NCERT Solutions
Class 10 Maths
Arithmetic Progressions
Ex. 5.3 Q.3
In an AP:
- given a = 5, d = 3, an = 50, find n and Sn.
- given a = 7, a13 = 35, find d and S13.
- given a12 = 37, d = 3, find a and S12.
- given a3 = 15, S10 = 125, find d and a10.
- given d = 5, S9 = 75, find a and a9.
- given a = 2, d = 8, Sn = 90, find n and an.
- given a = 8, an = 62, Sn = 210, find n and d.
- given an = 4, d = 2, Sn = –14, find n and a.
- given a = 3, n = 8, S = 192, find d.
- given l = 28, S = 144, and there are total 9 terms. Find a.
- Given that, a = 5, d = 3, a = 50
As an = a + (n − 1)d
=> a13 = a + (13 − 1)d
=> 50 = 5 + (n − 1)3
=> 45 = (n − 1)3
=> 15 = n − 1
=> n = 16
Now sum Sn = (n/2)*(a + an)
= (16/2)*{5 + 50}
= 8 * 55
= 440
- Given a = 7, a13 = 35
As an = a + (n − 1) d
=> a12 = a + (13 − 1) d
=> 35 = 7 + 12 d
=> 35 − 7 = 12d
=> 28 = 12d
=> d = 28/12
=> d = 2
Now sum Sn = (n/2)*(a + an)
= (13/2)*{7 + 35}
= (13 * 42)/2
= 13 * 21
= 273
(iii) Given a12 = 37, d = 3 As an = a + (n − 1)d, a12 = a + (12 − 1)3 37 = a + 33
a = 4
Now sum Sn = (n/2)*(a + an)
= (12/2)*{4 + 37}
= 6 * 41
= 246
(iv) Given that a3 = 15, S10 = 125 As an = a + (n − 1)d, a3 = a + (3 − 1)d
15 = a + 2d …………(1)
Sn = (n/2)[2a + (n - 1)d]
S10 = (10/2)[2a + (10 - 1)d]
125 = 5(2a + 9d)
25 = 2a + 9d ………….(2)
On multiplying equation (1) by 2, we obtain
30 = 2a + 4d ………(3)
On subtracting equation (3) from (2), we obtain
−5 = 5d d = −1
From equation (1),
15 = a + 2(−1) 15 = a − 2 a = 17 a10 = a + (10 − 1)d a10 = 17 + (9) (−1) a10 = 17 − 9 = 8
(v) Given d = 5, S9 = 75
Sn = (n/2)[2a + (n - 1)d]
S9 = (9/2)[2a + (9 - 1)5]
75 = (9/2) (2a + 40)
75 = 9 (a + 20)
25 = 3(a + 20)
25 = 3a + 60
3a = 25 – 60
3a = -35 a = -35/3 an = a + (n − 1)d a9 = a + (9 − 1) (5) a9 = -35/3 + 40 a9 = (-35 + 120)/3 a9 = 85/3
(vi) Given a = 2, d = 8, Sn = 90
Sn = (n/2)[2a + (n - 1)d]
90 = (n/2)[2*2 + (n - 1)8]
90 = n[2 + (n − 1)4]
90 = n[2 + 4n − 4]
90 = n(4n − 2)
90 = 4n2 − 2n
4n2 − 2n − 90 = 0
4n2 − 20n + 18n − 90 = 0
4n (n − 5) + 18 (n − 5) = 0
(n − 5) (4n + 18) = 0 Either n − 5 = 0 or 4n + 18 = 0 n = 5 or n = -18/9 = -2
However, n can neither be negative nor fractional. Therefore, n = 5 an = a + (n − 1)d
a5 = 2 + (5 − 1)8
= 2 + (4) (8)
= 2 + 32 = 34
(vii) Given a = 8, an = 62, Sn = 210
Sn = (n/2)[a + l]
210 = (n/2)[8 + 62]
210 = (n/2) * 70
210 = 35n n = 210/35 n = 6 an = a + (n − 1)d
62 = 8 + (6 − 1)d
62 − 8 = 5d 54 = 5d d = 54/5
(viii) Given 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 obtain a = 6 − 2n
= 6 − 2(7)
= 6 − 14
= −8
(ix) Given a = 3, n = 8, S = 192
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 l = 28, S = 144, and there are total 9 terms.
Sn = (n/2) [a + l]
144 = (9/2) [a + 28]
(144 * 2)/9 = a + 28
(16) * 2 = a + 28 32 = a + 28
a = 32 – 28 a = 4