This is the basic and conceptual question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise 9.6
In this question we have been given an arithmetic progression and we have to find out how many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636
CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 10(iii)
Given A.P. has first term(a) = 9,
Common difference(d) = 17 – 9 = 8 and sum(Sn) = 636.
We know sum of n terms of an A.P. is given by, Sn = n[2a + (n – 1)d] / 2.
=> 636 = n[2(9) + (n − 1)(8)]/2
=> 636 = n[18 + (8n − 8)]/2
=>1271 = 10n + 8n2
=> 8n2 + 10n − 1272 = 0
=> 4n2+ 5n − 636 = 0
=> 4n2 − 48n + 53n − 636 = 0
=> 4n(n − 12) + 53(n − 12) = 0
=> (4n + 53)(n − 12) = 0
=> n = −53/4 or n = 12
Ignoring n = −53/4 as number of terms cannot be fraction. So we get, n = 12.
Hence, the number of terms (n) is 12.