Question

In an AP given a_n = 4, d = 2, S_n = −14, find n and a.

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

Answer 1

Given that, a_n = 4, d = 2, S_n = −14

a_n = a + (n − 1)d

4 = a + (n − 1)2

4 = a + 2n − 2

a + 2n = 6

a = 6 − 2n       ...(i)

`S_n = n/2[a+a_n]`

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

−28 = n (a + 4)

−28 = n (6 − 2n + 4)      ...{From equation (i)}

−28 = n (− 2n + 10)

−28 = − 2n² + 10n

2n² − 10n − 28 = 0

n² − 5n −14 = 0

n² − 7n + 2n − 14 = 0

n (n − 7) + 2(n − 7) = 0

(n − 7) (n + 2) = 0

Either n − 7 = 0 or n + 2 = 0

n = 7 or n = −2

However, n can neither be negative nor fractional.

Therefore, n = 7

From equation (i), we obtain

a = 6 − 2n

a = 6 − 2(7)

a = 6 − 14

a = −8

Answer 2

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 - a²  + 24 - 4a = -56

-a² + 2a + 24 + 56 = 0

So, we get the following quadratic equation,

-a² + 2a + 80 = 0

a² - 2a - 80 = 0

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

a² - 2a - 80 = 0

a² - 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.