Selina Concise Class 10 Math Chapter 10 Arithmetic Progression Solutions
Exercise 10C
(Q1) Find the sum of the first 22 terms of the A.P. 8, 3, -2, ——
Solution:
Given arithmetic progression 8, 3, -2, ——
Here, t1 = a = 8
t2 – t1 = 3 – 8 = -5 = d1
t3 – t2 = -2 – 3 = -5 = d2
∴ d1 = d2 =d
First term (a) = 8 and common difference (d) = -5
We know that,
Sn = n/2 [2a + (n – 1) d]
To find first 22 terms
∴ n = 22
S22 = 22/2 [2×8 + (22 – 1) (-5)]
S22 = 11 [16 + (21) (-5)]
= 11 [16 – 105]
S22 = 11 [-89]
S22 = – 979
∴ The sum of the first 22 term is – 979.
(Q2) How many terms of the arithmetic progression: 24, 21, 18, —— must be taken so that their sum is 78?
Solution:
Given arithmetic progression is 24, 21, 18, —–
Here, t1 = a = 24
t2 – t1 = 21 – 24 = – 3 = d1
t3 – t2 = 18 – 21 = – 3 = d2
∴ d1 = d2 = d
First term (a) = 24 and common difference (d) = -3
Also given that the sum is 78,
We know that,
Sn = (n/2) [2a + (n – 1) d]
∴ Sn = 78
78 = n/2 [2×24 + (n – 1) (-3)]
= n/2 [48 + (-3n + 3)]
= n/2 [48 – 3n + 3]
= n/2 [-3n + 51]
78×2 = n [-3n + 51]
156 = -3n2 + 51n
3n2 – 51n + 156 = 0
Divide ‘3’ on both sides,
3n2/3 – 51n/3 + 156/3 = 0
n2 – 17n + 52 = 0 (Now middle term)
n2 – 13n – 4n + 52 = 0 (-17n we can write -13n -4n)
N (n – 13) – 4 (n – 13) = 0 (After common)
(n – 13) (n – 4) = 0
n – 13 = 0 or n – 4 = 0
n = 13 or n = 4
∴ The terms of the arithmetic progression are 13 and 4.
(Q3) Find the sum of 28 terms of an A.P whose nth term is 8n – 5
Solution:
Given that the nth term is 8n – 5
To find: Sum of 28 terms
First we have to find first term, so, we have to put n = 1
a = 8 (1) – 5
= 8 – 5
a = 3
Also given that, the last term is 28,
n = 28
l = 8 (28) – 5 = 224 – 5 = 219
We know that,
Sn = n/2 (a + l)
S28 = 28/2 (a + 219)
= 28/2 (3 + 219)
= 14 (3 + 219)
= 14 (222)
S28 = 3108
∴ The sum of 28 terms are 3108.
(Q4) (i) Find the sum of all odd natural numbers less than 50
Solution:
The all odd natural number less than 50 are – 1, 3, 5, 7, 9, 11, —- 49.
We cannot take 50 because in question tell only less than 50.
Here, t1 = a = 1 and l = 49.
Now, we have to find n
We know that,
l = a + (n – 1) d
49 = 1 + (n – 1) (d)
To find ‘d’ –
t2 – t1 = 3 – 1 = 2 = d1
t3 – t2 = 5 – 3 = 2 = d2
d1 = d2 = d
∴ 49 = 1 + (n – 1) (2)
= 1 + 2n – 2
49 = – 1 + 2n
49 + 1 = 2n
50 = 2n
50/2 = n
25 = n
∴ We know that,
S25 = n/2 (a + l)
S25 = 25/2 (1 + 49)
= 25/2 (50) (cut 50 by 2)
= 25 (25)
S25 = 625
∴ The sum of all odd natural numbers are 625.
(ii) Find the sum of first 12 natural numbers each of which is a multiple of 7.
Solution:
The sum of first 12 natural numbers which are multiple of 7 are –
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77 and 84
First term (a) = 7 and common difference = t2 – t1
= 14 – 7
= 7 = d1
t3 – t2 = 21 – 14
= 7 = d2
∴ d1 = d2 = d
∴ d = 7
l = 84 (Last term)
We know that,
Sn = n/2 (a + l)
S12 = 12/2 (7 + 84) (∵ n = 12)
= 6 (91)
S12 = 546
∴ The sum of first 12 natural numbers of divisible by 7 are 546.
(Q5) Find the sum of fist 51 terms of an arithmetic progression whose 2nd and 3rd terms are 14 and 18 respectively.
Solution:
Given that the sum of first 51 terms of an arithmetic progression
∴ n = 51.
Also given that, the second term is 14.
t2 = 14
And the third term is 18
t3 = 18.
We have to find common difference (d) –
t3 – t2 = 18 – 14
= 4
∴ d = 4
We have to find first term (a) –
We know that,
t2 = a + d
14 = a + (4) (∵ t2 = 14)
14 – 4 = a
10 = a
∴ The first term (t1) = a = 10
Now, we have to find sum of first 51 terms –
We know that,
Sn = n/2 [2a + (n – 1) d]
S51 = 51/2 [2×10 + (51 – 1) × 4]
S51 = 51/2 [20 + 50 × 4]
= 51/2 [20 + 200]
= 51/2 [220]
= 51 × 110
S51 = 5610
∴ The sum of first 51 terms are 5610