Question

heya!! 

Ques . If the first , second and last terms of AP are a , b and 2a respectively . Then its sum is 

A) ab/2(b-a)

B) ab/( b-a)

C) 3ab / 2 (b-a)

D) None of these 

Complete solution ✔✔✔✔✔

not just options otherwise reported ❎❎❎❎❎❎❎❎❎❎❎❎

Thanks ​

Answer 1

$\huge\red{heya!!}$

Here,

a1 = a

a2 = b

common difference d = a2- a1 =

= b-a

let n be the number of terms in series

an = 2a = a + (n-1)d

(n-1) = a /(b-a)

n = (a/(b-a)) + 1

n= b/(b-a)

: sum = n/2 ( a1 +an)

= [b/2(b-a)] (a + 2a)

= 3ab/2(b-a)

So option c is the answer

Thanks!!!

Answer 2

HI ROMEO,,,

here is your answer.

a1=a

a2=b

common difference d=a2-a1=b-a

Let n be the no.of terms in series :

an=2a=a+(n-1)d

(n-1)=a/(b-a)

n=(a/b-a)+1

n=b/(b-a)

Sum=n/2(a1+an)

=[b/2/(b-a)](a+2a)

=3ab/2(b-a)

So,it is clear that ans is 【option C】◆●◆●◆●◆●◆●●