The common difference d of an AP is equal to the common ratio r of a GP. I have been told that the sum of the first ten terms of the AP is equal to fifteen times the sum of the first three terms of the GP.
I have tried using the formulas $\frac{10}{2}[2a+9d]$ = $15[\frac{a(1-d^3)}{1-d}]$, however I am unsure how to continue.
Best Answer
You are going to need more information to get a clean answer.
If the AP starts at $a$ and the GP starts at $b,$ then the equation becomes:
$$\frac{10}{2}(2a+9d)=15b(1+d+d^2)$$
Then you get $$15bd^2+(15b-45)d+(15b-10a)=0$$
Or equivalently:
$$bd^2+(b-3)d+(b-2a/3)=0$$
This gives a quadratic formula:
$$d=\frac{(3-b)\pm\sqrt{9 - 6 b + 8 a b/3 - 3 b^2}}{2b}$$ if $b\neq 0.$
If $b=0,$ then $d=\frac{-2a}{9}.$
Given any $b$ and $d,$ we can solve:
$$a=\frac{3}2(bd^2+(b-3)d+b)$$
This shows that there are infinitely many integer trios, $a,b,d,$ specifically, when $b,d$ both even or both odd.
Assuming $a=b,$ you need $b^2+18b- 27\leq 0,$ or $(b+9)^2\leq 108.$ For a $b$ positive integer, there is only one value, $b=1.$ But the value of $d$ in this case is irrational, $d=1\pm 2\sqrt{2/3}.$
If $b=1,$ then $$d=1\pm \sqrt{2a/3}$$
So, for example, when $a=6, b=1,$ you get $d=3,-1.$ Then when $d=3$:
$$6+9+\cdots +33=195,\\1+3+3^3=13,$$ and $13\cdot 15=195.$
When $d=-1,$ $$6+5+\cdots+(-3)=15,\\1+(-1)+1=1.$$
If $b=3$ you want $$d=\pm\sqrt{2a/9-1}$$
You can get any odd integer $d$ from some integer $a,$ specifically, $a=\frac{9(d^2+1)}{2}.$
For $d=5,$ $a=117,$ and:
$$117+122+\cdots+162=1395=93\cdot 15,\\3+15+75=93.$$
This gives another answer for $d=3,$ with $a=45,b=3.$
$$45+48+\dots+72=585=15\cdot 39,\\3+9+27=39$$