Question

The sum of 4 consecutive terms of an AP is 52 and the ratio of the product of the first and last terms and the product of the 2 middle terms is 11:20. Find the numbers. Tum sbko Mata Rani ki ksm bey span mat krna ಠ_ಠ​

Answer 1

For solution please see the attachment

Answer 2

Answer:

The four consecutive numbers are$-14,4,22,40$

Step-by-step explanation:

Let the $4$ consecutive terms be $a-3d,a-d,a+d,a+3d$

Given:

The sum of $4$ consecutive terms of an AP is $52$

$a-3d+a-d+a+d+a+3d=52\\\\4a=52\\\\a=13$

The ratio of the product of the first and last terms and the product of the $2$ middle terms is $11:20$.

$\frac{(a-3d)(a+3d)}{(a-d)(a+d)}=\frac{11}{20} \\\\\frac{(a^2-9d^2)}{(a^2-d^2)}=\frac{11}{20} \\\\20a^2-180d^2=11a^2-11d^2$

$9a^2=169d^2\\\\d^2=\frac{9a^2}{169}\\\\d=\frac{9*13}{13} \\\\d=9$

The four numbers will be

$a-3d=13-3(9)=-14\\\\a-d=13-9=4\\\\a+d=13+9=22\\\\a+3d=13+3(9)=40$