Question

How many terms of the A.P. 17, 15, 13, ...... must be taken, so that their sum is 81?

Answer 1

Given:

$a_1=17,a_2=15,a_3=13$

Sum of $n$ terms : $S_n=81$

To find $n$

Common difference is given by: $d=a_n-a_{n-1}$

Using above, $d=a_2-a_1$

$d=15-17=-2$

Sum of $n$-terms of an AP with first terms $a_1$ and common difference $d$ is given by: $S_n=\frac{n}{2}(2a_1+(n-1\left)d\right)$

$81=\frac{n}{2}[2\times 17+(n-1)\times (-2\left)\right]$

$162=-2n^2+36n$

$n^2-18n+81=0$

$(n-9{)}^2=0$

$n=9$