In an A.P.
Given an = 4, d = 2, Sn = –14, find ‘n’ and ‘a’.
Here, an = 4, d = 2
Sn = –14
We know that,
an = a + (n -1)d
4 = a + (n -1) x 2
4 = a + 2n - 2
a + 2n = 6 a = 6 - 2 n ....(i)
And. Sn = [2a+ (n - 1)d]
- 14 =
- 28 = n[2a + 2 (n -1) ]
- 28 = 2n [a + (n -1)]
- 14 = n[a + (n - 1)] ......(ii)
Putting the value of (i) in (ii), we get,
– 14 = n [6 – 2n + n – 1]
⇒ – 14 = n[5 – n]
⇒ –14 = 5n – n2
⇒ n2 –5n – 14 = 0
⇒ (n – 7) (n + 2) = 0
⇒ n – 7 = 0 or n + 2 = 0
⇒ n = 7 or n = –2
But n cannot be –ve. ∴ n = 7.
Putting this value of V in equation (i), we get
a + 2 x 7 = 6
⇒ a = 6 – 14 = – 8
Hence, n = 7 and a = –8
The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.