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)

Given A.P. has first term(a) = –14,

Fifth term(a

_{5}) = 2 and sum(S_{n}) = 40.Let the common difference of the A.P. be d. We get,

Then, a

_{5}= a + (5 – 1)d=> 2 = -14 + 4d

=> 4d = 16

=> d = 4

We know sum of n terms of an A.P. is given by, S

_{n}= 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

=> 4n

^{2}– 32n – 80 = 0=> n

^{2 }– 8n – 20 = 0=> n

^{2}-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.