Question

The first second and last term of an AP are a,b,c respectively prove that its sum is (a+c) (b+c-2a)/2(b-a)

Answer 1

$\huge\bf\red{Solution:-}$

$\bf \green{A.P=a,b,...,c}$

$\bf \red{common \:difference=d}$

$\bf \purple{a_1=a}$

$\bf\purple{ a_{2}=a+d=b}$

$\bf \purple{a_{n}=c(given)}$

$\Large{\underline{\boxed{\pink{a_{n}=a+(n-1)×d}}}}$

$\bf \red{a_{n}=a+(n-1)(b-a)}$

$\bf n=\dfrac{(b+c-2a) }{(b-a)}$

$\Large{\underline{\boxed{\pink{s_{n}=\dfrac{n}{2}(a+a_{n})}}}}$

$\bf \implies \dfrac{(b+c-2a) }{2(b-a)}(a+c)$

$\bf{sum=\dfrac{(b+c-2a) (a+c)}{2(b-a)}}$