Question

Answer for the sum of the 3rd and 7th terms of an ap is 6 and their product is 8 find the sum of first16 terms of an ap

Answer 1

Let the first term be 'a'
Common difference be 'd'
$t3 + t7 = 6 \\ (a + 2d) + (a + 6d) = 6 \\ 2a + 8d = 6 \\ a + 4d = 3 - - - (1)$

$t3 \times t7 = 8 \\ (a + 2d) \times (a + 6d) = 8 \\(a +4d - 2d) \times (a + 4d + 2d) = 8 \\ substituting \: \: from \: \: (1) \\ (3- 2d) \times (3 + 2d) = 8 \\ 9 - 4 {d}^{2} = 8 \\ d = \frac{1}{2} \: \: or \: \: \frac{ - 1}{2} \\ from \: \: (1) \: \: a = 1 \: \: or \: \: 5$

$sum \: \: of \: \: 16 \: \: terms \\ s16 = \frac{16}{2} \times (2a + (16 - 1) \times d) \\ = 8 \times (2a + 15d) \\ for \: \: a = 1 \: \: and \: \: d = \frac{1}{2} \\ s16 = 8 \times (2 \times 1 + 15 \times \frac{1}{2} ) = 76 \\ for \: \: a = 5 \: \: and \: \: d = \frac{ - 1}{2} \\ s16 = 8 \times (2 \times 5 + 15 \times \frac{ - 1}{2} ) =20$