Question

How many terms are there in the A.P. whose first and fifth terms are −14 and 2 respectively and the sum of the terms is 40?

Answer 1

In the given problem, we have the sum of the certain number of terms of an A.P. and we need to find the number of terms.

Here, let us take the common difference as d.

So, we are given,

First term (a₁) = −14

Fifth term (a₅) = 2

Sum of terms (s_n) = 40

Now,

`a_5 = a_1 + 4d`

2 = -14 + 4d

2 + 14 = 4d

`d = 16/4`

d= 4

Further, let us find the number of terms whose sum is 40. For that, 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 (a₁) = −14

The sum of n terms (S_n) = 40

Common difference of the A.P. (d) = 4

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

`40 = n/2[2(-14) + (n -1) (4)]`

`40 = (n/2) [-28 = (4n -4)]`

`40 = (n/2)[-32 + 4n]`

`40(2)= -32n + 4n^2`

So, we get the following quadratic equation,

`4n^2 - 32n - 80 = 0`

`n^2 - 8n + 20 =  0`

On solving by splitting the middle term, we get,

`n^2 - 10n + 2n + 20 = 0`

`n(n - 10) + 2(n - 10)=0`

`(n + 2)(n - 10)=0`

Further,

n + 2 = 0

n = -2

Or

n - 10  = 0

n= 10

Since the number of terms cannot be negative. Therefore, the number of terms (n) is n = 10