Three numbers whose sum is $15$ are successive terms of an arithmetic series. If $1, 1$ and $4$ are added to these three numbers respectively, the resulting numbers are successive terms of a geometric series. Find the numbers.
I have found from using the sum that $d=5-a$
and from that deduced that the second term in the GP must be $ 6$ (as $T_2 = a + d + 1$) but where do I go from here?
Thanks!
Answer
Let the three numbers be $5-d,5, 5+d$
New numbers are $6-d,6,9+d$ are in G.P
Clearly $6^2=(9+d)(6-d)\Rightarrow d=3,-6$
Original numbers are $2,5,8$ or $11,5,-1$.
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