Question

Find the sum of $7+10\frac{1}{2}+14+...+84$.

• A. $\frac{2093}{2}$
• B. $\frac{1093}{2}$
• C. $\frac{2193}{2}$
• D. $\frac{3093}{2}$

Answer 1

Answer: (A)

$\frac{2093}{2}$

Clearly, terms of the given series form an AP with first term and last term $a=7$ & $l=a_n=84$ respectively and common difference $d=\frac{21}{2}-7=\frac{7}{2}$.

Let the given series has $n$ terms.

Therefore,
$a_n=84=a+(n-1)d$

$84=7+(n-1)\times \frac{7}{2}$
$(n-1)\times \frac{7}{2}=77$

$\therefore n=23$
Here we use the following formula to calculate the sum of terms:

$S_n=\frac{n}{2}(a+l)$

$\therefore$Required sum $=S_{23}=\frac{23}{2}(7+84)$

$\Rightarrow S_{23}=\frac{23\times 91}{2}=\frac{2093}{2}$