Consider an Arithmetic Progression(A.P) with the first term $a$, the commom difference $d$ and a Geometric Progression(G.P) with first term again as $a$ but common ratio $r$ such that $a,d,r>0$ and both these progressions have same number of terms and their last terms are also equal.
Show that the sum of all the terms of A.P is greater than the sum of all the terms of the G.P.
My Attempt:
The terms between first and last terms are the $(n-2)$ Arithmetic Means(A.M's) or the Geometric Means(G.M's). Can it be proved that each of the A.M's is greater than the corresponding G.M's.
Let the A.P. and G.P. terms be: $$a,a+d,...,a+d(n-1)\\ a,ar,...,ar^{n-1}$$ The last terms are equal $$a+d(n-1)=ar^{n-1}$$ The condition $d>0$ implies the A.P. is an increasing progression. The equality of last terms implies the G.P. is also increasing, hence $r>1$.
Note that $n>2$, otherwise for $2,6$ ($d=4,r=3$), the sums are equal.
Now it needs to be proved for $a,d>0,r>1,n>2$: $$\frac{2a+d(n-1)}{2}n>\frac{a(r^n-1)}{r-1}\iff \\ \frac{(a+ar^{n-1})n}{2}>\frac{a(r^n-1)}{r-1}\iff \\ \frac{(1+r^{n-1})n}{2}>\frac{r^n-1}{r-1}$$ WA answer.