Question

10.
In a set of four numbers, the first three are in GP & the last three are in A.P. with common difference
6. If the first number is the same as the fourth, find the four numbers.​

Answer 1

Answer:

Now correct way of doing this is given as:

Let the last three numbers be a,a+6,a+12a,a+6,a+12, so that the first three numbers are a+12,a,a+6a+12,a,a+6. As these are in G.P.

a2=(a+12)(a+6)⇒a=−4

a2=(a+12)(a+6)⇒a=−4

Common ratio is −2−2

But how I did this is

I took first term of G.P. as a1a1 and the first term of A.P. as a2a2. So the series is

a1,(a1r) or (a2),(a1r2) or (a2+6),a2+12

a1,(a1r) or (a2),(a1r2) or (a2+6),a2+12

So, a1=a2+12a1=a2+12 and r=a2+6a2r=a2+6a2 and a1r2−a1r=6a1r2−a1r=6

Solving these three equations I get Common ratio (r) = −0.5