Question

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

Options

• 5

• 10

• 12

• 14

• 20

Answer 1

In the given problem, we have an A.P.  3,7,11,15,....

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 = n/2[2a + (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 (S_n) = 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 = n/2 [ 2(3) + (n-1) (4) ] `

`406 = (n/2) [ 6 +(4n- 4)]`

`406 = (n/2) [ 2 + 4n]`

`406 = n  + 2n^2`

So, we get the following quadratic equation,

`2n^2 + n - 406 = 0`

On solving by splitting the middle term, we get,

`2n^2 - 28n + 29n - 406 = 0`

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

           `(2n - 29 ) ( n- 14) = 0`

Further,

2n - 29 = 0

`n = 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