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# Question 10(iii) How many terms of the A.P. 9, 17, 25, . . . must be taken so that their sum is 636?

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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)

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1. Given A.P. has first term(a) = 9,

Common difference(d) = 17 â€“ 9 = 8 and sum(Sn) = 636.

We know sum of n terms of an A.P. is given by, SnÂ = n[2a + (n â€“ 1)d] / 2.

=> 636 = n[2(9) + (n âˆ’ 1)(8)]/2

=> 636 = n[18 + (8n âˆ’ 8)]/2

=>1271 = 10n + 8n2

=> 8n2Â + 10n âˆ’ 1272 = 0

=> 4n2+ 5n âˆ’ 636 = 0

=> 4n2Â âˆ’ 48n + 53n âˆ’ 636 = 0

=> 4n(n âˆ’ 12) + 53(n âˆ’ 12) = 0

=> (4n + 53)(n âˆ’ 12) = 0

=> n = âˆ’53/4 or n = 12

Ignoring n = âˆ’53/4 as number of terms cannot be fraction. So we get, n = 12.

Hence, the number of terms (n) is 12.

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