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Question: If the \[(p + q)\] th term of a geometric series is \(m\) and the \((p - q)\) th term is \(n\) , the...

If the (p+q)(p + q) th term of a geometric series is mm and the (p−q)(p - q) th term is nn , then the pp th term is
A) mn\sqrt {mn}
B) mnmn
C) m+nm + n
D) m−nm - n

Explanation

Solution

A geometric series is a series for which the ratio of each two consecutive terms is a constant function. Formula for finding the nn th term in geometric sequence is Tn=a×rn−1{T_n} = a \times {r^{n - 1}}, where a=a = start term and r=r = common ratio. First we find the given terms then multiply them and calculate the multiplication after that we get the required result.

Complete step by step answer:
In the given data there are not given the start term and the common ratio of a geometric series
then we take a=a = start term and r=r = common ratio
First we find the (p+q)(p + q) th term in geometric series
Tp+q=a×rp+q−1{T_{p + q}} = a \times {r^{p + q - 1}}
From the given data , we get
Tp+q=a×rp+q−1=m{T_{p + q}} = a \times {r^{p + q - 1}} = m
Now we find the (p−q)(p - q) th term in geometric series
Tp−q=a×rp−q−1{T_{p - q}} = a \times {r^{p - q - 1}}
From the given data , we get
Tp−q=a×rp−q−1=n{T_{p - q}} = a \times {r^{p - q - 1}} = n
Multiply the (p+q)(p + q) th term and (p−q)(p - q) th term then we have
(Tp+q)×(Tp−q)=mn({T_{p + q}}) \times ({T_{p - q}}) = mn
⇒(arp+q−1)×(arp−q−1)=mn\Rightarrow (a{r^{p + q - 1}}) \times (a{r^{p - q - 1}}) = mn
We know if ra{r^a} multiply with rb{r^b} then it becomes ra+b{r^{a + b}} , we use this in above equation and we get
⇒a2×r(p+q−1)+(p−q−1)=mn\Rightarrow {a^2} \times {r^{(p + q - 1) + (p - q - 1)}} = mn
⇒a2×rp+q−1+p−q−1=mn\Rightarrow {a^2} \times {r^{p + q - 1 + p - q - 1}} = mn
⇒a2×r2p−2=mn\Rightarrow {a^2} \times {r^{2p - 2}} = mn
We take common 22 from the power of rr , we get
⇒a2×r2(p−1)=mn\Rightarrow {a^2} \times {r^{2(p - 1)}} = mn
Taking square root both sides of the above equation and get
⇒a2×r2(p−1)=mn\Rightarrow \sqrt {{a^2} \times {r^{2(p - 1)}}} = \sqrt {mn}
⇒arp−1=mn\Rightarrow a{r^{p - 1}} = \sqrt {mn} ………………………………..(i)
Now we find the pp th term of geometric series
Tp=a×rp−1{T_p} = a \times {r^{p - 1}}
From the equation (i) , we get the value of pp th term and we substitute this and get
Tp=arp−1=mn{T_p} = a{r^{p - 1}} = \sqrt {mn}
∴\therefore The pthp^{th} term of the geometric series is Tp=mn{T_p} = \sqrt {mn} . So, Option (A) is correct.

Note:
If you forget the formula you can establish at the time . In a geometric progression there are a=a = start term and r=r = common ratio then we get the members are a,ar,ar2,ar3,...,arn,...a,ar,a{r^2},a{r^3},...,a{r^n},... , then we need to find the ss th term then we take the sequence T1=a=ar(1−1){T_1} = a = a{r^{(1 - 1)}} ,T2=ar=ar(2−1){T_2} = ar = a{r^{(2 - 1)}} ,T3=ar2=ar(3−1){T_3} = a{r^2} = a{r^{(3 - 1)}} ,……., Tn=arn−1{T_n} = a{r^{n - 1}} ,…..
Ts={T_s} = a×rs−1a \times {r^{s - 1}} . This is the required formula and you can establish it easily when you need it.