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

Answer:

C) 3ab / 2(b-a)

Step-by-step explanation:

a, b, 2a are in AP

=>  b - a = 2a - b      [ the common difference ]

=>  3a = 2b

The sum of the three numbers is then

S = a + b + 2a = 3a + b = 2b + b = 3b

Writing A for the expression in (A), we have

A = ab / 2(b-a)

  = 3ab / 2(3b-3a)           [ multiply numerator and denominator by 3 ]

  = 2b² / 2(3b-2b)           [ use 3a = 2b ]

  = 2b² / 2b

  = b

and this is NOT equal to the sum S = 3b.

B = ab / (b-a) = 2A = 2b, is also NOT equal to the sum S = 3b.

C = 3A = 3b = S.

So the answer is (C).

Answer 2

$\huge\red{heya!!}$

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!!!