Question

2nd and 45th term of an arithmetic progression are 10 and 96 respectively find the first term and common difference and hence find sum of the first 15 terms

Answer 1

so a=8, d=2, S 25=330

Answer 2

Answer:

330

Step-by-step explanation:

Formula of nth term in A.P. = $a_n=a+(n-1)d$

Substitute n = 2

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

Since we are given that 2nd term is 10

So,$10=a+d$  ---1

Substitute n = 45

$a_{45}=a+(45-1)d$

$a_{45}=a+44d$

Since we are given that 45th term is 96

$96=a+44d$  --2

Subtract 1 from 2

$96 -10 =a+44d-a-d$

$86 =43d$

$2=d$

Substitute value of d in 1 to get value of a

$10=a+2$

$10-2=a$

$8=a$

Thus the first term is 8

Common difference is 2

Formula of sum of first n terms= $S_n=\frac{n}{2}(2a+(n-1)d)$

Put n =15

$S_{15}=\frac{15}{2}(2(8)+(15-1)2)$

$S_{15}=330$

Hence the sum of first 15 terms is 330