Question

If the first , second and last of an $A.P$ are $a,b$ and $2a$, respectively, then sum will be

• A. $\frac{ab}{b-a}$
• B. $\frac{ab}{2(b-a)}$
• C. $\frac{3ab}{2(b-a)}$
• D. $\frac{4ab}{2(b-a)}$

Answer 1

Answer: (C)

$\frac{3ab}{2(b-a)}$

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)];

$sum=\frac{3ab}{2(b-a)}$