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.

  • $\begingroup$ Can you give an example of such an A.P and G,P? $\endgroup$ Commented Oct 7, 2019 at 7:38
  • 1
    $\begingroup$ One such A.P is $2,4,6,8,10$ and the corresponding G.P is $2,2(5)^{1/4},2(5)^{2/4},2(5)^{3/4},10$ $\endgroup$
    – Maverick
    Commented Oct 7, 2019 at 7:52
  • $\begingroup$ Is it possible to show that each term of the A.P is $\geq$ each term of the G.P $\endgroup$ Commented Oct 7, 2019 at 7:55
  • $\begingroup$ in the series that I have mentioned each term of A.P is indeed greater than the corresponding term of the G.P. Has it got something to do with A.M>G.M inequality $\endgroup$
    – Maverick
    Commented Oct 7, 2019 at 8:02
  • $\begingroup$ counterexample: $2,6$. So, it is needed $n>2$. $\endgroup$
    – farruhota
    Commented Oct 7, 2019 at 9:02

3 Answers 3


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.


the terms of an AP can be thought of, on a graph by a straight line. Similarly, the GP can be visualized an exponential graph. Since their first and last terms are the same......which means graphically that the exponential and the straight line graph meet at two points. So, The area of graph under exponential graph is the sum of GP and that under the straight line graph is the sum of AP (If the domain of the terms is extended over real numbers). Since even in real numbers the Sum of GP is less than that of AP, then it cannot be greater than AP in a sequence where domain is integers.


Long comment, not answer:

Let $A_n$ be the sum of the $n$ first terms of the arithmetic progression and $G_n$ be the sum of the first $n$ terms of the geometric progression satisfying the conditions mentioned in the post. I generated some experimental data for different initial conditions $a,d$ and $n \le 10^{12}$. The experimental data shows that regardless of the choice of $a$, $d$ and hence $r$, we have an elegant asymptotic relationship

$$ \frac{A_n}{G_n} \sim \frac{\log n}{2} $$


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