An important and exam oriented question from arithmetic progression chapter as it was already asked in previous year paper of 2014 in which we are to divide 32 into four parts which are in A.P. such that the product of extremes is to the product of means is 7:15.
Book – RS Aggarwal, Class 10, chapter 5B, question no 12
Let the four parts be (a−3d),(a−d),(a+d) and (a+3d).
Then, Sum of the numbers =32
⟹(a−3d)+(a−d)+(a+d)+(a+3d)=32⟹4a=32⟹a=8
It is given that
(a−d)(a+3d)/(a+d)(a−3d)=157
⟹(a^2−9d^2)/a^2-d^2=7/15
⟹(64-9d^2)/64-d^2=7/15⟹128d^2=512
⟹d2=4
⟹d=±2
Thus, the four parts are a−3d,a−d,a+d and 3d, i.e. 2,6,10,14.