Question

If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is
(a) $\frac{ab}{2(b-a)}$
(b) $\frac{ab}{(b-a)}$
(c) $\frac{3ab}{2(b-a)}$
(d) none of these

Answer 1

Answer:

The sum is 3ab/2(b - a)

Among the given options option (d) 3ab/2(b - a) is a correct answer.

Step-by-step explanation:

Given :  

First term = a, second term =  b and last term (an), l =  2a are in A.P  

Common difference ,d = (b – a)

By using the formula nth term ,an = a + (n - 1)d

2a = a (n - 1) (b - a)

2a - a =  (n - 1) (b - a)

a =  (n - 1) (b - a)

a/(b - a) = n - 1

n - 1 = a/(b - a)

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

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

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

 

By using the formula ,Sum of nth terms , Sn = n/2 [a + l]

Sn = ½ [b/(b - a)] [a + 2a]

[From eq 1]

Sn = ½ [b/(b - a)] [3a]

Sn = [b/2(b - a)] {3a}

Sn = 3ab/2(b - a)

Hence, the sum is 3ab/2(b - a)

HOPE THIS ANSWER WILL HELP YOU….

 

 

Answer 2

Given AP,

a,b,.....................,2a

Last term,l=2a

Common Difference: Second term - First term

:b-a

First term,a:a

★First of all,we need to find the no.of terms in the given AP

Thus,

Last term:2a

→a+(n-1)d=2a

→a+(n-1)(b-a)=2a

→n-1=a/(b-a)

$\implies \: \sf{n = \frac{a}{b - a} + 1 } \\ \\ \implies \: \sf{n = \frac{a + (b - a)}{b - a} } \\ \\ \implies \: \boxed{\sf{n = \frac{b}{b - a} }}$

Now,

Sum of the terms:

Using the formula,

$\sf{ \frac{n}{2}(a + l) }$

Substituting the values,

$\sf{ = \frac{ \frac{b}{(b - a)} }{2}(a + 2a) } \\ \\ = \: \sf{ \frac{b}{b - a} \times \frac{1}{2} \times 3a} \\ \\ = \huge{\sf{\frac{3ab}{2(b - a)}}}$

The correct option is (c)