Question

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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
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Answer 1

Answer:

Step-by-step explanation:

Heya!!!!

Its means that option is C one (correct)

What is your question?

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

​

Solution-

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)

correct answer

Hope u like it

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!!!