The numbers $a,b,c$ are in an arithmetic progression with $a+b+c=124$
If $a,b,c$ are the third, thirteenth and fifteenth terms of an arithmetic progression, find $a,b,c$
There are no hints in the book regarding this question. A friend of mine suggests that the problem is worded wrongly and the first sentence should be:
The numbers $a,b,c$ are in a geometric progression with $a+b+c=124$
If I consider $a,b,c$ as being in a geometric progression as the initial information I can solve the problem. I get the result $r = 1/5$, $a=100$, $b=20$, $c=4$.
I can't solve the other way around (if $a,b,c$ are in an arithmetic progression as the first sentence). Something doesn't really click in my brain so that's why I am here for help.
To sum it all up, I am interested in finding out:
a. if the exercise is wrong and it should be geometric instead of arithmetic in the first sentence
or b. a hint to solve the original exercise if it turns out to be correct
The exercise is wrong: both ways of interpreting the question lead to unexpected answers.
Suppose "In arithmetic progression" means that $b-a = c-b$. Using the general term of an arithmetic progression:
$$T(n) = x + (n-1)d$$
where $x$ is the first term and $d$ the common difference, we obtain
$$a=x+2d, b=x+12d, c=x+14d \leadsto b-a = 10d=c-b=2d$$
giving $d=0, a=b=c=\frac{124}3$, which is weird to say the least.
Now suppose "In arithmetic progression" gives no extra information beyond the "3rd, 13th, 15th term" condition. Then we have:
$$x+2d+x+12d+x+14d = 3x+28d = 124$$
which is a single equation with two unknowns, which has an infinite number of solutions, and a unique solution cannot be obtained without additional information (e.g. they are actually in geometric progression).
With the extra information, we can write
$$\frac {10d}{x+2d} = \frac{2d}{x+12d} \leadsto5(x+12d) = x+2d \leadsto 4x+58d=0$$
which combined with $3x+28d=124$ gives the solution you found ($x=116, d= -8$).