Question

How many terms of the A.P. $-6,-\frac{11}{2},-5,\dots \dots$ are needed to give the sum -25 ?

Answer 1

Let the number of terms to be added to the series is n
Now a = -6 and d = 0.5
Therefore
$-25=\frac{n}{2}\left[2\right(-6)+(n-1\left)\right(0.5\left)\right]-50=n[-12+0.5n-0.5]=-12.5n+0.5n^2+50=0n^2-25n+100=0\Rightarrow n^2-20n-5n+100\Rightarrow n(n-20)-5(n-20)\Rightarrow (n-20)(n-5)n=20,5$
$\therefore$ Value of n will be either 20 or 5.