Solution:
an = a + (n – 1) d4 = a + (n – 1) 2
a + 2n = 6
a = 6 – 2n
–14 = 5n – n2
n2 – 5n + 14 = 0
n(n – 7) + 2(n – 7) = 0
(n + 2) (n – 7) 20
n = –2, 7
n can not be negative
n = 7
a = 6 – 2 × 7
= 6 – 14
= –8
Hence n is 7 and a is –8
MathematicsIf an AP given an = 4, d = 2, sn = –14 then n is
If an AP given an = 4, d = 2, sn = –14 then n is
4 = a + (n – 1) 2
a + 2n = 6
a = 6 – 2n
–14 = 5n – n2
n2 – 5n + 14 = 0
n(n – 7) + 2(n – 7) = 0
(n + 2) (n – 7) 20
n = –2, 7
n can not be negative
n = 7
a = 6 – 2 × 7
= 6 – 14
= –8
Hence n is 7 and a is –8
