Question

Find the sum of first 15 terms of an A.P. whose 5th and 9th terms are 26 and 42 respectively

Answer 1

Given an AP with :

Fifth term, $a_5=26$

Ninth term, $a_9=42$

We know that $n^{th}$ term of AP is given by: $a_n=a+(n-1)d$

$a_5=26=a+(5-1)d.........\left(i\right)$

$a_9=42=a+(9-1)d...........\left(ii\right)$

Subtracting (i) from (ii), we get

$16=4d$

$d=4$

Substituting value of $d$ in (i), we get

$26=a+4\times 4$

$a=10$

Now, sum of first $n$ terms is given by:

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

Substituting values, we get sum of fifteen terms as

$S_{15}=\frac{15}{2}(2\times 10+(15-1)\times 4)$
$S_{15}=\frac{15}{2}(20+56)$
$S_{15}=15\times 38$

$S_{15}=570$