The number of terms of the A.P. 3, 7, 11, 15, ... to be taken so that the sum is 406 is

In the given problem, we have an A.P. 

Here, we need to find the number of terms n such that the sum of n terms is 406.

So here, we will use the formula,

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

Where; a = first term for the given A.P.

d = common difference of the given A.P.

n = number of terms

The first term (a) = 3

The sum of n terms (Sn) = 406

Common difference of the A.P. $(d)=a_{2}-a_{1}$

$=7-3$

$=4$

So, on substituting the values in the formula for the sum of n terms of an A.P., we get,

$406=\frac{n}{2}[2(3)+(n-1)(4)]$

$406=\left(\frac{n}{2}\right)[6+(4 n-4)]$

$406=\left(\frac{n}{2}\right)[2+4 n]$

$406=n+2 n^{2}$

So, we get the following quadratic equation,

$2 n^{2}+n-406=0$

On solving by splitting the middle term, we get,

$2 n^{2}-28 n+29 n-406=0$

$2 n(n-14)-29(n-14)=0$

$(2 n-29)(n-14)=0$

Further,

$2 n-29=0$

$n=\frac{29}{2}$

Or,

$n-14=0$

$n=14$

Since, the number of terms cannot be a fraction, the number of terms $(n)$ is $n=14$

Hence, the correct option is (d).