Prove that sum of AP is greater than sum of GP if first and last terms of both series are equal, all terms are positive and number of terms is equal

Question

Question: An AP and a GP with positive terms have the same number of terms and their first and last terms are equal. Prove that the sum of GP cannot exceed sum of AP.

Attempt:

• Let sum of AP be $A$ and that of GP be $G$. 'r' is the common ratio.
• let first and last terms of both series be a and b.

$A=\frac{n}{2}(a+b)=\frac{na}{2}(1+r^{n-1})$

$\frac{G}{A}=\frac{2(1+r+r^2+....+r^{n-1})}{n(1+r^{n-1})}$

since all terms are positive, $r$ is positive and $1+r+r^2+r^{n-1}$ is greater than $1+r^{n-1}$ for all positive r.

But this wont help me get the inequality. I did not understand the solution given in the book.So i am tried to solve it myself.

Answer 1

one possible approach

suppose $A_k$ and $G_k$ are the terms of the series, with $A_0=G_0=a$ and $A_n=G_n=b$

then $$ A_k = a + k\frac{b-a}{n} $$ and $$ G_k = a \left(\frac{b}{a} \right)^{\frac{k}{n}} $$ set $\lambda=\frac{b}{a}-1$. then $$ D_k=\frac{A_k-G_k}{a} = 1 + \frac{k}{n}\lambda - (1+ \lambda)^{\frac{k}{n}} $$ to obtain the result you require, it is sufficient to show that for $\alpha \in (0,1)$ and for $\lambda \gt 0$ the following inequality holds: $$ (1+\lambda)^{\alpha} \le 1 + \lambda \alpha $$

Answer 2

This is an unusual method for me. But anyway, this is the answer given in the book: