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# Question 10(ii) How many terms are there in the A.P. whose first and fifth terms are –14 and 2 respectively and the sum of the terms is 40?

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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 asked to find how many terms are there in the A.P.

whose first and fifth terms are –14 and 2 respectively and the sum of the terms is 40

This is the basic and conceptual question.

CBSE DHANPAT RAI publications
Class:- 10th
Solutions of CBSE Mathematics
Question 10(ii)

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1. Given A.P. has first term(a) = –14,

Fifth term(a5) = 2 and sum(Sn) = 40.

Let the common difference of the A.P. be d. We get,

Then, a5 = a + (5 – 1)d

=> 2 = -14 + 4d

=> 4d = 16

=> d = 4

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

=> 40 = n[2(-14) + (n – 1)(4)]/2

=> 40 = n[-28 + (4n – 4)]/2

=> 40 = n[-32 + 4n]/2

=> 4n2 – 32n – 80 = 0

=> n– 8n – 20 = 0

=> n2 -10n + 2n – 20 = 0

=> n(n -10) + 2(n – 10 ) = 0

=> (n + 2)(n – 10) = 0

=> n = -2 or n = 10

Ignoring n = -2 as number of terms cannot be negative. So we get, n = 10.

Hence, the number of terms (n) is 10.

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