Question

The sum of four consecutive number in Ap is 37 and the ratio of the product of the first and the last term of the product of two middle term is 7:15. Find the number

Answer 1

Let the consecutive numbers be
$a - 3d \: \: \: a - d \: \: \: a + d \: \: \: a + 3d$
Sum is given as 37
$4a = 37 \: \: \: \: \: a = \frac{37}{4}$
Given ratio of products as 7:15
$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15}$
$15 \times ( {a}^{2} - 9 {d}^{2} ) = 7 \times ( {a}^{2} - {d}^{2} )$
${a}^{2} = 16 {d}^{2}$
$d = \frac{a}{4} \: \: or \: \: - \frac{a}{4}$
$d = \frac{37}{16} \: \: or \: \: - \frac{37}{16}$
Therefore the numbers (by taking positive value for d) are
$\frac{37}{16} \: \: \: \frac{111}{16 } \: \: \: \frac{185}{16} \: \: \: \frac{259}{16}$
Note: AP may be formed by reversing the numbers also (by taking negative value for d)