The question is: Find all Arithmetic progression of natural numbers starting with $3$ whose sum is a $3$ digit number whose digits are in non constant GP.
I tried that the sum could be $124, 421, 139, 931, 469, 964, 248, 842 but then $3+(n-1)d$ equals the above numbers would give lot of boundary conditions. Not sure how to go further.
Pl help.
The formula for the sum of an AP of $n$ terms beginning at $a$ and with a common difference $d$ is
$S=\frac{n(2a + (n-1)d)}{2}$. The last term is $a+(n-1)d$
You have made a good start by listing the possible sums. I would then list AP's of 2 terms eg (3, 121), then (3, 418) etc. AP's of 3 terms can only add to sums which are divisible by 3, so there aren't any. APs of 4 terms must have an even sum. There will quite a lot of answers, but not a ridiculous numbers, and finding the factors of your sums will help.
