Question

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.

Answer 1

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.