Question

The first term and the c.d of an AP are 13 and 5 respectively. Find the sum from the sixth term to the 20th term *

1 point

a) 576

b) 146

c) 1095

d) 1080

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Answer 1

ANSWER :–

Required sum = 1,095

EXPLANATION :–

GIVEN :–

• First term of A.P. = 13

• Common difference (c.d.) = 5

TO FIND :–

☞ The sum from the sixth term to the 20th term.

SOLUTION :–

• We know that sum of n terms of A.P. –

☞ ${ \boxed { \bold { s_{n} = \dfrac{n}{2}[2a + (n - 1)d]}}}$

• First we have to find –

☞ Sum of first 5 terms –

${ \bold { \implies (S_{5}) = \dfrac{5}{2}[2(13) + (5 - 1)5]}}$

$\\ { \bold { \implies S_{5} = \dfrac{5}{2}[26 + (4)5]}}$

$\\ { \bold { \implies S_{5} = \dfrac{5}{2}[26 + 20]}}$

$\\ { \bold { \implies S_{5} = \dfrac{5}{2}(46)}}$

$\\ { \bold { \implies S_{5} = 5 \times 23=115 }}$

• Now we have to find –

☞ Sum of first 20 terms –

${ \bold { \implies (S_{20}) = \dfrac{20}{2}[2(13) + (20 - 1)5]}}$

$\\ { \bold { \implies S_{20} = (10)[26 + (19)5]}}$

$\\ { \bold { \implies S_{20} = (10)[26 + 95]}}$

$\\ { \bold { \implies S_{20} = (10)(121)}}$

$\\ { \bold { \implies S_{20} = 1210}}$

• Now –

☞ Sum from the sixth term to the 20th term = ${ \bold { S_{20} - S_{5}}}$

=> Required sum = 1210 - 115 = 1095

☞ Hence, option (c) is correct