To find the last term in a sequence, each term formed by adding similar indexed term from an AP and a GP.
A four-term sequence is formed by adding each term of a four-term arithmetic sequence of positive integers to the corresponding term of a four-term geometric sequence of positive integers. The first three terms of the resulting four-term sequence are 57, 60 and 91. What is the fourth of this sequence?
Use the standard forms of the terms of the progressions to obtain a system of equations.
Try to reduce the number of variables from the system by subtracting two subsequent equations at a time.
Let $a,ar,ar^2,ar^3$, be the first three terms of the geometric progression, and $b,b+d,b+2d,b+3d$ be the corresponding terms of the arithmetic progression.
We are given, that
$$a+b=57$$
$$ar+b+d=60$$
$$ar^2+b+2d=91.$$
These are 3 non-linear equations in 4 variables, so we can't directly conclude anything. Notice that if we subtract the first two equations we get, discarding $b$ $$3=a(r-1)+d$$ and similarly
$$31=ar(r-1)+d.$$
Each of these equations contain, the same variable. So subtracting again, we get
$$28=ar^2-2ar+a=a(r-1)^2.$$
Now since we're dealing with sequences of positive integers, then we can only equate $(r-1)^2$ to either $4$ or $1$.
Then we can conclude that either $a=28$ and $r=2$ or $a=7$ and $r=3$.
If $a=28$, then we get $b=57-28=29$ and $d=-25$. But that makes the arithmetic progression $29,4,-21,-46$, which is a contradiction since the sequence is of positive integers. With $a=7$, $b=50$, and $d=-11$ we get following progressions $50,39,28,17$ and $7,21,63,189$.
The desired number is then
$$17+189=206$$
