Question
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 term is twice the (q+1)th terms. If it's (3p+1)th term is λ times the (p+q+1)th term, then λ is equal to:
A. 2
B. 21
C. 3
D.None of these.
Solution
Here, we will first find the (p+1)th term and (q+1)th term using the formula of nthterm 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 term and (p+q+1)th term using the formula nthterm 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 nthterm of an A.P. is given by tn=a+(n−1)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., (p+1)th term is twice the (q+1)th term.
Now, we will find the (p+1)th term and (q+1)thterm using the formula tn=a+(n−1)d.
Substituting n=p+1 in the formula, we get (p+1)th term of an A.P. as:
tp+1=a+(p+1−1)d
Subtracting the terms in the bracket, we get
⇒tp+1=a+(p)d
Substituting n=q+1 in the formula, we get (q+1)th term of an A.P. as:
tq+1=a+(q+1−1)d
Subtracting the terms in the bracket, we get
⇒tq+1=a+(q)d
Since, it is given that (p+1)th term is twice the (q+1)th term, so we get
tp+1=2tq+1
Substituting the values of (p+1)th term and (q+1)th term, we get
⇒a+pd=2(a+qd)
Now, multiplying the terms using the distributive property, we get
⇒a+pd=2a+2qd
Rewriting the above equation, we get
⇒2a−a=pd−2qd
Simplifying the equation, we get
⇒a=(p−2q)d
Since, it is given that (3p+1)th term is λtimes the (p+q+1)th term, so
t3p+1=λtp+q+1
Using the formula of The nthterm of an A.P. is given by tn=a+(n−1)d, we get
⇒a+(3p+1−1)d=λ[a+(p+q+1−1)d]
By simplifying the equation, we get
⇒a+(3p)d=λ[a+(p+q)d]
By substituting a=(p−2q)d in the above equation, we get
⇒(p−2q)d+(3p)d=λ[(p−2q)d+(p+q)d]
By simplifying and rewriting the equation, we get
⇒pd−2qd+3dp=λ[pd−2qd+pd+qd]
Adding and subtracting the like terms, we get
⇒4pd−2qd=λ[2pd−qd]
⇒2(2pd−qd)=λ(2pd−qd)
By dividing (2pd−qd) on both sides, we get
⇒λ=(2pd−qd)2(2pd−qd)
⇒λ=2
Therefore, the value of λ 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.