How many terms are there in the A.P. whose first and fifth terms are 14 and 2 respectively and the sum of the terms is 40?
a5= a + (n -1) d
2 = -14 + 4d
16 = 4d
d = 4
Sn = 
[2a + (n – 1) d]
40 = 
[2(-14) + (n -1) (4)]
40 = 12n2 – 16n
2n2 – 16n -40 = 0
n2 – 8n -20 = 0
n2 – 10n + 2n – 20 = 0
(n – 10) (n + 2) = 0
n = 10 and n = -2 (not possible)
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