(i) Given a = 8, a_n = 62, S_n = 210, find n and d.
(ii) Given a_n = 4, d = 2, S_n = − 14, find n and a.

Solutions:

(i) Provided that, a = 8, a_n = 62, S_n = 210

In AP, the sum of n terms is

S_n = n/2 (a + a_n)

210 = n/2 (8 +62)

⇒ 35n = 210

⇒ n = 210/35 = 6

Now, 62 = 8+5d

⇒ 5d = 62-8 = 54

⇒ d = 54/5 = 10.8

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

As we know from the nth term formula in an AP,

a_n = a+(n −1)d,

Putting the given values, as a result we get,

4 = a+(n −1)2

4 = a+2n−2

a+2n = 6

a = 6 − 2n …………………………………………. (i)

The sum of n terms, as we all know, 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 fractional nor negative.

Therefore, n = 7

From equation (i), we get

a = 6−2n

a = 6−2(7)

= 6−14

= −8