NCERT Solutions 10th Maths Chapter 5 Arithmetic Progressions Exercise 5.3

NCERT Solutions For Class 10 Maths Chapter 5 Exercise 5.3 we will restart our exploration of the world of Arithmetic Progressions. Thus, we are providing you Chapter 5 Arithmetic Progressions NCERT Class 10 Maths Solutions that will help in achieving more marks. You don't have to wander and waste your precious time in finding best CBSE NCERT Solutions.

Exercise 5.3
1. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Answer
Let there be n terms of this A.P.
For this A.P., a = 9
d = a2 - a1 = 17 - 9 = 8
As Sn = n/2 [2a + (n - 1)d]
636 = n/2 [2 × a + (8 - 1) × 8]
636 = n/2 [18 + (n- 1) × 8]
636 = n [9 + 4n - 4]
636 = n (4n + 5)
4n2 + 5n - 636 = 0
4n2 + 53n - 48n - 636 = 0
n (4n + 53) - 12 (4n + 53) = 0
(4n + 53) (n - 12) = 0
Either 4n + 53 = 0 or n - 12 = 0
n = (-53/4) or n = 12
n cannot be (-53/4). As the number of terms can neither be negative nor fractional, therefore, n = 12 only.
2. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Answer
Given that,
a = 5
l = 45
Sn = 400
Sn = n/2 (a + l)
400 = n/2 (5 + 45)
400 = n/2 (50)
n = 16
l = a + (n - 1) d
45 = 5 + (16 - 1) d
40 = 15d
d = 40/15 = 8/3
3. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Answer
Given that,
a = 17
l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n - 1) d
350 = 17 + (n - 1)9
333 = (n - 1)9
(n - 1) = 37
n = 38
Sn = n/2 (a + l)
S38 = 13/2 (17 + 350)
= 19 × 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.
4. Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Answer
d = 7
a22 = 149
S22 = ?
an = a + (n - 1)d
a22 = a + (22 - 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
Sn = n/2 (a + an)
= 22/2 (2 + 149)
= 11 × 151
= 1661
5. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
Answer
Given that,
a2 = 14
a3 = 18
d = a3 - a2 = 18 - 14 = 4
a2 = a + d
14 = a + 4
a = 10
Sn = n/2 [2a + (n - 1)d]
S51 = 51/2 [2 × 10 + (51 - 1) × 4]
= 51/2 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610