Question

How many terms of the series $12,16,20,.....$ must be taken to make the sum equal to $208$?

Answer 1

The given series is $12,16,20,.....$

Let the number of terms in the series be $n$ , such that the sum of $n$ terms in the series is $208$

The first term in the series be $a=12$ and the common difference be $d=16-12=4$

The sum of $n$ terms in the series is $\frac{n}{2}(2a+(n-1\left)d\right)$ , given that it is equal to $208$

$\Rightarrow \frac{n}{2}(2a+(n-1\left)d\right)=208$

$\Rightarrow \frac{n}{2}(24+(n-1)\times 4)=208$

$\Rightarrow n(n+5)=104=8\times 13$

$\Rightarrow n=8$

Therefore we must take $8$ terms in the series in order to make their sum equal to $208$