Let there be an A.P. with the first term '*a*', common difference'. If *a*_{n} a denotes in *n*^{th} term and *S*_{n} the sum of first *n* terms, find.

n and a, if a_{n} = 4, d = 2 and S_{n}_{ }= −14.

#### Solution

Here, we have an A.P. whose n^{th} term (a_{n}), the sum of first n terms (S_{n}) and common difference (d) are given. We need to find the number of terms (n) and the first term (a).

Here

Last term (`a_n`) = 4

Common difference (*d*) = 2

Sum of *n* terms (*S*_{n}) = −14

So here we will find the value of *n* using the formula, `a_n = a + (n - 1)d`

So, substituting the values in the above-mentioned formula

4 = a + (n - 1)2

4 = a + 2n - 2

4 + 2 = a + 2n

`n = (6 - a)/2` ....(1)

Now, here the sum of the *n* terms is given by the formula,

`S_n = (n/2) (a + l)`

Where, *a* = the first term

l = the last term

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

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

-14(2) = n(a + 4)

`n = (-28)/(a + 4)` ...(2)

Equating (1) and (2), we get,

`(6 - a)/2 = (-28)/(a + 4)`

(6 - a)(a + 4) = -28(2)

`6a - a^2 + 24 - 4a = -56`

`-a^2 + 2a + 24 + 56 = 0`

So, we get the following quadratic equation,

`-a^2 + 2a + 80 = 0`

`a^2 - 2a - 80= 0`

Further, solving it for *a* by splitting the middle term,

`a^2 - 2a - 80 = 0`

`a^2 - 10a + 8a - 80 = 0`

a(a - 10) + 8(a - 10) = 0

(a - 10)(a + 8) = 0

So, we get,

a - 10 = 0

a = 10

or

a + 8 = 0

a = -8

Substituting, a = 10 in (1)

`n = (6 - 10)/2`

`n = (-4)/2`

n = -2

Here, we get* n *as negative, which is not possible. So, we take a = -8

`n = (6 - (-8))/2`

`n = (6 + 8)/2`

`n = 14/2`

n = 7

Therefore, for the given A.P n = 7 and a = -8