Question

Sum $\frac{3}{4},\frac{2}{3},\frac{7}{12}$, .... to $19$ terms.

Answer 1

Given series is $\frac{3}{4},\frac{2}{3},\frac{7}{12},......$ to 19 terms

Given series is equivalent to $\frac{9}{12},\frac{8}{12},\frac{7}{12},.....$ to 19 terms

Given series is equivalent to $\frac{1}{12}(9,8,7,........)$ to 19 terms

Observe that $9,8,7,.......$ is in AP with common difference $-1$ and first term as $9$

Therefore, sum of first $19$ terms in that series is $\frac{1}{12}\left(\frac{19}{2}\right(2\times 9+(19-1)\times -1\left)\right)=0$