NCERT MATHS CLASS - X
Arithmetic Progressions
Exercise 5.3
Question 4:
How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?
Sol.
Let the sum of n terms be 636.
Here, a = 9
d = a2 − a1 = 17 − 9 = 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
Since the number of terms can neither be negative nor fractional, therefore, n =12
Question 5:
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.
Given that,
a = 5, l = 45, Sn = 400
n = 16
Again,
l = a + (n − 1) d
45 = 5 + (16 − 1) d
40 = 15d
Question 6:
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?
Given that,
a = 17
l = 350
d = 9
Let the no. of terms in this A.P be n.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
Thus, this A.P has 38 terms and the sum of the terms of this A.P. is 6973.
Question 7:
Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Given
d = 7
a22 = 149
S22 = ?
an = a + (n − 1)d
a22 = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
