Question

the sum of four consecutive numbers in AP is 32 and the ratioof the product of the first and last term to the product of two middle terms is 7:15.find the numbers.​

Answer 1

Step-by-step explanation:

thank me

marks me as brainlist

Answer 2

Answer:

$\mathfrak{ANSWER}$

let the four consecutive numbers be $\tt{\green{(a - 3d) , ( a - d ) , ( a + d ) and (a + 3d)}}$

Now , according to question ,

$\tt{ a - 3d + a - d + a + 3d + a - d = 32}$

$\tt{ 4a = 32}$

$\tt{a = 8-----------------(1)}$

$\tt{And , }$

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

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

$\tt{15a^2 - 135d^2 = 7a^2 - 7d^2}$

$\tt{ 8a^2 = 128d^2}$

$\tt{a^2 = 16d^2}$

$\tt{ 64 = 16d^2 --------------from 1st}$

$\tt{d^2 = 4 }$

$\tt{\green{ d = 2}}$

Therefore, the numbers are :

$\tt{\red{ a - 3d = 8 - 6 →2}}$

$\tt{\red{ a - d = 8 - 2 → 6}}$

$\tt{\red{a + 3d = 8 + 6 → 14}}$

$\tt{\red{a + d = 8 + 2 → 10}}$

$\tt{\huge{May Help♡}}$