Question

$\huge\red{heya!!}$

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

$\huge\red{heya!!}$

Let the AP series be x, x+d, x+2d…x+(n-1)d, where;

x is the first term of the sequence;

d is the arithmetic difference;

n is the number of terms.

According to the question;

first term=x=a;……(1)

second term=x+d=b;

d=b-a;……………….(2)

last term=x+(n-1)d=2a;

Substituting the value of d from equation 2 in the above equation, we get;

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

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

n=b/(b-a)…………..(3);

Hence as we know sum of terms of an AP=n/2[2*first term+(n-1)d];

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

sum=3ab/2(b-a). ANS