In an AP
(i) Given a = 2, d = 8, Sn = 90, find n and an.
(ii) Given a = 8, an = 62, Sn = 210, find n and d.
(iii) Given an = 4, d = 2, Sn = -14, find n and a.
(i) Given that, a = 2, d = 8, Sn = 90
Sn = n/2 [2a +(n -1)d]
90 = n/2 [2a +(n -1)d]
⇒ 180 = n(4+8n -8)
= n(8n-4) = 8n2-4n
⇒ 8n2-4n –180 = 0
⇒ 2n2–n-45 = 0
⇒ 2n2-10n+9n-45 = 0
⇒ 2n(n -5)+9(n -5) = 0
⇒ (n-5)(2n+9) = 0
So, n = 5 (as n only be a positive integer)
∴ a5 = 8+5×4 = 34
(ii) Given that, a = 8, an = 62, Sn = 210
Sn = n/2 (a + an)
210 = n/2 (8 +62)
⇒ 35n = 210
⇒ n = 210/35 = 6
Now, 62 = 8+5d
⇒ 5d = 62-8 = 54
⇒ d = 54/5 = 10.8
(iii) Given that, nth term, an = 4, common difference, d = 2, sum of n terms, Sn = −14.
an = a+(n −1)d,
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;
Sn = n/2 (a+an)
-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