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Question: In an A.P., \[{\left( {p + 1} \right)^{th}}\] term is twice the \[{\left( {q + 1} \right)^{th}}\] te...

In an A.P., (p+1)th{\left( {p + 1} \right)^{th}} term is twice the (q+1)th{\left( {q + 1} \right)^{th}} terms. If it's (3p+1)th{\left( {3p + 1} \right)^{th}} term is λ\lambda times the (p+q+1)th{\left( {p + q + 1} \right)^{th}} term, then λ\lambda is equal to:
A. 22
B. 12\dfrac{1}{2}
C. 33
D.None of these.

Explanation

Solution

Here, we will first find the (p+1)th{\left( {p + 1} \right)^{th}} term and (q+1)th{\left( {q + 1} \right)^{th}} term using the formula of nth{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 (3p+1)th{\left( {3p + 1} \right)^{th}} term and (p+q+1)th{\left( {p + q + 1} \right)^{th}} term using the formula nth{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 nth{n^{th}}term of an A.P. is given by tn=a+(n1)d{t_n} = a + \left( {n - 1} \right)d, where aa is the first term and dd is the common difference.

Complete step-by-step answer:
We will consider an A.P. with the first term as ‘ aa’ and the common difference as ‘ dd’.
We are given that in an A.P., (p+1)th{\left( {p + 1} \right)^{th}} term is twice the (q+1)th{\left( {q + 1} \right)^{th}} term.
Now, we will find the (p+1)th{\left( {p + 1} \right)^{th}} term and (q+1)th{\left( {q + 1} \right)^{th}}term using the formula tn=a+(n1)d{t_n} = a + \left( {n - 1} \right)d.
Substituting n=p+1n = p + 1 in the formula, we get (p+1)th{\left( {p + 1} \right)^{th}} term of an A.P. as:
tp+1=a+(p+11)d{t_{p + 1}} = a + \left( {p + 1 - 1} \right)d
Subtracting the terms in the bracket, we get
tp+1=a+(p)d\Rightarrow {t_{p + 1}} = a + \left( p \right)d
Substituting n=q+1n = q + 1 in the formula, we get (q+1)th{\left( {q + 1} \right)^{th}} term of an A.P. as:
tq+1=a+(q+11)d{t_{q + 1}} = a + \left( {q + 1 - 1} \right)d
Subtracting the terms in the bracket, we get
tq+1=a+(q)d\Rightarrow {t_{q + 1}} = a + \left( q \right)d
Since, it is given that (p+1)th{\left( {p + 1} \right)^{th}} term is twice the (q+1)th{\left( {q + 1} \right)^{th}} term, so we get
tp+1=2tq+1{t_{p + 1}} = 2{t_{q + 1}}
Substituting the values of (p+1)th{\left( {p + 1} \right)^{th}} term and (q+1)th{\left( {q + 1} \right)^{th}} term, we get
a+pd=2(a+qd)\Rightarrow a + pd = 2\left( {a + qd} \right)
Now, multiplying the terms using the distributive property, we get
a+pd=2a+2qd\Rightarrow a + pd = 2a + 2qd
Rewriting the above equation, we get
2aa=pd2qd\Rightarrow 2a - a = pd - 2qd
Simplifying the equation, we get
a=(p2q)d\Rightarrow a = \left( {p - 2q} \right)d
Since, it is given that (3p+1)th{\left( {3p + 1} \right)^{th}} term is λ\lambda times the (p+q+1)th{\left( {p + q + 1} \right)^{th}} term, so
t3p+1=λtp+q+1{t_{3p + 1}} = \lambda {t_{p + q + 1}}
Using the formula of The nth{n^{th}}term of an A.P. is given by tn=a+(n1)d{t_n} = a + \left( {n - 1} \right)d, we get
a+(3p+11)d=λ[a+(p+q+11)d]\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
a+(3p)d=λ[a+(p+q)d]\Rightarrow a + \left( {3p} \right)d = \lambda \left[ {a + \left( {p + q} \right)d} \right]
By substituting a=(p2q)da = \left( {p - 2q} \right)d in the above equation, we get
(p2q)d+(3p)d=λ[(p2q)d+(p+q)d]\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
pd2qd+3dp=λ[pd2qd+pd+qd]\Rightarrow pd - 2qd + 3dp = \lambda \left[ {pd - 2qd + pd + qd} \right]
Adding and subtracting the like terms, we get
4pd2qd=λ[2pdqd]\Rightarrow 4pd - 2qd = \lambda \left[ {2pd - qd} \right]
2(2pdqd)=λ(2pdqd)\Rightarrow 2\left( {2pd - qd} \right) = \lambda \left( {2pd - qd} \right)
By dividing (2pdqd)\left( {2pd - qd} \right) on both sides, we get
λ=2(2pdqd)(2pdqd)\Rightarrow \lambda = \dfrac{{2\left( {2pd - qd} \right)}}{{\left( {2pd - qd} \right)}}
λ=2\Rightarrow \lambda = 2
Therefore, the value of λ\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.