Question

Question 32
How many terms of the AP –15, –13, –11,. … are needed to make the sum –55?

Answer 1

Let n be the number of terms needed to make the sum 55.
Here, first term (a) = - 15, common difference (d) = - 13 + 15 = 2
Sum of n terms of an AP, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow -55=\frac{n}{2}\left[2\right(-15)+(n-1\left)2\right][\because S_n=-55(given\left)\right]$
$\Rightarrow -55=-15n+n(n-1)$
$\Rightarrow n^2-16n+55=0$
$\Rightarrow n^2-11n-5n+55=0\left[byfactorizationmethod\right]$
$\Rightarrow n(n-11)-5(n-11)=0$
$\Rightarrow (n-11)(n-5)=0$
$\therefore n=5,11$
Hence, either 5 or 11 terms are needed to make the sum - 55.