In an AP
(i) Given a = 5, d = 3, an = 50, find n and Sn.
(ii) Given a = 7, a13 = 35, find d and S13.
(iii) Given a12 = 37, d = 3, find a and S12.
(iv) Given a3 = 15, S10 = 125, find d and a10.
(v) Given d = 5, S9 = 75, find a and a9.
(vi) Given a = 2, d = 8, Sn = 90, find n and an.
(vii) Given a = 8, an = 62, Sn = 210, find n and d.
(viii) Given an = 4, d = 2, Sn = − 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x) Given l = 28, S = 144 and there are total 9 terms. Find a
(i) Given that, a = 5, d = 3, an = 50
As an = a + (n − 1)d,
⇒ 50 = 5 + (n - 1) × 3
⇒ 3(n - 1) = 45
⇒ n - 1 = 15
⇒ n = 16
Now, Sn = n/2 (a + an)
Sn = 16/2 (5 + 50) = 440
(ii) Given that, a = 7, a13 = 35
As an = a + (n − 1)d, ⇒ 35 = 7 + (13 - 1)d
⇒ 12d = 28
⇒ d = 28/12 = 2.33
Now, Sn = n/2 (a + an)
S13 = 13/2 (7 + 35) = 273
(iii)Given that, a12 = 37, d = 3 As an = a + (n − 1)d,
⇒ a12 = a + (12 − 1)3
⇒ 37 = a + 33
⇒ a = 4
Sn = n/2 (a + an)
Sn = 12/2 (4 + 37)
= 246
(iv) Given that, a3 = 15, S10 = 125
As an = a + (n − 1)d,
a3 = a + (3 − 1)d
15 = a + 2d ... (i)
Sn = n/2 [2a + (n - 1)d]
S10 = 10/2 [2a + (10 - 1)d]
125 = 5(2a + 9d)
25 = 2a + 9d ... (ii)
On multiplying equation (i) by (ii), we get
30 = 2a + 4d ... (iii)
On subtracting equation (iii) from (ii), we get
−5 = 5d
d = −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a = 17
a10 = a + (10 − 1)d
a10 = 17 + (9) (−1)
a10 = 17 − 9 = 8
(v) Given that, d = 5, S9 = 75
As Sn = n/2 [2a + (n - 1)d]
S9 = 9/2 [2a + (9 - 1)5]
25 = 3(a + 20)
25 = 3a + 60
3a = 25 − 60
a = -35/3
an = a + (n − 1)d
a9 = a + (9 − 1) (5)
= -35/3 + 8(5)
= -35/3 + 40
= (35+120/3) = 85/3
(vi) Given that, a = 2, d = 8, Sn = 90
As 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
⇒ (2n - 9)(2n + 9) = 0
So, n = 5 (as it is positive integer)
∴ a5 = 8 + 5 × 4 = 34
(vii) Given that, a = 8, an = 62, Sn = 210
As 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
(viii) Given that, an = 4, d = 2, Sn = −14
an = a + (n − 1)d
4 = a + (n − 1)2
4 = a + 2n − 2
a + 2n = 6
a = 6 − 2n ... (i)
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
(ix) Given that, a = 3, n = 8, S = 192
As Sn = n/2 [2a + (n - 1)d]
192 = 8/2 [2 × 3 + (8 - 1)d]
192 = 4 [6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6
(x) Given that, l = 28, S = 144 and there are total of 9 terms.
Sn = n/2 (a + l)
144 = 9/2 (a + 28)
(16) × (2) = a + 28
32 = a + 28
a = 4
