Question

the sum of four consecutive numbers in an ap is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. find the numbers. (cbse 2018)

Answer 1

Hi,

2,6,10,14

Explanation:

Let's assume a the first term and r the commun difference.

The four consecutive numbers in the ap are: a,a+r,a+2r,a+3r

and their sum is 4a+6r=32 or 2a=16-3r

`(a*(a+3r))/((a+r)*(a+2r)) =(7)/(15) \\\\15(a^2+3ra)=7(a^2+3ar+2r^2)\\\\8a^2+24a-14r^2=0\\\\4a^2+12ar-7r^2=0\\(2a)^2+6*(2a)*r-7r^2=0\\`

we substitue 2a with its value:

`(16-3r)^2+6r*(16-3r)-7r^2=0\\\\256-96r+9r^2+96r-18r^2-7r^2=0\\\\256-16r^2=0\\\\r^2-16=0\\(r+4)(r-4)=0\\\\if\ r=4\ then\ a=8-3*2=2, numbers\ are\ 2,6,10,14\\\\if r=-4\ then\ a=8+3*2=14, numbers\ are\ 14,10,6,2.\\`