Question

the sum offour consecutive numbers in an ap 32 and the ratio of hte product of the first and last term to the product of two middle term is7:15 find the numbers

Answer 1

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Answer 2

Let the four terms be a-3d, a-d, a+d, a+3d

a - 3d + a - d + a + d + a + 3d = 32

4a = 32

a = 8

$\frac{(a-3d)(a+3d)}{(a-d)(a+d)}$ = $\frac{7}{15}$

$\frac{a^2 - 9d^2}{a^2 - d^2}$ = $\frac{7}{15}$

$15a^2 - 135d^2 = 7a^2 - 7d^2$

$8a^2 = 128d^2$

$\frac{a^2}{d^2}$ = $\frac{16}{1}$

$\frac{a}{d}$ = $\frac{4}{1}$

$a = 4d$

$d = \frac{a}{4}$

$d = \frac{8}{4}$

d = 2

Therefore, four terms = 2, 6, 10, 16

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