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

Question:

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

(a) 5

(b) 10

(c) 12

(d) 14

(e) 20

Solution:

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.

= 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).

 

 

 

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