Question

the sum of four consecutive terms which are in arithmetic progression is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15 find the number

Answer 1

here we are taking d =2d
let 4 terms = a-3d, a-d, a+d,a+3d
sum of all = 32
a-3d+a-d+a+d+a+3d=32
4a=32
a=8
$\frac{(a - 3d) \times (a + 3d)}{(a - d) \times (a + d)} = \frac{7}{15} \\ \frac{ {a}^{2} - {(3d)}^{2}}{ {a}^{2} - {d}^{2} } = \frac{7}{15} \\ \frac{{a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15} \\ 15 {a}^{2} - {135}^{d} = 7 {a}^{2} - 7 {d}^{2} \\ 8 {a}^{2} - 128 {d}^{2} = 0 \\ {a}^{2} - 16 {d}^{2} = 0 \\ 64 = 16 {d}^{2} \\ 4 = {d}^{2} \\ \sqrt{4} = d$
so d =±2
so if d=2 AP = 2,6,10,14
if d =-2 AP=14,10,6,2