Given that, n^th term, a_n = 4, common difference, d = 2, sum of n terms, S_n = −14.

As we know, from the formula of the n^th term in an AP,

a_n = a+(n −1)d,

Therefore, putting the given values, we get,

4 = a+(n −1)2

4 = a+2n−2

a+2n = 6

a = 6 − 2n …………………………………………. (i)

As we know, the sum of n terms is;

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 get

a = 6−2n

a = 6−2(7)

= 6−14

= −8