Question

Find the sum of $A.P.$ whose first and last term is $13$ and $216$ respectively & common difference is $7$.

• A. $3434$
• B. $3435$
• C. $1545$
• D. $3456$

Answer 1

The correct option is B $3435$

Let ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{d}^{\prime }$ be the first term and the common difference of the given AP respectively.

Let there are ${}^{\prime }{n}^{\prime }$ terms in the given AP.

Given $a=13,d=7$ and $a_n=216$

$\Rightarrow a+(n-1)d=216$ [$\because n^{th}$ term of an AP is given by $a_n=a+(n-1)d$]

$\Rightarrow 13+(n-1)\times 7=216$

$\Rightarrow 13+(n-1)\times 7=216$

$\Rightarrow (n-1)=\frac{216}{7}$

$\Rightarrow n=29+1=30$

Therefore, there are $30$ terms in the given AP.

We know that the Sum of ${}^{\prime }{n}^{\prime }$ terms is given by $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

Now, Required sum of $30$ terms $=S_{30}=\frac{30}{2}[2\times 13+(30-1)\times 7]$

$=15\times [26+29\times 7]$

$=15\times 229$

$=3435$
Hence, Required sum is $3435$.