Answer to Question #255766 in Algebra for roshan

Question

There are four numbers such that first three of them from an A.P. and the last three form a G.P. The sum of the first and third number is 2 and that of second and fourth is 26. What are these numbers ?

Answer 1

1. let the 1st term be (a-d), a, (a+d), where a is the 1st term and d be the common difference a-d+a+d=2, then 2a=2 $\therefore$ a=2. Given a+ar²=26 ,factor out a $\to$ a(1+r²)=26 but a=1 $\therefore$ r²=25$\to$ r=5 or -5 2nd term=a,3rd term=ar=5 or -5,4th term =ar²=25. d=a₃-a₂=5-1=4 or -5-1=-6, $\therefore$ we know that a₁=a₂-d=1-4=-3 or1-(-6)=7 $\therefore$ there are two sets of numbers {-3,1,5,25} and {7,1,-5,25}