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 1

We know that,
\(a_n = a + (n − 1) d\)
\(a_3 = a + (3 − 1) d\)
\(a_3 = a + 2d\)
Similarly, \(a_7 = a + 6d\)
Given that,
\(a_3 + a_7 = 6\)
\((a + 2d) + (a + 6d) = 6\)
\(2a + 8d = 6\)
\(a + 4d = 3\)
\(a = 3 − 4d\)        ..…… (i)
Also, 
it is given that \((a_3) × (a_7) = 8\)
\((a + 2d) × (a + 6d) = 8\)
From equation (i),
\((3-4d+2d)(3-4d+6d) = 8\)
\((3-2d)(3+2d) = 8\)
\(9-4d^2 = 8\)
\(4d^2 = 9-8\)
\(4d^2 = 1\)
\(d^2 = \frac 14\)
\(d = ±\frac 12\)
\(d = \frac 12\) or \(-\frac 12\)
From equation number (i),
(Where \(d = \frac 12\))
\(a = 3-4d\)
\(a = 3-4(\frac 12)\)
\(a = 3-2\)
\(a = -1\)
(Where \(d =- \frac 12\))
\(a = 3-4(-\frac 12)\)
\(a = 3+2\)
\(a = 5\)
\(S_n = \frac n2[2a+(n-1)d]\)
(Where \(a = 1\) and \(d =\frac  12\))
\(S_{16} = \frac {16}{2}[2(1)+(16-1)(\frac 12)]\)

\(S_{16} = 8[2+\frac {15}{2}]\)
\(S_{16} = 4 \times 19\)
\(S_{16} = 76\)
(where \(a = 5 \)and \(d =- \frac 12\))
\(S_{16} = \frac {16}{2}[2(5)+(16-1)(-\frac 12)]\)

\(S_{16} = 8[10+15(-\frac 12)]\)
\(S_{16}= 8 \times \frac 52\)
\(S_{16} = 20\)