Question

Find the sum of the first $111$ terms of an A.P. whose $56$th term is $\frac{5}{37}$

Answer 1

We know that the formula for the nth term is $t_n=a+(n-1)d$, where $a$ is the first term, $d$ is the common difference.

It is given that the $56$th term is $t_{56}=\frac{5}{37}$, therefore,

$\frac{5}{37}=a+(56-1)d\Rightarrow a+55d=\frac{5}{37}.......\left(1\right)$

We also know the sum of $n$ terms of an A.P with first term $a$ and the common difference $d$ is:

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

Therefore, using equation 1, the sum of first $111$ terms is:

$S_{111}=\frac{111}{2}\left[\right(2\times a)+(111-1\left)\right(d\left)\right]=\frac{111}{2}(2a+110d)=\frac{111\times 2}{2}(a+55d)=111\times \frac{5}{37}=3\times 5=15$

Hence, the sum is $15$.