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Question
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.
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Solution
We know that,
an = a + (n − 1) d
a3 = a + (3 − 1) d
a3 = a + 2d
Similarly, a7 = a + 6d
Given that, a3 + a7 = 6
(a + 2d) + (a + 6d) = 6
2a + 8d = 6
a + 4d = 3
a = 3 − 4d ...(i)
Also, it is given that (a3) × (a7) = 8
⇒ (a + 2d) × (a + 6d) = 8
From equation (i),
(3 - 4d + 2d) × (3 - 4d + 6d) = 8
(3 - 2d) × (3 + 2d) = 8
9 - 4d2 = 8
4d2 = 9 - 8
4d2 = 1
d2 = `1/4`
d = `± 1/2`
d = `1/2 or 1/2`
From the equation (i)
`("When" "d" 1/2)`
a = 3 - 4d
a = `3 -4(1/4)`
a = 3 - 2
a = 1
When d is − `1/2`
a = `3 - 4(-1/2)`
a = 3 + 2
a = 5
`S_n = n/2 [2a(n-1)s]`
(When a is 1 and d is `1/2`)
`S_16 = (16/2)[2"a" + (16 - 1)"d"]`
`S_16 = 8[2 xx 1 + 15(1/2)]`
`S_16 = 8(2 + 15/2)`
`S_16 = 8 xx 19/2`
S16 = 76
(When a is 5 and d is `-1/2`)
a = `3 - 4(-1/2)`
a = `3 + 4/2`
a = 5
S16 = `(16/2)[2a + (n - 1)d]`
= 8`[2(5) + 15(-1/2)]`
= `8[10 - 15/2]`
= `8 xx 5/2`
= 20
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