Question

If first, second and last terms are respectively a, b and $2a$ of an arithmetic progression, then prove that sum of progression will be $\frac{3ab}{2(b-a)}$.

Answer 1

Let the number of terms of the progression is $n$ and common difference between the terms is $r$

given first term$=a$

second term$=a+r=b$

last term$=a+(n-1)r=2a$

now $r=b-a$

$(n-1)r=a$

so $n-1=\frac{a}{b-a}n=\frac{a}{b-a}+1n=\frac{b}{b-a}$

sum of the progression$\left(S\right)=\frac{n}{2}$$($first term$+$last term$)$

$S=\frac{n}{2}(a+2a)=\frac{b}{b-a}\left(3a\right)=\frac{3ab}{b-a}$

(Hence proved)