒ a = {46 } $$

Therefore, First term = a = 46

Question 1 (iv)

(iv) Given, a = -18.9 , d = 2.5 , n = ? , $ a_n = 3.6 $

Solution :

According to the question

Given, First term, a = -18.9

Common difference, d = 2.5

Number of terms, n = ?

Nth term, $a_n = 3.6 $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ 3.6 = { -18.9 + ( n - 1)×(2.5) } $$

$$ ⇒ 3.6 = -18.9 + 2.5n - 2.5 $$

$$ ⇒ 3.6 = -21.4 + 2.5n $$

$$ ⇒ 3.6 + 21.4 = 2.5n $$

$$ ⇒ 2.5n = 25 $$

$$ ⇒ n = 10 $$

Therefore, Number of terms, n = 10

Question 1 (v)

(v) Given, a = 3.5, d = 0, n = 105, $ a_n = ? $

Solution :

According to the question

Given, First term, a = 3.5

Common difference, d = 0

Number of terms, n = 105

Nth term, $a_n = ? $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ a_n = 3.5 + (105 - 1)0 $$

$$ ⇒ 3.5 + 0 $$

$$ a_n = 3.5 $$

Therefore, Number of terms, n = 3.5

Question 2 (i)

Choose the correct choice in the following and justify:
(i) 30th term of AP: 10, 7, 4, ..., is
A) 97
B) 77
C) – 77
D) – 87

Solution :

According to the question

The AP is 10, 7, 4, …

First term = $ { a_1 } $ = 10

Common difference d = $ { a_2 - a_1 } $

Common difference d = $ 7 - 10 = -3 $

Number of terms, n = 30

$ N^{th}$ term term, $a_n = ? $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ a_{30} = { 10 + ( 30 - 1) (-3) } $$

$$ a_{30} ={ 10 - 87 } $$

$$ a_{30} = { - 77 } $$

Therefore,The correct option is c
the 30th term of the A.P. is = -77

Question 2 (ii)

Choose the correct choice in the following and justify:
(ii) $ 11^{th}$ term of the AP: – 3, $ { -1 \over 2 } $ , 2, ... is
A) 28
B) 22
C) – 38
D) $ – 48{ 1 \over 2 } $

Solution :

According to the question

The AP is – 3, $ { -1 \over 2 } $ , 2, ...

First term = $ { a_1 } $ = -3

Common difference d = $ { a_2 - a_1 } $

⇒ d = $ { -1 \over 2 } $ - (-3)

⇒ d = $ { -1 \over 2 } $ + 3

⇒ d = $ { -1 + 6} \over 2 $

⇒ d = $ {5 \over 2 } $

Number of terms, n = 11

$ N^{th}$ term, $a_n = ? $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ ⇒ a_{11} = { -3 + ( 11 - 1) ({5 \over 2 }) } $$

$$ ⇒ a_{11} = { -3 + 10× ({5 \over 2 }) } $$

$$ ⇒ a_{11} = { -3 + 25} $$

$$ ⇒ a_{11} = { 22} $$

Therefore,The correct option is b
the 11th term of the A.P. is = 22

Question 3 (i)

Choose the correct choice in the following and justify:
In the following APs, find the missing terms in the boxes:
(i) 2 , $ { \Box } $ , 26 ...

Solution :

According to the question

First term = $ { a_1 } $ = 2

Common difference d = ?

$ a_3 = 26 $

$ N^{th}$ term, $ a_2 = ? $

We know that

$$ a_n = { a + ( n - 1) d } $$

We can find the common difference (d ), The third term can be expressed

$$ a_3 = 2 + (3 - 1)d $$

$$ ⇒ 26 = 2 + 2d $$

$$ ⇒ 26 - 2 = 2d $$

$$ ⇒ { 24 \over 2 } = d $$

$$ ⇒ { 12 } = d $$

Now, to find the second term a₂

$$ a_2 = 2 + (2 - 1)12 $$

$$ ⇒ a_2 = 2 + 12 $$

$$⇒ a_2 = 14 $$

The missing term in the box is 14 .

Question 3 (ii)

Choose the correct choice in the following and justify:
In the following APs, find the missing terms in the boxes:
(ii) $ { \Box } $ , 13 , $ { \Box } $ , 3 , ...

Solution :

According to the question

First term = $ { a_1 } $ = ?

$ a_2 = 13 $

$ { a_3 } $ = ?

$ a_4 = 3 $

Common difference d = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence $$ a_2 = a + (2 - 1)d $$

$$ 13 = a + d .....(i) $$

And

$$ a_4 = a + (4 - 1)d $$

$$ 3 = a + 3d .....(ii) $$

After subtracting euqation (i) from (ii)

(a + 3d) - (a + d) = 3 - 13

⇒ 2d = - 10

⇒ d = $ -10 \over 2 $

⇒ d = - 5

Next, we find the first term by substituting the value of d = -5 in euqation (i)

13 = a + d

⇒ 13 = a + (-5)

⇒ a = 18

Finally, we find the third term, We can do this by adding the common difference to the second term:

$ a_3 $ = a + 2d

⇒ $ a_3 $ = 18 + 2(-5)

⇒ $ a_3 $ = 8

The missing terms in the boxes are 18 and 8 respectively .

Question 3 (iii)

Choose the correct choice in the following and justify:
In the following APs, find the missing terms in the boxes:
(iii) 5, $ { \Box } $ ,$ { \Box } $ ,$ 9{ 1\over 2 } $ , ...

Solution :

According to the question

First term = $ { a_1 } $ = 5

$ a_2 = ? $

$ { a_3 } $ = ?

$ a_4 = 9{ 1\over 2 } = { 19\over 2 } $

Common difference d = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence$$ a_4 = a + (4 - 1)d $$

$$ ⇒{ 19\over 2 } = 5 + (4 - 1)d $$

$$ ⇒ { 19\over 2 } = 5 + 3d $$

$$ ⇒ 19 = (5 + 3d)2 $$

$$ ⇒ 19 = 10 + 6d $$

$$ ⇒ 19 - 10 = 6d $$

$$ ⇒ { 9\over 6 } = d $$

$$ ⇒ { 3\over 2} = d $$

Next, we find the Second term by substituting the value of d = $ 3\over 2 $

$ a_2 $ = a + d

⇒ $ a_2 $ = 5 + ${ 3\over 2}$

⇒ $ a_2 $ = ${ 13\over 2}$

Finally, we find the third term, We can do this by adding the common difference to the second term:

$ a_3 $ = a + 2d

⇒ $ a_3 $ = 5 + $2({ 3\over 2}) $

$ a_3 $ = 8

The missing terms in the boxes are $ 13\over 2 $ and 8 respectively .

Question 3 (iv)

Choose the correct choice in the following and justify:
In the following APs, find the missing terms in the boxes:
(iv) -4, $ { \Box } $ ,$ { \Box } $ ,$ { \Box } $ ,$ { \Box } $ ,6 , ..

Solution :

According to the question

First term = $ { a_1 } $ = −4

$ a_2 = ? $ , $ { a_3 } $ = ? , $ a_4 = ? $ , $ { a_5 } $ = ?

$ a_6 = 6 $

Common difference d = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence$$ a_6 = a + (6 - 1)d $$

$$ ⇒ 6 = - 4 + (6 - 1)d $$

$$ ⇒ 6 = - 4 + 5d $$

$$ ⇒ 6 + 4 = 5d $$

$$ ⇒ { 10\over 5 } = d $$

$$ d = 2 $$

Next, we find the Second term by substituting the value of d = $2 $

∴ 2nd term = $ a_2 $ = $ a_1 $ + d = - 4 + 2 = -2

3rd term = $ a_3 $ = $ a_2 $ + d = -2 + 2 = 0

4th term = $ a_4 $ = $ a_3 $ + d = 0 + 2 = 2

5th term = $ a_5 $ = $ a_4 $ + d = 2 + 2 = 4

The missing terms in the boxes are -2 , 0 , 2 and 4 respectively .

Question 3 (v)

Choose the correct choice in the following and justify:
In the following APs, find the missing terms in the boxes:
(v) $ { \Box } $ , 38 , $ { \Box } $ ,$ { \Box } $ ,$ { \Box } $ ,-22 , ...

Solution :

According to the question

a = ? , $ a_2 = 38 $ , $ a_3 = ? $, $ a_4 = ? $ , $ a_5 = ? $ , $ a_6 = -22 $

Common difference d = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence $$ a_2 = a + (2 - 1)d $$

$$ 38 = a + d .....(i) $$

And

$$ a_6 = a + (6 - 1)d $$

$$ -22 = a + 5d .....(ii) $$

After subtracting euqation (i) from (ii)

$$ (a + 5d) - (a + d) = -22 - 38 $$

$$ 4d = - 60 $$

$$ d = { -60 \over 4 } $$

$$ d = - 15 $$

Putting value of d = -15 in euqation (i)

$$ 38 = a + d $$

$$ 38 = a + (-15) $$

$$ a = 53 $$

Next, we find the remaining term by substituting the value of d = -15

3rd term = $ a_3 $ = $ a_2 $ + d = 38 + (- 15) = 23

4th term = $ a_4 $ = $ a_3 $ + d = 23 + (- 15) = 8

5th term = $ a_5 $ = $ a_4 $ + d = 8 + (- 15) = -7

The missing terms in the boxes are 53 , 23 , 8 and -7 respectively .

Question 4

Which term of the AP: 3, 8, 13, 18, ……… is 78?

Solution :

Given, a = 3 , $ a_2 = 8 $ , $ a_3 = 13 $, $ a_n = 78 $

Common difference, d = 8 - 3 = 5

Number of terms, n = ?

Nth term, $a_n = 78 $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ ⇒ 78 = 3 + ( n - 1) 5 $$

$$ ⇒ 78 - 3 = ( n - 1) 5 $$

$$ ⇒ {75 \over 5 } = ( n - 1) $$

$$ ⇒ 15 + 1 = n $$

$$ n = 16 $$

Therefore, 78 is the 16 ^th term of AP.

Question 5 (i)

Find the number of terms in each of the following APs
(i) 7, 13, 19, ..., 205

Solution :

Given, a = 7 , $ a_2 = 13 $ , $ a_3 = 19 $, $ a_n = 205 $

Common difference, d = 13 - 7 = 6

Number of terms, n = ?

Nth term, $a_n = 205 $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ ⇒ 205 = 7 + ( n - 1) 6 $$

$$ ⇒ 205 - 7 = ( n - 1) 6 $$

$$ ⇒ {198 \over 6 } = ( n - 1) $$

$$ ⇒ 33 + 1 = n $$

$$ n = 34 $$

Therefore, 205 is the 34^th term of AP.

Question 5 (ii)

Find the number of terms in each of the following APs
(ii) 18, $15{ 1\over 2}$ ,13, ..., -47

Solution :

Given, a = 18 , $ a_2 = 15{ 1\over 2} $ , $ a_3 = 13 $

Common difference, d = ${ 31\over 2}$ - 18

d = ${ 31 - 36 }\over 2$ = $ - 5\over 2$

Number of terms, n = ?

Nth term, $a_n = -47 $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ ⇒ -47 = 18 + ( n - 1)(- {5\over 2}) $$

$$ ⇒ -47 - 18 = ( n - 1)(- {5\over 2}) $$

$$ ⇒ -65 = ( n - 1)(-{ 5\over 2}) $$

$$ ⇒ {{-65× 2} \over -5} = ( n - 1) $$

$$ ⇒ {{-130 }\over - 5} = ( n - 1) $$

$$ ⇒ 26 = ( n - 1) $$

$$⇒ n = 26 + 1 $$

$$ n = 27 $$

Therefore, -47 is the 27^th term of AP.

Question 6

Find the number of terms in each of the following APs
Check whether – 150 is term of the AP: 11, 8, 5, 2,...

Solution :

Given, a = 11 , $ a_2 = 8 $ , $ a_3 = 5 $,

Common difference, d = 8 - 11 = -3

Number of terms, n = ?

Nth term, $ a_n = -150 $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ ⇒ -150 = 11 + ( n - 1) (-3) $$

$$ ⇒ -150 - 11 = ( n - 1) (-3) $$

$$ ⇒ -161 = -3n + 3 $$

$$ ⇒ -164 - 3 = -3n $$

$$ ⇒ -164 = -3n $$

$$ ⇒ {{164 } \over 3} = n $$

n = $ 164\over 3 $ which is a fraction.

Hence given number is not a term of AP because ‘n’ is not positive integer.

Question 7

Find the 31^st term of an AP whose 11^th term is 38 and the 16^th term is 73

Solution :

Given, a = ? , $ a_{11} = 38 $ , $ a_{16} = 73 $

Common difference, d = ?

Number of terms, n = 31

Nth term, $ a_{31} = ? $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ a_{11} = a + (11 - 1)d $$

$$ 38 = a + 10d .....(i) $$

And

$$ a_{16} = a + (16 - 1)d $$

$$ 73 = a + 15d .....(ii) $$

Now, by subtracting equation (i) from equation (ii) we get

$$ (a + 15d) - (a + 10d) = 73 - 38$$

$$⇒ 5d = 35 $$

$$⇒ d = {35 \over 5 } $$

$$ d = 7 $$

Putting value of d = 7 in euqation (i)

$$ 38 = a + 10d $$

$$ ⇒ 38 = a + 10 × 7 $$

$$ ⇒ a = 38-70 $$

$$ a = -32 $$

Now 31st term can be calculated as follows:= $$ a_{31} = a + 30d $$

$$ ⇒ a_{31} = -32 + 30×(7) $$

$$ ⇒ a_{31} = -32 + 210 $$

$$ a_{31} = 178 $$

Hence 31^st term is 178.

Question 8

An AP consists of 50 terms of which 3^rd term is 12 and the last term is 106 . Find the 29^th term

Solution :

Given, a = ? , $ a_{3} = 12 $ , $ a_{50} = 106 $

Common difference, d = ?

Number of terms, n = 29

Nth term, $ a_{29} = ? $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ a_{3} = a + (3 - 1)d $$

$$ 12 = a + 2d .....(i) $$

And

$$ a_{50} = a + (50 - 1)d $$

$$ 106 = a + 49d .....(ii) $$

Now, by subtracting equation (i) from equation (ii) we get

$$ (a + 49d) - (a + 2d) = 106 - 12$$

$$⇒ 47d = 94 $$

$$⇒ d = { 94 \over 47 } $$

$$ d = 2 $$

Putting value of d = 2 in euqation (i)

$$ 12 = a + 2d $$

$$ ⇒ 12 = a + 2 × (2) $$

$$ ⇒ a = 12-4 $$

$$ a = 8 $$

Now 29th term can be calculated as follows:= $$ a_{29} = a + 28d $$

$$ ⇒ a_{29} = 8 + 28×(2) $$

$$ ⇒ a_{29} = 8 + 56 $$

$$ a_{29} = 64 $$

Hence 29^th term is 64

Question 9

If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?

Solution :

Given, a = ? , $ a_{3} = 4 $ , $ a_{9} = -8 $

Common difference, d = ?

Number of terms, n = ?

Nth term, $ a_n = 0 $

We know that

$$ a_n = { a + ( n - 1) d } $$

Hence, Substituting values we get

$$ a_{3} = a + (3 - 1)d $$

$$ 4 = a + 2d .....(i) $$

And

$$ a_{9} = a + (9 - 1)d $$

$$ -8 = a + 8d .....(ii) $$

Now, by subtracting equation (i) from equation (ii) we get

$$ (a + 8d) - (a + 2d) = -8 - 4 $$

$$⇒ 6d = -12 $$

$$⇒ d = { -12 \over 6 } $$

$$ d = -2 $$

Putting value of d = -2 in euqation (i)

$$ 4 = a + 2d $$

$$ ⇒ 4 = a + 2 × (-2) $$

$$ ⇒ a = 4 + 4 $$

$$ a = 8 $$

Let nth term of the A.P. be zero. $$ ⇒ 0 = 8 + ( n - 1) (-2)$$

$$ ⇒ 0 = 8 -2n + 2 $$

$$ ⇒ 2n = 10$$

$$ n = 5 $$

Hence 5^th term will be 0.

Question 10

The 17^th term of an AP exceeds its 10^th term by 7. Find the common difference

Solution :

Common difference, d = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

According to the question :: 17th term of an AP exceeds its 10th term by 7, we get :

$$ a_{17} = a_{10} + 7 $$

$$ a + 16d = a + 9d + 7 $$

$$⇒ a + 16d - a - 9d = 7 $$

$$⇒ 7d = 7 $$

$$ d = 1 $$

Hence Common difference = d = 1

Question 11

Which term of the AP: 3, 15, 27, 39,… will be 132 more than Its 54^th term?

Solution :

Given, a = 3, $ a_{2} = 15 $ , $ a_{3} = 27$

Common difference, d = 15 - 3 = 12

Nth term, $ a_n = ? $

We know that

$$ a_n = { a + ( n - 1) d } $$

Given that nth term of AP is 132 more than its 54^th term $$ a_{n} = a_{54} + 132 $$

$$ a + ( n - 1) d = 3 + ( 54 - 1) 12 + 132 $$

$$⇒ 3 +( n - 1) 12 = 3 +( 53) × 12 + 132 $$

$$ ⇒ 3 + 12n - 12 = 639 + 132 $$

$$ ⇒ 12n - 9 = 771 $$

$$ ⇒ 12n = 771 + 9 $$

$$ ⇒ n = {780 \over 12 } $$

$$ ⇒ n = 65 $$

Thus, the required term is 65th term.

Question 12

Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their l000th terms?

Solution :

Let the two APs are a₁,a₂,a₃, ....and b₁,b₂,b₃, ....

and the Common difference of these A.P’s be d

For first A.P.

$$ a_{100} = a_1 + (100 - 1)d $$

$$ a_{100} = a_1 + 99d $$

$$ a_{1000} = a_1 + (1000 - 1)d $$

$$ a_{1000} = a_1 + 999d $$

And for second A.P.

$$ b_{100} = b_1 + (100 - 1)d $$

$$ b_{100} = b_1 + 99d $$

$$ b_{1000} = b_1 + (1000 - 1)d $$

$$ b_{1000} = b_1 + 999d $$

According to the question

difference between 100th term of these A.P.s = 100

$$ ( a_1 + 99d ) - (b_1 + 99d ) = 100 $$

$$ ⇒ a_1 - b_1 = 100 .....(i) $$

Now, difference between 1000th term is

$$ ( a_1 + 999d ) - (b_1 + 999d ) = a_1 - b_1 $$

Putting value from euqation (i)

$$ ⇒ a_1 - b_1 = 100 $$

$$ ⇒ 100 $$

Therefore, difference between 1000th term is 100

Question 13

How many three digit numbers are divisible by 7 ?

Solution :

We know that the first three digit number is 100 , which is when divided by 7 gives a remainder = 2

Thus, 100+5 = 105 is the first three-digits number divisible by 7

Next number = 105 + 7 = 112

Therefore , the series becomes 105, 112, 119 ....

Now, the largest three-digits number = 999

When 999 is divided by 7 it gives a remainer = 5

Means 999 - 5 = 994 is the largest three-digits number divisible by 7

Let there are 'n' three digit numbers divisible by 7

Therefore ,

a = 105, $ a_{2} = 112 $ , d = 7 , $ a_{n} = 994 $, n = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

$$ ⇒ 994 = { 105 + ( n - 1) 7 } $$

$$ ⇒ 994 = 105 + 7n - 7 $$

$$ ⇒ 994 - 105 + 7 = 7n $$

$$ ⇒ 896 = 7n $$

$$ ⇒ n = {896 \over 7 } $$

$$ ⇒ n = 128 $$

Therefore, 128 three-digit numbers are divisible by 7.

Question 14

How many multiples of 4 lie between 10 and 250?

Solution :

By Observation, First multiple of 4 that is greater than 10 is 12.

Next number = 16

Therefore , the series becomes 12, 16, 20 ....

When 250 is divided by 4 it gives a remainer = 2

Means 250 - 2 = 248 is the largest multiple of 4 within 250

Let there are 'n' multiple of 4

Therefore ,

a = 12, $ a_{2} = 16 $ , d = 4 , $ a_{n} = 248 $, n = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

$$ ⇒ 248 = { 12 + ( n - 1) 4} $$

$$ ⇒ 248 = 12 + 4n - 4 $$

$$ ⇒ 248 - 12 + 4 = 4n $$

$$ ⇒ 240 = 4n $$

$$ ⇒ n = {240 \over 4 } $$

$$ ⇒ n = 60 $$

Therefore, there are 60 numbers divisible by 4 lie between 10 and 250.

Question 15

For what value of n, are the nth terms of two APs: 63, 65, 67,… and 3, 10, 17,… equal?

Solution :

Given in 1st AP,63, 65, 67,…

a = 63, $ a_{2} = 65 $ , $ a_{3} = 67 $

d = 65 - 63 = 2

Given in Second AP,3, 10, 17,……

a = 3, $ a_{2} = 10 $ , $ a_{3} = 17 $

d = 10 - 3 = 7

We know that

$$ a_n = { a + ( n - 1) d } $$

Therefore , For Ist AP:

$$ a_n = { 63 + ( n - 1) 2 } ....(i) $$

For Second AP:

$$ a_n = { 3 + ( n - 1) 7 } ....(ii) $$

Let n^th term of both the AP's are equal

$$ { 63 + ( n - 1) 2 } = { 3 + ( n - 1) 7 }$$

$$ ⇒ 63 + 2n − 2 = 3 + 7n − 7 $$

$$ ⇒ 61 + 4 = 7n − 2n $$

$$ ⇒ 65 = 5n $$

$$ ⇒ n = {65 \over 5 } $$

$$ ⇒ n = 13 $$

Therefore, The 13^th term of both the AP's are equal .

Question 16

Determine the AP whose third term Is 16 and the 7th term exceeds the 5th term by 12.

Solution :

Let $ a$ be the first term and $ d $ the common difference.

According to the question

Given that, $ a_{3} = 16 $ ,and $ a_{7} - a_{5}= 12 $

We know that

$$ a_n = { a + ( n - 1) d } $$

Therefore ,

$$ 16 = { a + ( 3 - 1) d } $$

$$ 16 = { a + 2d } ....(i) $$

And, 7th term - 5th term = 12

$$ 12 = a_{7} - a_{5} $$

$$ ⇒ 12 ={ a + ( 7 - 1) d } - ({ a + ( 5 - 1) d })$$

$$⇒ 12 = { a + 6d - a - 4d }$$

$$⇒ 12 = {2d }$$

$$ ⇒ 6 = {d }$$

Put the value of d in equation (i) we will get

$$ 16 = a + 2 (6) $$

$$ ⇒ 16 – 12 = a $$

$$ ⇒ a = 4 $$

Now, AP with first term = 4 and common difference = 6 is

4,10,16,22,..

Question 17

Find the 20^th term from the last term of the AP: 3, 8, 13, …., 253

Solution :

According to the question

Given that, a = 3 , $ a_{2} = 8 $ ,and $ a_{3} = 13 $ , $ a_{n} = 253 $ , n = ?

Common difference (d) = 8 - 3 = 5.

We know that

$$ a_n = { a + ( n - 1) d } $$

Therefore, Let suppose there are 'n' terms in the AP

$$ 253 = { 3 + ( n - 1)5 } $$

$$ ⇒ 253 = { 3 + 5n - 5 } $$

$$ ⇒ 253 = { -2 + 5n } $$

$$ ⇒ 253 + 2 = {5n } $$

$$ ⇒ n = ({255 \over 5 }) $$

$$ ⇒ n = 51 $$

So, there are 51 terms in the given AP

And 20th term from the last will be (51 – 19 ) = 32th term from the starting,
Therefore,

$$ a_{32} = { 3 + ( 32 - 1)5} $$

$$ ⇒ a_{32} = { 3 + 155} $$

$$ ⇒ a_{32} = { 158} $$

Therefore, 20^th term from the last term is 158

Question 18

The sum of the 4^th and 8^th terms of an AP is 24 and the sum of the 6^th and 10^th terms is 44. Find the first three terms of the AP.

Solution :

Let 'a' be the first term and 'd' the common difference.

According to the question

Given that,

$$ a_{4} + a_{8} = 24 $$

$$⇒ ({ a + 3d }) + ({ a + 7d }) = 24 $$

$$ ⇒ { 2a + 10d} = 24 $$

divide by 2

$$⇒ a + 5d = 12 ....(i) $$

Also

$$ a_{6} + a_{10} = 44 $$

$$⇒ ({ a + 5d }) + ({ a + 9d }) = 44 $$

$$ ⇒ { 2a + 14d} = 44 $$

divide by 2

$$⇒ a + 7d = 22 ....(ii) $$

On solving equation (i) and (ii) by subtracting we will get

$$ (a + 7d)- (a + 5d) = 22 - 12 $$

$$ ⇒ 2d = 10 $$

$$ ⇒ d = 5 $$

Put the value of d in equation (i) we will get

$$ a + 5d = 12 $$

$$ ⇒ a + 5(5) = 12 $$

$$ ⇒ a = 12 - 25 $$

$$ ⇒ a = - 13 $$

first term = -13 and common difference = 5

Therefore,first three terms of AP are , – 13, -8, -3

Question 19

Subba Rao started work in 1995 at the annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000.

Solution :

According to the question

The annual salary of Subba Rao in 1995 = Rs. 5000

The annual salary of Subba Rao in 1996 = Rs. 5200

∴ AP = 5000, 5200, 5400,…, 7000.

Given that, a = 5000 , $ a_{2} = 5200 $ ,and d = 200 , $ a_{n} = 7000 $ , n = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

$$ a_n = { a + ( n - 1) d } $$

$$ 7000 = { 5000 + ( n - 1) 200 } $$

$$⇒ 7000 = { 5000 + 200n - 200 } $$

$$⇒ 7000 = { 4800 + 200n } $$

$$⇒ 7000 - 4800 = 200n $$

$$⇒ 2200 = 200n $$

$$⇒ n = ({ 2200 \over 200 })$$

$$⇒ n = 11 $$

∴ The annual salary of Subba Rao will be Rs 7000 after 11 years, starting from 1995,

So we have to add 10 in 1995, because these numbers are in years.

$⇒ 1995 + 10 =2005$. Hence, his salary reached at Rs. 7000 in 2005.

Question 20

Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week her weekly savings become RS 20.75, find n.

Solution :

According to the question

The savings of Ramkali in the first week is = Rs. 5

In the consecutive week = Rs. 5 + 1.75 = Rs. 6.75

∴ AP = 5, 6.75, 8.5, …………. 20.75

Given that, a = 5 , $ a_{2} = 6.75 $ ,and d = 1.75 , $ a_{n} = 20.75 $ , n = ?

We know that

$$ a_n = { a + ( n - 1) d } $$

$$⇒ 20.75 = { 5 + 1.75n - 1.75 } $$

$$⇒ 20.75 = { 3.25 + 1.75n } $$

$$⇒ 20.75 - 3.25 = 1.75n $$

$$⇒ 17.50 = 1.75n $$

$$⇒ n = ({ 17.50 \over 1.75 })$$

$$⇒ n = 10 $$

Therefore, after 10 weeks her saving will become Rs 20.75