Answer

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**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.

**Complete step-by-step answer:**

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 $S\left( n \right) = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$.

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