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.
Answer:
Let
first term of AP = a
common difference of AP = d
Given 3rd term + 7th term = 6
=> a+2d +a+6d = 6
=> 2a + 8d = 6
=> a + 4d = 3.......................1
Again given 3rd term * 7th term = 8
=> (a+2d) * (a+6d) = 8
=> a2 + 8ad + 12d2 = 8.............2
from equation 1
a + 4d =3
=> a = 3 - 4d ...................3
Put value of a in equation 2, we get
(3 - 4d)2 + 8*(3-4d)*d + 12d2 = 8
=> 9 + 16d2 - 24d + 24d - 32d2 + 12 d2 = 8
=> 9 - 4d2 = 8
=> 4d2 = 9 - 8
=> 4d2 = 1
=> d2 = 1/4
=> d = 1/2, -1/2
Now put this value in equation 3
=> a = 3 - 4*(1/2) and a = 3 - 4*(-1/2)
=> a = 3 - 2 and a = 3 + 2
=> a =1, 5
Case 1: when a=1, d = 1/2
Sum of first 16 terms = (n/2) *{2a + (n-1)*d}
= (16/2)*{2*1 + (16-1)*1/2}
= 8*(2 + 15/2)
= 8*(2 + 7.5)
= 8*9.5
= 76
Case 2: when a=5, d = -1/2
Sum of first 16 terms = (n/2) *{2a + (n-1)*d}
= (16/2)*{2*5 + (16-1)*(-1/2)}
= 8*(10 - 15/2)
= 8*(10 - 7.5)
= 8*2.5
= 20
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