the sum of third and seventh term of an AP is 6 and their product is 8 find sum of first sixteen term of AP
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So using formula for n th term
a_n=a+(n-1)d
a = first term
n = no. of terms
d = common difference
So given
a + (3-1)d + a (7-1)d = 6
2a + 8d = 6
a + 4d = 3
a = 3 - 4d
(a + 2d)(a + 6d) = 8
(3 - 2d)( 3 + 2d) = 8 {substituting a = 3 - 2d}
9 - 4d² = 8
d = +1/2 and - 1/2
so when d = +1/2 then a = 3 - 2 = 1 and when d = -1/2 a = 3 + 2 = 5
so using formula for
S_n=\frac{n}{2}[2a+(n-1)d]
when d = +1/2
S_n=\frac{16}{2}[2*1+(16-1)d]
S_n=76
Similarly when d = -1/2
S_n=20
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