Question

Let $a_1,a_2,a_3,....,a_n,.....$ be in A.P. If $a_3+a_7+a_{11}+a_{15}=72$, then the sum of its first $17$ terms is equal to:

• A. $306$
• B. $204$
• C. $153$
• D. $612$

Answer 1

The correct option is A $306$

Given: $a_1,a_2,a_3,....,a_n$ are in A.P

Then, $a_3+a_7+a_{11}+a_{15}=72$

$\Rightarrow a+2d+a+6d+a+10d+a+14d=72[\because a_n=a+(n-1\left)d\right]$

$\Rightarrow 4a+32d=72$

$\Rightarrow a+8d=18...\left(i\right)$

Now, $S_{17}=\frac{17}{2}[2a+(17-1\left)d\right][\because S_n=\frac{n}{2}[2a+(n-1)d\left]\right]$

$=\frac{17}{2}[2a+16d]$

$=\frac{17}{2}\times 2[a+8d]$

$=17\times 18\left[\text{using}\right(i\left)\right]$

$=306$

So, $\text{A}$ is the correct option.