Question

the sum of the third and the Seven terms of an ap is 6 and their product is 8 find the sum of 16 terms of an ap

Answer 1

$a3 + a7 = 6 \\ a + 2d + a + 6d = 6 \\ 2a + 8d = 6 \\ a + 4d = 3 \\ a = 3 - 4d\\ a3.a7 = 8 \\ (a + 2d)(a + 6d ) = 8 \\ (3 - 4d + 2d)(3 - 4d + 6d) = 8 \\ (3 - 2d)(3 + 2d) = 8 \\ 9 - 4 {d}^{2} = 8 \\ 4 {d}^{2} = 1 \\ d = + - \frac{1}{2} \\ a = 1 \: or \: 5\\ s16 = \frac{16}{2}(2 \times 1 + 15 \times \frac{1}{2} ) \: or \: \frac{16}{2} (2 \times 5 - 15 \times \frac{1}{2} ) \\ = 8 \times \frac{19}{2} \: or \: 8 \times \frac{5}{2} \\ = 76 \: or \: 20$