Question

How many terms of the $AP:15,13,11,.......$ are needed to make the sum 55?

Answer 1

In the given $AP:15,13,11,...$
$a=15$, $d=13-15=-2$

Let the number of termswhose sum is 55 be $n$

$S_n=\frac{n}{2}[2a+(n-1\left)\right]d=55$

$\Rightarrow \frac{n}{2}[2\times 15+(n-1)\times -2]=55$
$\Rightarrow \frac{n}{2}[30-2n+2]=55$

$\Rightarrow n[-2n+32]=110$
$\Rightarrow -2n^2+32n=110$

$\Rightarrow 2n^2-32n+110=0$

$\Rightarrow n^2-16n+55=0$
$\Rightarrow n^2-5n-11n+55=0$
$\Rightarrow n(n-5)-11(n-5)=0$

$\Rightarrow (n-5)(n-11)=0$

$\Rightarrow n=5,11$

The values of $n$ are positive so they are valid. We get the double answer because the sum of $6^{th}$ to ${11}^{th}$ terms is zero as some terms are negative and some are positive.