Arithmetic Progressions ⇒⇒ Exercise 5.3

Question 1 (i)

Find the sum of the following APs:
(i) 2, 7, 12, ....., to 10 terms

Solution :

According to the question

Given, First term, a = 2

Common difference, d = 7-2 = 5

Number of terms, n = 10

We have to find the Sum of first 10 terms $ S_{10} = ? $

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

$$ ⇒ S_{10} = {10\over 2}[ { 2 × 2 + (10 - 1) 5 } ] $$

$$ ⇒ S_{10} = {5}[ { 4 + 45 } ] $$

$$ ⇒ S_{10} = { 5 } ×{ 49 } $$

$$ ⇒ S_{10} = 245 $$

∴ Sum of first 10 terms of given AP = 245.

Question 1 (ii)

Find the sum of the following APs:
(ii) Find the sum of the following APs: -37,-33,-29,..., to 12 terms

Solution :

According to the question

Given, First term, a = -37

Common difference, d = -33- (-37) = 4

Number of terms, n = 12

We have to find the Sum of first 12 terms $ S_{12}= ? $

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ ⇒ S_{12} = {12\over 2}[ { 2 × (-37) + (12 - 1) 4 } ] $$

$$ ⇒ S_{12} = {6}[ { -74 + 11 × 4 } ] $$

$$ ⇒ S_{12} = {6} [ { -74 + 44 } ] $$

$$ ⇒ S_{12} = {6} × [ { -30 } ] $$

$$ ⇒ S_{12} = - 180 $$

∴ Sum of first 12 terms of given AP = - 180.

Question 1 (iii)

Find the sum of the following APs:
(iii) Find the sum of the following APs: 0.6, 1.7, 2.8,……… , to 100 terms

Solution :

According to the question

Given, First term, a = 0.6

Common difference, d = 1.7 – 0.6 = 1.1

Number of terms, n = 100

We have to find the Sum of first 100 terms $ S_{100} = ?$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ ⇒ S_{100} = {100\over 2}[ { 2 ×(0.6) + (100 - 1) 1.1 } ] $$

$$ ⇒ S_{100} = {50}[ { 1.2 + 99 × 1.1 } ] $$

$$ ⇒ S_{100} = {50} ×[ 1.2 + 108.9 ] $$

$$ ⇒ S_{100} = 50 ×[ 110.1] $$

$$ ⇒ S_{100} = 5505 $$

∴ Sum of first 100 terms of given AP = 5505.

Question 1 (iv)

Find the sum of the following APs:
(iii) Find the sum of the following APs: $ {1\over 15} , {1\over 12} , {1\over 10 } ,$ ……… , to 11 terms.

Solution :

According to the question

Given, First term, a = $ {1\over 15} $

Common difference, d = ${1\over 12} – {1\over 15} = {1\over 60} $

Number of terms, n = 11

We have to find the Sum of first 11 terms $ S_{11} = ?$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ S_{11} = {11\over 2} × [ { 2({1\over 15} ) + (11 - 1) ({1\over 60}) } ] $$

$$ S_{11} = {11\over 2} × [ { {2\over 15} + 10 ×({1\over 60}) } ] $$

$$ ⇒ S_{11} = {11\over 2} × [ {2\over 15} +{1\over 6} ] $$

$$ ⇒ S_{11} = {11\over 2} × [ {{4+5}\over 30} ] $$

$$ ⇒ S_{11} = {11\over 2} × [ {{9}\over 30} ] $$

$$ ⇒ S_{11} = {99\over 60} = {33\over 20} $$

∴ Sum of first 11 terms of given AP = ${33\over 20} $.

Question 2 (i)

Find the sums given below::
(i) Find the sum of the following APs: 7 + 10 $ {1\over 2} $ + 14 + ...84

Solution :

According to the question

Given, First term, a = 7

Common difference, d = $ 10 {{1\over 2}} – 7 = 3{1\over 2} = {7 \over 2} $

Last term, $ a_n = l = 84 $

Number of terms, n = ?

We have to find the Sum of first n terms $ S_{n} = ?$

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 84 = 7 + ( n - 1){7\over 2} $$

$$ ⇒ 84 -7 = ( n - 1)×{7\over 2} $$

$$ ⇒ 77 = ( n - 1)×{7\over 2} $$

$$ ⇒ {{77 × 2}\over 7} = ( n - 1) $$

$$ ⇒ 22 = ( n - 1) $$

$$ ⇒ 22 + 1 = n $$

$$ ⇒ 23 = n $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ S_{23} = {23\over 2} × [ { 2(7 ) + (23 - 1) {7 \over 2} } ] $$

$$ ⇒ S_{23} = {23\over 2} × [ {14 + (22) × {7 \over 2} } ] $$

$$ ⇒ S_{23} = {23\over 2} × [ {14 + 77 } ] $$

$$ ⇒ S_{23} = {23\over 2} × 91 $$

$$ ⇒ S_{23} = {2093\over 2} $$

$$ ⇒ S_{23} = 1046{1\over 2} $$

Hence, sums of a given AP is $ 1046{1\over 2} $

Question 2 (ii)

Find the sums given below::
(ii) Find the sum of the following APs: 34 + 32 + 30 + .... + 10

Solution :

According to the question

Given, First term, a = 34

Common difference, d = 32 – 34 = -2

Last term, $ a_n = l = 10 $

Number of terms, n = ?

We have to find the Sum of first n terms $ S_{n} = ?$

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 10 = 34 + ( n - 1)(-2) $$

$$ ⇒ 10 = 34 -2n + 2 $$

$$ ⇒ 10 - 36 = -2n $$

$$ ⇒ - 26 = -2n $$

$$ ⇒ n = {26 \over 2 } $$

$$ ⇒ n = 13 $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ S_{13} = {13\over 2} × [ { 2(34 ) + (13 - 1) {- 2} } ] $$

$$ S_{13} = {13\over 2} × [ {68 + 12 × {- 2} } ] $$

$$ S_{13} = {13\over 2} × [ {68 - 24} ] $$

$$ ⇒ S_{13} = {13\over 2}×(44 ) $$

$$ ⇒ S_{13} = {13 × 22} $$

$$ ⇒ S_{13} = 286 $$

Hence, sums of a given AP is $ 286 $

Question 2 (iii)

Find the sums given below::
(iii) Find the sum of the following APs: -5 + (- 8) + (- 11) +……… + (- 230)

Solution :

According to the question

Given, First term, a = -5

Common difference, d = -8 – (-5) = -3

Last term, $ a_n = l = -230 $

Number of terms, n = ?

We have to find the Sum of first n terms $ S_{n} = ?$

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ -230 = -5 + ( n - 1)(-3) $$

$$ ⇒ -230 = -5 -3n + 3 $$

$$ ⇒ -230 + 2 = -3n $$

$$ ⇒ - 228 = -3n $$

$$ ⇒ n = {228 \over 3 } $$

$$ ⇒ n = 76 $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ S_{76}= {76\over 2} × [ { 2( -5 ) + (76 - 1) {- 3} } ] $$

$$⇒ S_{76}= 38 × [ { -10 + 75 × {- 3} } ] $$

$$⇒ S_{76}= 38 × [ { -10 - 225 } ] $$

$$ ⇒ S_{76}= 38 × ( - 235) $$

$$ ⇒ S_{76}= -8930 $$

Hence, sums of a given AP is $ -8930 $

Question 3 (i)

In an AP:
(i) given a = 5, d = 3, $ a_{n} = 50 $ , find ‘n’ and $ S_{n} $

Solution :

According to the question

Given, First term, a = 5

Common difference, d = 3

Last term, $ a_n = l = 50 $

Number of terms, n = ?

We have to find the Sum of first n terms $ S_{n} = ?$

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 50 = 5 + ( n - 1)(3) $$

$$ ⇒ 50 = 5 + 3n - 3 $$

$$ ⇒ 50 = 2 + 3n $$

$$ ⇒ 50 - 2 = 3n $$

$$ ⇒ n = {48 \over 3 } $$

$$ ⇒ n = 16 $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ ⇒ S_{16}= {16\over 2} × [ { 2( 5 ) + (16 - 1) ×{ 3} } ] $$

$$⇒ S_{16}= 8 × [ { 10 + 15 × 3 } ] $$

$$⇒ S_{16}= 8 × [ { 10 + 45 } ] $$

$$ ⇒ S_{16}= 8 × ( 55) $$

$$ ⇒ S_{16}= 440 $$

Hence, sums of a given AP is $440 $

Question 3 (ii)

In an AP:
(ii) given a = 7, $ a_{13} = 35 $ , find 'd' and $ S_{13} $

Solution :

According to the question

Given, First term, a = 7

Common difference, d = ?

Last term, $ a_{13} = l = 35 $

Number of terms, n = 13

We have to find the Sum of first n terms $ S_{13} = ?$

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 35 = 7 + ( 13 - 1)(d) $$

$$ ⇒ 35 = 7 + 12d $$

$$ ⇒ 35 - 7 = 12d $$

$$ ⇒ 28 = 12d $$

$$ ⇒ d = {28 \over 12 } $$

$$ ⇒ d = {7 \over 3 } $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ ⇒ S_{13} = {13\over 2} × [ { 2( 7 ) + (13 - 1) × {7 \over 3 } } ] $$

$$⇒ S_{13} = {13\over 2} × [ { 14 + 12 × {7 \over 3 } } ] $$

$$⇒ S_{13} = {13\over 2} × [ { 14 + 28 } ] $$

$$ ⇒ S_{13} = {13\over 2} × ( 42) $$

$$ ⇒ S_{13} = 13 × 21 $$

$$ ⇒ S_{13}= 273 $$

Hence, sums of a given AP is $273 $

Question 3 (iii)

In an AP:
(iii) given $ a_{12} = 37 $ , d = 3, find 'a' and $ S_{12} $

Solution :

According to the question

Given, First term, a = ?

Common difference, d = 3

Last term, $ a_{13} = l = 35 $

Number of terms, n = 12

We have to find the Sum of first n terms $ S_{12} = ?$

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 37 = a + ( 12 - 1)×(3) $$

$$ ⇒ 37 = a + 11 × (3) $$

$$ ⇒ 37 = a + 33 $$

$$ ⇒ 37 - 33 = a $$

$$ ⇒ a = 4 $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ ⇒ S_{12} = {12\over 2} × [ { 2( 4 ) + (12 - 1) × {3 } } ] $$

$$⇒ S_{12} = 6 × [ { 8 + 11 × 3 } ] $$

$$⇒ S_{12} = 6 × [ { 8 + 33} ] $$

$$ ⇒ S_{12} = 6 × ( 41) $$

$$ ⇒ S_{12} = 246 $$

Hence, sums of a given AP is $246 $

Question 3 (iv)

In an AP:
(iv) given $ a_{3} = 15 $ and $ S_{10} = 125 $, find 'd' and $ a_{10} $

Solution :

According to the question

Given, $ a_{3} = 15 $

$ S_{10} = 125 $

Common difference, d = ?

We have to find the $ a_{10} = ?$

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 15 = a + ( 3 - 1)(d) $$

$$ ⇒ 15 = a + 2d $$

$$ ⇒ 15 - 2d = a $$

$$ ⇒ a = 15 - 2d ... (1) $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substittuting the value of 'a' from euqation 1

$$ ⇒ S_{10} = {10\over 2}[2 (15 - 2d) + ( 10 - 1) d ] $$

$$ ⇒ 125 = {5}[30 - 4d + 9d ] $$

$$ ⇒ 125 = {5}[30 + 5d ] $$

$$ ⇒ 125 = {150 + 25d } $$

$$ ⇒ 125 - 150 = 25d $$

$$ ⇒ -25 = 25d $$

$$ ⇒ d = -{25\over 25} $$

$$ ⇒ d = -1 $$

Substittuting the value of d = -1 in euqation 1

$$ a = 15 - 2d $$

$$ ⇒ a = 15 - 2(-1) $$

$$ ⇒ a = 15 + 2 $$

$$ ⇒ a = 17 $$

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

$$ a_{10} = a + 9d $$

$$⇒ a_{10} = 17 + 9(-1) $$

$$⇒ a_{10} = 17 - 9 $$

$$⇒ a_{10} = 8 $$

Therefore, the value of d = -1 and $ a_{10} $ = 8 .

Question 3 (v)

In an AP:
(v) given d = 5 , $ S_{9} = 75 $, find 'a' and $ a_{9} $

Solution :

According to the question

Given, a = ?

$ S_{9} = 75 $

Common difference, d = 5

We have to find the $ a_{9} = ?$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

$$ ⇒ S_{9} = {9\over 2}[2( a ) + ( 9 - 1)× 5 ] $$

$$ ⇒ 75 = {9\over 2} × [2( a ) + 40 ] $$

$$ ⇒ 75 × 2 = {9} × [2a + 40 ] $$

$$ ⇒ 150 = 18a + 360 $$

$$ ⇒ 150 - 360 = 18a $$

$$ ⇒ -210 = 18a $$

$$ ⇒ a = -{210\over 18} $$

$$ ⇒ a = -{35\over 3} $$

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

$$ ⇒ a_9 = { -{35\over 3} + ( 9 - 1) 5 } $$

$$ ⇒ a_9 = { -{35\over 3} + 40 } $$

$$ ⇒ a_9 = { {-35 + 120 }\over 3} $$

$$ ⇒ a_9 = {85 \over 3} $$

Therefore, the value of the value of a = $ -{35\over 3} $ and $ a_{9} = {85 \over 3 } $ .

Question 3 (vi)

In an AP:
(vi) given a = 2 , d = 8 , $ S_{n} = 90, $ find 'n' and $ a_{n} $

Solution :

According to the question

Given, a = 2

$ S_{n} = 90, $

Common difference, d = 8

We have to find the n = ? and $ a_{n} = ?$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

$$ ⇒ S_{n} = {n\over 2}[2( 2 ) + ( n - 1) 8 ] $$

$$ ⇒ 90 = {n\over 2}×[4 + 8n - 8 ] $$

$$ ⇒ 90 = {n\over 2}×[ 8n -4 ] $$

$$ ⇒ 90 × 2 = {n}×[8n -4 ] $$

$$ ⇒ 180 = {8n^2} -4n $$

$$ ⇒ {8n^2} -4n - 180 = 0 $$

$$ ⇒ 4({2n^2} -n - 45 )= 0 $$

$$ ⇒ {2n^2} -n - 45 = 0 $$

By expanding middle term. Here - n can be written as – 10n + 9n

$$ ⇒ {2n^2} – 10n + 9n – 45 = 0 $$

$$ ⇒ 2n(n – 5) + 9(n – 5) = 0 $$

$$ ⇒ (2n + 9)(n – 5) = 0 $$

Set each factor equal to zero :
Now, If n – 5 = 0 then

$ n = 5 $

And, if 2n + 9 = 0

$$ ⇒ 2n = -9 $$

$$ ⇒ n = { {-9}\over 2} $$

However, n can neither be negative nor fractional.

Therefore, the value of n = 5

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

$$ ⇒ a_5 = { 2 + ( 5 - 1) 8 } $$

$$ ⇒ a_5 = { 2 + 4 × 8 } $$

$$ ⇒ a_5 = { 34 } $$

Therefore, n = 5 and $ a_5 = { 34 } $.

Question 3 (vii)

In an AP:
(vii) given a = 8 , $ a_{n} = 62 $ , $ S_{n} = 210 $ find 'd' and 'n'

Solution :

According to the question

Given, a = 8

$ S_{n} = 210, $

$ a_{n} = 62 $

We have to find the d = ? and n = ?

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 62 = 8 + (n - 1)d $$

$$ ⇒ (n - 1)d = 54 .... ... (1) $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values we get from euqation 1

$$ ⇒ 210 = {n\over 2}× [2 × 8 + ( n - 1) d ] $$

$$ ⇒ 210 = {n\over 2}× [2 × 8 + 54 ] $$

$$ ⇒ 210 = {n\over 2}× [16 + 54 ] $$

$$ ⇒ 210 = {n\over 2}× 70 $$

$$ ⇒ 210 = 35n $$

$$ ⇒ n = {210 \over 35} = 6 $$

We know that

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

$$ ⇒ a_6 = { 8 + ( 6 - 1) d } $$

$$ ⇒ 62 = { 8 + 5d } $$

$$ ⇒ 62 - 8 = 5d $$

$$ ⇒ d = {54 \over 5} $$

∴ The value of d = ${54 \over 5}$ and n = 6

Question 3 (viii)

In an AP:
(viii) given $ a_{n} = 4 $ , d = 2 , $ S_{n} = -14 $ find 'a' and 'n'

Solution :

According to the question

Given, d = 2

$ S_{n} = -14 $

$ a_{n} = l = 4 $

We have to find the a = ? and n = ?

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 4 = { a + ( n - 1)× 2 } $$

$$ ⇒ 4 = { a + 2n - 2 } $$

$$ ⇒ 4 + 2 = { a + 2n } $$

$$ ⇒ 6 - 2n = a . ... (i) $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values we get from euqation 1

$$ ⇒ -14 = {n\over 2}× [2 × (6 - 2n) + ( n - 1) × 2 ] $$

$$ ⇒ -14 = {n\over 2}× [ 12 - 4n + 2n - 2 ] $$

$$ ⇒ -14 = {n\over 2}[ 10 - 2n] $$

$$ ⇒ -14 × 2 = {n}[ 10 - 2n] $$

$$ ⇒ -28 = 10n - 2n^2 $$

$$ ⇒ 2n^2 -10n -28 = 0 $$

$$ ⇒ n^2 - 5n - 14 = 0 $$

By expanding middle term. Here - 5n can be written as – 7n + 2n

$$ ⇒ n^2 – 7n + 2n – 14 = 0 $$

$$ ⇒ n(n – 7) + 2 (n – 7) = 0 $$

$$ ⇒ (n – 7) (n + 2) = 0 $$

Set each factor equal to zero :

Now, if n – 7 = 0

$$ ⇒ n = 7 $$

and , if n + 2 = 0

$$ ⇒ n = -2 $$

However, n can neither be negative nor fractional.

Therefore, the value of n = 7

From equation 1

$$ a = 6 - 2n $$

$$ ⇒ a = { 6 - 2( 7) } $$

$$ ⇒ a = { 6 - 14 } $$

$$ ⇒ a = { -8 } $$

∴ The value of a = -8 and n = 7

Question 3 (ix)

In an AP:
(ix) given a = 3, n = 8, S = 192, d =?

Solution :

According to the question

Given, a = 3

$ S_{8} = 192 $

n = 8

We have to find the d = ?

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting the values

$$ ⇒ 192 = {8\over 2}×[ { 2(3) + ( 8 - 1) × d } ] $$

$$ ⇒ 192 = {4}×[ { 6 + 7d } ] $$

$$ ⇒ 192 = {24 + 28d } $$

$$ ⇒ 192 - 24 = 28d $$

$$ ⇒ 168 = 28d $$

$$ ⇒ d = { 168 \over 28 } $$

$$ ⇒ d = { 6 } $$

∴ The value of d = 6

Question 3 (x)

In an AP:
(x) given l = 28, S = 144 and there are total 9 terms. Find a.

Solution :

According to the question

Given, a = ?

$ l = 28 $

$ n = 9 $ , and $ S_{9} = 144 $

We know that

$$ S_n = {n\over 2}[ First term + Last term ] $$

$$ ⇒ 144 = {9\over 2}[ a + 28 ] $$

$$ ⇒ {{144 × 2}\over 9 }= [ a + 28 ] $$

$$ ⇒ {{288}\over 9 }= [ a + 28 ] $$

$$ ⇒ 32 = a + 28 $$

$$ ⇒ 32 - 28 = a $$

$$ ⇒ a = 4 $$

∴ The value of a = 4

Question 4

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

Solution :

According to the question

Given, a = 9

$ S_{n} = 636 $

d = 17-9 = 8

$ n = ? $

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values we get from euqation 1

$$ ⇒ 636 = {n\over 2}× [2 × (9) + ( n - 1) × 8 ] $$

$$ ⇒ 636 = {n\over 2}× [ 18 + 8n - 8 ] $$

$$ ⇒ 636 = {n} × [ 9 + 4n - 4 ] $$

$$ ⇒ 636 = 9n + 4n^2 - 4n $$

$$ ⇒ 4n^2 + 5n - 636 = 0 $$

By expanding middle term. Here 5n can be written as +53n −48n

$$ ⇒ 4n^2 +53n − 48n - 636 = 0 $$

$$ ⇒ n(4n + 53) - 12 (4n + 53) = 0 $$

$$ ⇒ (4n + 53) (n - 12) = 0 $$

Set each factor equal to zero :

Now, if 4n + 53 = 0

$$ ⇒ n = {-53 \over 4} $$

and , if n - 12 = 0

$$ ⇒ n = 12 $$

However, n can neither be negative nor fractional.

Therefore, the value of n = 12

Question 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 :

According to the question

Given, First term = 5

Last term = l = 45

$ S_{n} = 400 $

$ n = ? $

We know that

$$ S_n = {n\over 2}[ First term + Last term ] $$

$$ ⇒ 400 = {n\over 2}[ 5 + 45 ] $$

$$ ⇒ 400 = {n\over 2}× [ 50 ] $$

$$ ⇒ 400 = 25n $$

$$ ⇒ n = {400\over 25} $$

$$ ⇒ n = 16 $$

We know that

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

$$ ⇒ a_16 = { 5 + ( 16 - 1) d } $$

$$ ⇒ 45 = { 5 + 15d } $$

$$ ⇒ 45 - 5 = 15d $$

$$ ⇒ d = {40 \over 15} $$

$$ ⇒ d = {8 \over 3} $$

∴ The value of d = ${8 \over 3}$ and n = 16

Question 6

The first and the last terms 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 :

According to the question

Given, First term = 17

Last term = l = 350

$ S_{n} = ? $

$ d = 9 $

We know that

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

$$ ⇒ 350 = { 17 + ( n - 1)× 9} $$

$$ ⇒ 350 = { 17 + 9n - 9} $$

$$ ⇒ 350 = {8 + 9n } $$

$$ ⇒ 350 - 8 = 9n $$

$$ ⇒ 342 = 9n $$

$$ ⇒ n = {342 \over 9} $$

$$ ⇒ n = 38 $$

We know that

$$ S_n = {n\over 2}[ First term + Last term ] $$

$$ ⇒ S_{38} = {38\over 2}[ 17 + 350] $$

$$ ⇒ S_{38} = 19 × 367 $$

$$ ⇒ S_{38} = 6973 $$

Therefore, the value of n = 38 and Sum od A.P. = 6973

Question 7

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

Solution :

According to the question

Given, First term = ?

Last term = l = $ a_{22} = $ 149

$ S_{n} = ? $

$ d = 7 $

We know that

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

$$ ⇒ 149 = { a + ( 22 - 1) × 7} $$

$$ ⇒ 149 = { a + ( 21) × 7} $$

$$ ⇒ 149 = { a + (147} $$

$$ ⇒ 149 -147 = a $$

$$ ⇒ a = 2 $$

We know that

$$ S_n = {n\over 2}[ First term + Last term ] $$

$$ ⇒ S_{22} = {22\over 2}[ 2 + 149] $$

$$ ⇒ S_{22} = 11 × 151 $$

$$ ⇒ S_{22} = 1661 $$

Therefore, the value of a = 2 and Sum od A.P. = 1661

Question 8

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

Solution :

According to the question

Given, First term = ?

$ a_{2} = $ 14

$ a_{3} = $ 18

$ S_{51} = ? $

$ d = ? $

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 14 = a + (2 - 1)d $$

$$ ⇒ 14 = a + d .... ... (1) $$

And

$$ ⇒ 18 = a + (3 - 1)d $$

$$ ⇒ 18 = a + 2d .... ... (2) $$

Subtracting equation (1) from (2), we get :

$$ ⇒ a + 2d - (a + d) = 18 - 14 $$

$$ ⇒ a - a + 2d - d = 4 $$

$$ ⇒ d = 4. $$

Hence, Substituting value of d in equation 1 we get :

$$ ⇒ 14 = a + d $$

$$ ⇒ 14 = a + 4 $$

$$ ⇒ a = 14 -4 $$

$$ ⇒ a = 10 $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{51} = {51\over 2}× [2 × 10 + ( 51 - 1) × 4 ] $$

$$ ⇒ S_{51} = {51\over 2}× [20 + ( 50) × 4 ] $$

$$ ⇒ S_{51} = {51\over 2}× [20 + 200 ] $$

$$ ⇒ S_{51} = {51\over 2}× [220 ] $$

$$ ⇒ S_{51} = 51 × 110 $$

$$ ⇒ S_{51} = 5610 $$

Thus, sum of first 51 terms = 5610

Question 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 :

According to the question

Given, First term = ?

$ S_{7} = 49 $

$ S_{17} = $ 289

$ S_{n} = ? $

and d = ?

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Hence, Substituting values we get :

$$ ⇒ 49 = {7\over 2}×[ { 2a + ( 7 - 1) d } ] $$

$$ ⇒ 49 = {7\over 2}×[ { 2a + 6d } ] $$

$$ ⇒ 49 = {7\over 2}× 2[ {a + 3d } ] $$

$$ ⇒ {49\over 7} = a + 3d $$

$$ ⇒ 7 = a + 3d $$

$$ ⇒ 7 - 3d = a ........(1)$$

And

$$ ⇒ 289 = {17\over 2}×[ { 2a + ( 17 - 1) d } ] $$

$$ ⇒ 289 = {17\over 2}×[ { 2a + 16d } ] $$

$$ ⇒ 289 = {17\over 2}× 2[ {a + 8d } ] $$

$$ ⇒ {289\over 17} = a + 8d $$

$$ ⇒ 17 = a + 8d $$

$$ ⇒ 17 - 8d = a ........(2)$$

Subtracting equation (1) from (2), we get :

$$ ⇒ 17 - 8d - ( 7 - 3d) = a - a $$

$$ ⇒ 17 - 8d - 7 + 3d = 0$$

$$ ⇒ 10 -5d =0 $$

$$ ⇒ 10 = 5d $$

$$ ⇒ d = 2 $$

Hence, Substituting value of d in equation 1 we get :

$$ ⇒ 7 - 3d = a$$

$$ ⇒ 7 - 3 × 2 = a $$

$$ ⇒ 7 - 6 = a $$

$$ ⇒ a = 1 $$

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{n} = {n\over 2}× [2 × 1 + ( n - 1) × 2 ] $$

$$ ⇒ S_{n} = {n\over 2}× [2 + 2n - 2 ] $$

$$ ⇒ S_{n} = {n\over 2} × 2n $$

$$ ⇒ S_{n} = n × n $$

$$ ⇒ S_{n} = n^2 $$

Thus, sum of first n terms = $ S_{n} = n^2 $

Question 10 (i)

Show that $a_{1}$, $a_{2}$,......., an,........ form an AP where an is defined as below :
(i) $a_{n}$ = 3 + 4n . Also find the sum of the first 15 terms in each case.

Solution :

According to the question

Given, $a_{n}$ = 3 + 4n

Hence, Substituting values we get :

First term = $a_{1}$ = 3 + 4 × 1

$a_{1}$ = 3 + 4 = 7

Second term = $a_{2}$ = 3 + 4 × 2

$a_{2}$ = 3 + 8 = 11

And

Third term = $a_{3}$ = 3 + 4 ×3

$a_{3}$ = 3 + 12 = 15

Fourth term = $a_{4}$ = 3 + 4 × 4

$a_{2}$ = 3 + 16 = 19

Thus, list of number formed : 7, 11 , 15, 19....

here $ a_2 - a_1 = 11 - 7 = 4,$

and $ a_3 - a_2 = 15 - 11 = 4,$

Since, $ a_3 - a_2 = a_2 - a_1,$

Thus, given list of number is in Arithmetic Progression(AP) with common difference = 4 and first term a = 7

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{15} = {15\over 2}× [2 × 7 + ( 15 - 1) × 4 ] $$

$$ ⇒ S_{15} = {15\over 2}× [14 + ( 14) × 4 ] $$

$$ ⇒ S_{15} = {15\over 2}× [14 + 56 ] $$

$$ ⇒ S_{15} = {15\over 2}× 70 $$

$$ ⇒ S_{15} = 15 × 35 $$

$$ ⇒ S_{15} = 525 $$

Thus, sum of first n terms = $ S_{15} = 525 $

Question 10 (ii)

Show that $a_{1}$, $a_{2}$,......., an,........ form an AP where an is defined as below :
(i) $a_{n}$ = 9 - 5n . Also find the sum of the first 15 terms in each case.

Solution :

According to the question

Given, $a_{n}$ = 9 - 5n

Hence, Substituting values we get :

First term = $a_{1}$ = 9 -5 × 1

$a_{1}$ = 9 -5 = 4

Second term = $a_{2}$ = 9 -5 × 2

$a_{2}$ = 9 - 10 = -1

And

Third term = $a_{3}$ = 9 -5 ×3

$a_{3}$ = 9 - 15 = -6

Fourth term = $a_{4}$ = 9 -5 × 4

$a_{2}$ = 9 -20 = -11

Thus, list of number formed : 4, -1 , -6, -11...

here $ a_2 - a_1 = 1 - 4 = -5,$

and $ a_3 - a_2 = -6 - (-1) = -5,$

Since, $ a_3 - a_2 = a_2 - a_1,$

Thus, given list of number is in Arithmetic Progression(AP) with common difference = -5 and first term a = 4

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{15} = {15\over 2}× [2 × 4 + ( 15 - 1) × (-5) ] $$

$$ ⇒ S_{15} = {15\over 2}× [8 + ( 14) × (-5 )] $$

$$ ⇒ S_{15} = {15\over 2}× [8 - 70 ] $$

$$ ⇒ S_{15} = {15\over 2}× (-62) $$

$$ ⇒ S_{15} = 15 × (-31) $$

$$ ⇒ S_{15} = -465 $$

Thus, sum of first n terms = $ S_{15} = - 465 $

Question 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 3rd, the 10th and the nth terms.

Solution :

According to the question

Given, Sum of first n terms $S_{n}$ = $ 4n – n^2 $

$ a_2 = ? , a_3 = ? $ and $ a_{10} = ?$ and $ S_2 = ?$

Hence, Substituting values we get :

Sum of First term = $S_{1}$ = $ 4 × 1 – (1)^2 $

$S_{1}$ = (4 - 1) = 3

Sum of Second term = $S_{2}$ = $ 4 × 2 – (2)^2 $

$S_{2}$ = (8 - 4) = 4

Sum of Third term = $S_{3}$ = $ 4 × 3 – (3)^2 $

$S_{3}$ = (12 - 9) = 3

Sum of nineth term = $S_{9}$ = $ 4 × 9 – (9)^2 $

$S_{9}$ = (36 - 81) = - 45

Sum of Tenth term = $S_{10}$ = $ 4 × 10 – (10)^2 $

$S_{10}$ = (40 - 100 )= -60

Sum of (n-1)th term = $S_{n-1}$ = $ 4 × (n-1) – (n-1)^2 $

$S_{n-1}= 4n -4 -(n^2 + 1 - 2n) $

$S_{n-1}= 4n -4 - n^2 - 1 + 2n $

$S_{n-1}= 6n - 5 - n^2 $

Now we know by the formula that: ($S_{n}$ - $S_{n-1}$ ) = $ a_n $

Thus, Second term $ a_2 $ = ($S_{2}$ - $S_{1}$ ) = ( 4 - 3 )= 1

Thus, Third term $ a_3 $ = ( $S_{3}$ - $S_{2}$ ) = (3 - 4) = -1

Thus, Tenth term $ a_{10} $ = ( $S_{10}$ - $S_{9}$ ) = (-60 -(-45)) = -15

Thus, nth term $ a_n $ = ( $S_{n}$ - $S_{n-1}$ ) = ($ 4n – n^2 $)- ($6n - 5 - n^2 $)

= ($ 4n – n^2 - 6n + 5 + n^2$)

= ($ 5 - 2n $)

Therefore, $S_{1}$ = 3 , $S_{2}$ = 4, and $ a_2 $ = 1, $ a_3 $ = -1 and $ a_{10} $ = -15, $ a_n $ = ( 5 - 2n)

Question 12

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

Solution :

The smallest positive integer which is divisible by 6 is 6 itself

Thus, here, First term = $a_{1}$ = 6

A.P. = 6,12,18,24,……… up to 40 term

Common difference, d = 6

Number of terms n = 40

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{40} = {40\over 2}× [2 × 6 + ( 40 - 1) × 6 ] $$

$$ ⇒ S_{40} = {20 }× [12 + ( 39) × 6 ] $$

$$ ⇒ S_{40} = {20 }× [ 12 + 234] $$

$$ ⇒ S_{40} = 20 × 246 $$

$$ ⇒ S_{40} = 4920 $$

Thus, sum of first 40 terms = $ S_{40} = 4920 $

Question 13

Find the sum of the first 15 multiples of 8.

Solution :

First multiple of 8 is 8 itself

Thus, here, First term = $a_{1}$ = 8

A.P. = 8,16,24,……… up to 15 term

Common difference, d = 8

Number of terms n = 15

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{15} = {15\over 2}× [2 × 8 + ( 15 - 1) × 8 ] $$

$$ ⇒ S_{15} = {15\over 2}× [16 + ( 14) × 8 ] $$

$$ ⇒ S_{15} = {15\over 2}× [16 + 112 ] $$

$$ ⇒ S_{15} = {15\over 2}× 128 $$

$$ ⇒ S_{15} = 15 × 64 $$

$$ ⇒ S_{15} = 960 $$

Thus, sum of first 15 multiples of 8 is = 960

Question 14

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

Solution :

First odd number is 1 itself

Thus, here, First term = $a_{1}$ = 1

A.P. = 1,3,5,,……… 49

Common difference, d = 2

$a_{n}$ = 49

Number of terms can be calculated as follows, We know that ,

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

Hence, Substituting values we get :

$$ ⇒ 49 = 1 + (n - 1)× 2 $$

$$ ⇒ 49 = 1 + 2n - 2 $$

$$ ⇒ 49 = -1 + 2n $$

$$ ⇒ 50 = 2n $$

$$ ⇒ n = 25 $$

Number of terms n = 25

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{25} = {25\over 2}× [2 × 1 + ( 25 - 1) × 2 ] $$

$$ ⇒ S_{25} = {25\over 2}× [2 + ( 24) × 2 ] $$

$$ ⇒ S_{25} = {25\over 2}× [2 + 48 ] $$

$$ ⇒ S_{25} = {25\over 2}× 50 $$

$$ ⇒ S_{25} = {25}× 25 $$

$$ ⇒ S_{25} = 625 $$

Thus, sum of the odd numbers between 0 and 50 = 625

Question 15

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being ₹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 :

First day penalty = Rs. 200

Thus, here, First term = $a_{1}$ = 200

A.P. = 200 , 250 , 300 , ......... upto 30 term

Common difference, d = 50

$a_{n}$ = 30

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{30} = {30\over 2}× [2 × 200 + ( 30 - 1) × 50 ] $$

$$ ⇒ S_{30} = {15}× [400 + ( 29) × 50 ] $$

$$ ⇒ S_{30} = {15}× [400 + 1450 ] $$

$$ ⇒ S_{30} = {15}× 1850 $$

$$ ⇒ S_{30} = 27,750 $$

Thus, total penalty = Rs. 27,750

Question 16

A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.

Solution :

First term = ?

Sum of first 7 terms $ S_7$ = 700

Common difference, d = 20

total number of terms = 7

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values

$$ ⇒ S_{7} = {7\over 2}× [2a + ( 7 - 1) × 20 ] $$

$$ ⇒ 700 = {7\over 2}× [2a + ( 6) × 20 ] $$

$$ ⇒ 700 = {7\over 2}× [2a + 120 ] $$

$$ ⇒ 700 = {7}× [a + 60 ] $$

$$ ⇒ {700\over 7} = a + 60 $$

$$ ⇒ {100} = a + 60 $$

$$ ⇒ a = 100 - 60 $$

$$ ⇒ a = 40 $$

We know that

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

Hence, Substituting values we get :

Second term = $$ a_{2} = { 40 + ( 2 - 1) × 20 } $$

$$ a_{2} = { 40 + 1 × 20 } = 60 $$

Third term = $$ a_{3} = { 40 + (3 - 1) × 20 } $$

$$ a_{3} = { 40 + 2 × 20 } = 80 $$

And, Fourth term = $$ a_{4} = { 40 + (4 - 1) × 20 } $$

$$ a_{4} = { 40 + 3 × 20 } = 100 $$

Fifth term = $$ a_{5} = { 40 + ( 5 - 1) × 20 } $$

$$ a_{5} = { 40 + 4 × 20 } = 120 $$

Sixth term = $$ a_{6} = { 40 + (6 - 1) × 20 } $$

$$ a_{3} = { 40 + 5 × 20 } = 140 $$

Seventh term = $$ a_{7} = { 40 + (7 - 1) × 20 } $$

$$ a_{4} = { 40 + 6 × 20 } = 160 $$

Thus, values of each of the prizes are, 160, 140, 120, 100, 80, 60, and 40 respectively

Question 17

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution :

First there are 12 classes and each class has 3 sections

Since each section of class 1 will plant 1 tree, so 3 trees will be planted by 3 sections of class 1

Therefor, number of trees planted by Class I = $ 1 × 3 = 3 $

Similarly, number of trees planted by Class II = $ 2 × 3 = 6 $

Similarly, number of trees planted by Class III = $ 3 × 3 = 9 $

Its clearly an AP with first term $a_{1}$ = 3 , d = 3 and n =12

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values for sum of first 12 terms

$$ ⇒ S_{12} = {12\over 2}× [2 × 3 + ( 12 - 1) × 3 ] $$

$$ ⇒ S_{12} = {6}× [ 6 + ( 11) × 3 ] $$

$$ ⇒ S_{12} = {6} × [ 6 + 33 ] $$

$$ ⇒ S_{12} = {6} × 39 $$

$$ ⇒ S_{12} = 234 $$

Therefore, total number of trees planted by students will be = 234

Question 18

A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ....... as shown in the fig. below. What is the total length of such a spiral made up of thirteen consecutive semicircles? ( Take π = 22/7)

Solution :

The spiral is made of 13 semicircles

The length of a semicircle is given by the formula L = π × r, where r is the radius.

First radius = $ r_{1}$ = 0.5 cm

Second radius = $ r_{2}$ = 1.0 cm

Third radius = $ r_{3}$ = 1.5 cm

Thirteenth radius = $ r_{13}$ = 6.5 cm

Its clearly , The radii form an arithmetic series: 0.5 + 1.0 + 1.5 + .....+6.5

n = number of terms = 13

first term = $ a_{1}$ = 0.5

last term = $ a_{13}$ = 6.5

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { a + l }] $$

Substituting values for sum of first 12 terms

$$ ⇒ S_{13} = {13\over 2}× [0.5 + 6.5] $$

$$ ⇒ S_{13} = {13\over 2}× 13 $$

$$ ⇒ S_{13} = 13 × 3.5 $$

$$ ⇒ S_{13} = 45.5 cm $$

Therefore, Total Length = π × (sum of radius)

$$ ⇒ {22\over 7}× 45.5 $$

$$ ⇒ 22 × 6.5 $$

$$ ⇒ 143 cm $$

Total Length of spiral = 143 cm

Question 19

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution :

$S_n$ = 200

first term = $ a_{1}$ = (logs in the bottom row) = 20

common difference = d = -1

$ n = ? $

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values we get

$$ ⇒ 200 = {n\over 2}× [2 × (20) + ( n - 1) × (-1) ] $$

$$ ⇒ 200 = {n\over 2}× [40 - n + 1 ] $$

$$ ⇒ 200 × 2 = n × [41 - n ] $$

$$ ⇒ 400 = 41n -n^2 $$

$$ ⇒ 41n -n^2 - 400 = 0 $$

$$ ⇒ n^2 - 41n + 400 = 0 $$

By expanding middle term. Here -41n can be written as -16n −25n

$$ ⇒ n^2 -16n −25n + 400 = 0 $$

$$ ⇒ n(n - 16) - 25 (n - 16) = 0 $$

$$ ⇒ (n - 25) (n - 16) = 0 $$

Set each factor equal to zero :

Now, if (n - 25) = 0

$$ ⇒ n = 25 $$

and , if n - 16 = 0

$$ ⇒ n = 16 $$

This gives two possible solutions for the number of rows: n=16 or n=25.

To find the correct number of rows, we check which solution gives a realistic number of logs in the top row. The number of logs in any row must be a positive integer. We know that

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

Hence, Substituting values we get :

$$ a_{25} = { 20 + (25 - 1) × (-1)} $$

$$ a_{25} = { 20 - 24 } = -4 $$

A negative number of logs is not possible, so n=25 is not a valid solution.

$$ a_{16} = { 20 + (16 - 1) × (-1)} $$

$$ a_{16} = { 20 + (-15) } = 5 $$

This is a valid solution.

Therefore, the number of rows is 16, and the number of logs in the top row is 5.

Question 20

In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?

Solution :

Identify the Distances Run

First Potato: The potato is 5 m away

Distance = 2 × 5 = 10 m

Second Potato: The potato is 5 + 3 = 8 m away.

Distance = 2 × 8 = 16 m

Third Potato: The potato is 8 + 3 = 11 m away

Distance = 2 × 11 = 22 m

The distances form an arithmetic progression: 10, 16, 22, ....

first term = $ a_{1}$ = 10

common difference = d = 16 − 10 = 6 m

$ n = 10 $

We know that , Sum of first n terms

$$ S_n = {n\over 2}[ { 2a + ( n - 1) d } ] $$

Substituting values we get

$$ ⇒ S_{10} = {10\over 2}× [2 × (10) + ( 10 - 1) × 6 ] $$

$$ ⇒ S_{10} = 5 × [20 + ( 9) × 6 ] $$

$$ ⇒ S_{10} = 5 × [20 + 54 ] $$

$$ ⇒ S_{10} = 5 × 74 $$

$$ ⇒ S_{10} = 370 $$

The total distance the competitor has to run is 370 m.