Question

The number of terms of the AP (Arithmetic Progression) $3,7,11,15,...$ to be taken so that the sum is $406$ is

• A. $5$
• B. $10$
• C. $12$
• D. $14$

Answer 1

The correct option is D

$14$

Step 1: Apply the formula of the sum of term

The given AP (Arithmetic Progression) is $3,7,11,15,...$.

Here, the first term $a=3$

And, the common difference $d=7-3=4$

Let, the required number of terms be $n$.

Also, it is given that the sum of $n$ terms $S_n=406$

We know that the sum of $n$ terms of an AP is calculated by,

${\mathit{S}}_{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}}{\mathbf{2}}\left[2a+\left(n-1\right)d\right]$ $...\left(\mathrm{i}\right)$

Here, $a$ is the first term, $n$ is the number of terms and $d$ is the common difference.

Substituting the values of $a,d$ and $S_n$ in the equation $\left(\mathrm{i}\right)$, we get,

$\Rightarrow$ $406=\frac{\mathrm{n}}{2}[2\times 3+\left(\mathrm{n}-1\right)4]$

$\Rightarrow$ $406=\frac{n}{2}\times 2[3+\left(\mathrm{n}-1\right)2]$

$\Rightarrow$ $406=n[3+\left(\mathrm{n}-1\right)2]$

$\Rightarrow$ $406=2n^2+n$

$\Rightarrow$ $0=2n^2+n-406$

Step 2: Calculate the number of terms

Factorising the above quadratic equation,

$\because$ $2n^2+n-406=0$

$\Rightarrow$ $2n^2+29n-28n-406=0$

$\Rightarrow$ $n\left(2n+29\right)-14\left(2n+29\right)=0$

$\Rightarrow$ $\left(2n+29\right)\left(n-14\right)=0$

$\Rightarrow$ $n=-\frac{29}{2}$ or $n=14$

Since the number of terms cannot be negative.

Then, $n=14$

Thus, the sum of $14$ terms of the AP $3,7,11,15,...$ is $406$.

Hence, option (D) is the correct option.