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# Question 56. Let there be an A.P. with first term a and common difference d. If an denotes its nth term and Sn is the sum of first n terms, then find:(ii) n and a, if an = 4, d = 2 and Sn = –14

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This is an arithmetic progression based question from Chapter name- Arithmetic Progression
Chapter number- 9
Exercise – 9.6

In this question we have been given that  there be an A.P.

first term a and common difference d.

Also, If an denotes its nth term and Sn is the sum of first n terms,

Now we have to find n and a,

Where an = 4, d = 2 and Sn = –14

CBSE DHANPAT RAI PUBLICATIONS
Understanding CBSE Mathematics
Class :- 10th
Question no 56(ii)

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1. Given A.P. has d = 2, an = 4 and Sn = –14.

By using the formula of nth term of an A.P.

an = a + (n – 1)d

So,

=> 4 = a + (n – 1)2

=> 4 = a + 2n – 2

=> a = 6 – 2n  . . . . (1)

Now by using the formula of sum of n terms of an A.P.

Sn = n[a + an] / 2

So,

=> –14 = n[a + 4] / 2

=> n[a + 4] = –28

=> n[6 – 2n + 4] = –28   [Using eq(1)]

=> 10n – 2n2 = –28

=> n– 5n – 14 = 0

=> n– 7n + 2n – 14 = 0

=> n(n – 7) + 2(n – 2) = 0

=> (n – 7) (n + 2) = 0

=> n = 7 or n = –2

Ignoring n = –2 as number of terms cannot be negative. So, we get n = 7.

On putting n = 7 in (1), we get, a = 6 – 2(7) = –8

Hence, the value of n is 7 and a is –8.

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