Question

# The general form of A.P. is $a$, $a + d$, ________(a) $a + 2d$(b) $a - 2d$(c) $a + d$(d) $a - d$

Hint: In this type of question we will take the nth term of a A.P. (arithmetic progression)
that is, $T\left( n \right) = a + \left( {n - 1} \right)d$, where T is the nth term of the A.P. having first term a and d be difference between any two consecutive terms.

Here the given arithmetic progression is $a$, $a + d$, __
Now we will consider the nth term of A.P.
$T\left( n \right) = a + \left( {n - 1} \right)d$ -(1)
where T is the nth term of the A.P. having first term a and d be difference between any two
consecutive terms.
In this question we are given the first two terms of the A.P. that are $a$ and $a + d$.
And from these two consecutive terms we can see that their difference is
$= a + d - a \\ = d \\$
And the first term is $a$
According to the question we need to find the third term so $n = 3$.
Now putting all these values in (1), we get,
$T\left( n \right) = a + \left( {n - 1} \right)d \\ T\left( 3 \right) = a + \left( {3 - 1} \right)d \\ {\text{ = a + 2d}} \\$
Hence, the third term of the is ${\text{a + 2d}}$.
Therefore, option (a) is the correct answer.

Note: The arithmetic progression is a sequence of numbers which differ from each other by common difference. And the general A.P. is $a$, $a + d$, $a + 2d$ ,____, so on.
The formula for nth term is $T\left( n \right) = a + \left( {n - 1} \right)d$ and sum of n terms is $S\left( n \right) = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$.