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

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Understanding CBSE Mathematics

Class :- 10th

Question no 56(ii)

Given A.P. has d = 2, a

_{n}= 4 and S_{n}= –14.By using the formula of nth term of an A.P.a_{n}= a + (n – 1)dSo,=> 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.S_{n}= n[a + a_{n}] / 2So,=> –14 = n[a + 4] / 2

=> n[a + 4] = –28

=> n[6 – 2n + 4] = –28 [Using eq(1)]

=> 10n – 2n

^{2}= –28=> n

^{2 }– 5n – 14 = 0=> n

^{2 }– 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.