Question

How many term of the $AP,17,15,13$, must be added to get the sum $72$? Explain the double answer.

Answer 1

The series is given as $17,15,13,\cdots$.

The first term of the series is 17 and the common difference is $-2$.

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

$72=\frac{n}{2}\left[2\left(17\right)+\left(n-1\right)\left(-2\right)\right]$

$144=n\left(34-2n+2\right)$

$144=n\left(36-2n\right)$

$144=36n-2n^2$

$2n^2-36n+144=0$

$n^2-18n+72=0$

$n^2-12n-6n+72=0$

$n\left(n-12\right)-6\left(n-12\right)=0$

$\left(n-6\right)\left(n-12\right)=0$

$n=6,12$

For $n=6$, the series will be $17,15,13,12,11,9$ whose sum is 72.

For $n=12$, the series will be $17,15,13,11,9,7,5,3,1,-1,-3,-5$ whose sum is also 72 because the last six terms will be cancelled from each other due to their opposite signs.

Therefore, the double answer is valid.