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**Hint:**Here, we will first find the \[{\left( {p + 1} \right)^{th}}\] term and \[{\left( {q + 1} \right)^{th}}\] term using the formula of \[{n^{th}}\]term of an A.P. Then we will find the value of the first term. Then we will use the given condition and find the \[{\left( {3p + 1} \right)^{th}}\] term and \[{\left( {p + q + 1} \right)^{th}}\] term using the formula \[{n^{th}}\]term of an AP. We will then substitute the value of the first term in the equation to get the value of the unknown variable.

**Formula Used:**

The \[{n^{th}}\]term of an A.P. is given by \[{t_n} = a + \left( {n - 1} \right)d\], where \[a\] is the first term and \[d\] is the common difference.

**Complete step-by-step answer:**We will consider an A.P. with the first term as ‘ \[a\]’ and the common difference as ‘ \[d\]’.

We are given that in an A.P., \[{\left( {p + 1} \right)^{th}}\] term is twice the \[{\left( {q + 1} \right)^{th}}\] term.

Now, we will find the \[{\left( {p + 1} \right)^{th}}\] term and \[{\left( {q + 1} \right)^{th}}\]term using the formula \[{t_n} = a + \left( {n - 1} \right)d\].

Substituting \[n = p + 1\] in the formula, we get \[{\left( {p + 1} \right)^{th}}\] term of an A.P. as:

\[{t_{p + 1}} = a + \left( {p + 1 - 1} \right)d\]

Subtracting the terms in the bracket, we get

\[ \Rightarrow {t_{p + 1}} = a + \left( p \right)d\]

Substituting \[n = q + 1\] in the formula, we get \[{\left( {q + 1} \right)^{th}}\] term of an A.P. as:

\[{t_{q + 1}} = a + \left( {q + 1 - 1} \right)d\]

Subtracting the terms in the bracket, we get

\[ \Rightarrow {t_{q + 1}} = a + \left( q \right)d\]

Since, it is given that \[{\left( {p + 1} \right)^{th}}\] term is twice the \[{\left( {q + 1} \right)^{th}}\] term, so we get

\[{t_{p + 1}} = 2{t_{q + 1}}\]

Substituting the values of \[{\left( {p + 1} \right)^{th}}\] term and \[{\left( {q + 1} \right)^{th}}\] term, we get

\[ \Rightarrow a + pd = 2\left( {a + qd} \right)\]

Now, multiplying the terms using the distributive property, we get

\[ \Rightarrow a + pd = 2a + 2qd\]

Rewriting the above equation, we get

\[ \Rightarrow 2a - a = pd - 2qd\]

Simplifying the equation, we get

\[ \Rightarrow a = \left( {p - 2q} \right)d\]

Since, it is given that \[{\left( {3p + 1} \right)^{th}}\] term is \[\lambda \]times the \[{\left( {p + q + 1} \right)^{th}}\] term, so

\[{t_{3p + 1}} = \lambda {t_{p + q + 1}}\]

Using the formula of The \[{n^{th}}\]term of an A.P. is given by \[{t_n} = a + \left( {n - 1} \right)d\], we get

\[ \Rightarrow a + \left( {3p + 1 - 1} \right)d = \lambda \left[ {a + \left( {p + q + 1 - 1} \right)d} \right]\]

By simplifying the equation, we get

\[ \Rightarrow a + \left( {3p} \right)d = \lambda \left[ {a + \left( {p + q} \right)d} \right]\]

By substituting \[a = \left( {p - 2q} \right)d\] in the above equation, we get

\[ \Rightarrow \left( {p - 2q} \right)d + \left( {3p} \right)d = \lambda \left[ {\left( {p - 2q} \right)d + \left( {p + q} \right)d} \right]\]

By simplifying and rewriting the equation, we get

\[ \Rightarrow pd - 2qd + 3dp = \lambda \left[ {pd - 2qd + pd + qd} \right]\]

Adding and subtracting the like terms, we get

\[ \Rightarrow 4pd - 2qd = \lambda \left[ {2pd - qd} \right]\]

\[ \Rightarrow 2\left( {2pd - qd} \right) = \lambda \left( {2pd - qd} \right)\]

By dividing \[\left( {2pd - qd} \right)\] on both sides, we get

\[ \Rightarrow \lambda = \dfrac{{2\left( {2pd - qd} \right)}}{{\left( {2pd - qd} \right)}}\]

\[ \Rightarrow \lambda = 2\]

Therefore, the value of \[\lambda \] is 2.

**Thus, Option (A) is correct.**

**Note:**In order to solve this question we need to keep in mind the concept of arithmetic progression. An arithmetic progression is a sequence of numbers where the difference between the two consecutive numbers is constant. We might make a mistake by using the formula of geometric progression instead of an arithmetic progression. There is quite a large difference between Geometric progression and an Arithmetic progression. Geometric progression is a sequence or series in which there is a common ratio between two consecutive terms.

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