Question

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

Answer 1

Hint:
We will first calculate the common difference by subtracting the second term from the first term of the given sequence. Then, calculate the number of terms using the last term of the sequence and applying the formula, ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term of the sequence and $d$ is the common difference of the sequence. Next, calculate the sum of terms by substituting the values in the formula, $\dfrac{n}{2}\left( {A + {A_n}} \right)$, where $A$ is the first term and ${A_n}$ is the last term.

Complete step by step solution:
We are given that the first term of the A.P is $a$ and the second term is $b$.
As it is known that the terms of an A.P have a common difference. The common difference of the A.P. can be calculated by subtracting the first term from the second term.
Then, $d = b - a$
Also, we are given that the last term is $2a$.
We have to calculate the sum of all the terms, but for that, we need to find the number of terms first.
The number of terms of the A.P. can be calculated by using the last term of the given A.P.
We know that the ${n^{th}}$ term of the A.P. is ${a_n} = a + \left( {n - 1} \right)d$, where $a$ is the first term of the sequence and $d$ is the common difference of the sequence.
Hence, ${a_n} = 2a$ and $d = b - a$
Then, we have,
$
  2a = a + \left( {n - 1} \right)\left( {b - a} \right) \\
   \Rightarrow a = \left( {n - 1} \right)\left( {b - a} \right) \\
   \Rightarrow \dfrac{a}{{b - a}} = n - 1 \\
   \Rightarrow n = \dfrac{a}{{b - a}} + 1 \\
   \Rightarrow n = \dfrac{{a + b - a}}{{b - a}} \\
   \Rightarrow n = \dfrac{b}{{b - a}} \\
 $
Hence, the number of terms for the A.P is $\dfrac{b}{{b - a}}$
Now, we can calculate the sum using the formula, $\dfrac{n}{2}\left( {A + {A_n}} \right)$, where $A$ is the first term and ${A_n}$ is the last term
Hence, the sum of the terms is $\dfrac{{\dfrac{b}{{b - a}}}}{2}\left( {a + 2a} \right)$
On simplifying, we will get,
$\dfrac{b}{{2\left( {b - a} \right)}}\left( {3a} \right) = \dfrac{{3ab}}{{2\left( {b - a} \right)}}$

Hence, the option (c) is correct.

Note:
We can alternatively find the sum of A.P. using the formula, \[\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\], where $a$ is the first term, $d$ is the common difference and $n$ is the number of terms. Here, we used the formula, $\dfrac{n}{2}\left( {a + {a_n}} \right)$, where $a$ is the first term and ${a_n}$ is the last term because we were given the last term of the sequence.