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

Answer 1

The correct option is C $306$

It is given that $a_1,a_2,a_3,.......,a_n$ are in A.P and we know that:

The general form of an Arithmetic Progression is $a,a+d,a+2d,a+3d$ and so on. Thus $n$th term of an A.P series is $a_n=a_1+(n-1)d$, where $a_n=n$th term and $a=$ first term. Here $d=$ common difference = $a_n-a_{n-1}$.

Now consider,

$a_3+a_7+a_{11}+a_{15}=72\Rightarrow [a_1+(3-1\left)d\right]+[a_1+(7-1\left)d\right]+[a_1+(11-1\left)d\right]+[a_1+(15-1\left)d\right]=72\Rightarrow a_1+2d+a_1+6d+a_1+10d+a_1+14d=72\Rightarrow 4a_1+32d=72$

$\Rightarrow 4(a_1+8d)=72\Rightarrow a_1+8d=\frac{72}{4}\Rightarrow a_1+8d=18.......\left(1\right)$

We also know that the sum of first $n$ terms of an A.P is $S_n=\frac{n}{2}[2a_1+(n-1\left)d\right]$, therefore, the sum of first $17$ terms of the given A.P is as follows:

$S_{17}=\frac{17}{2}[2a_1+(17-1\left)d\right]=\frac{17}{2}(2a_1+16d)=\frac{17}{2}\times 2(a_1+8d)=17(a_1+8d)=17\times 18\left(Fromeqn\right(1\left)\right)=306$

Hence, the sum of first $17$ terms is $306$.