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, 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
=> n2 – 5n – 14 = 0
=> n2 – 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.