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# In an A.P., ${\left( {p + 1} \right)^{th}}$ term is twice the ${\left( {q + 1} \right)^{th}}$ terms. If it's ${\left( {3p + 1} \right)^{th}}$ term is $\lambda$ times the ${\left( {p + q + 1} \right)^{th}}$ term, then $\lambda$ is equal to:A. $2$ B. $\dfrac{1}{2}$ C. $3$ D.None of these.

Last updated date: 19th Sep 2024
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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.

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.