Question

In G.P, $$\left( p+q\right)^{th} $$ term is m, $$\left( p-q\right)^{th} $$ term is n, then $$p^{th}$$ term is
A) nm
B) $\sqrt{nm}$
C) $$\dfrac{m}{n}$$
D) $$\sqrt{\dfrac{m}{n} }$$

Answer 1

Hint: In this question it is given that in G.P, $$\left( p+q\right)^{th} $$ term is m and $$\left( p-q\right)^{th} $$ term is n, then we have to find $$p^{th}$$ term. So to find the solution we need to know that, if a be the first term of an Geometric progression and r be the common ratio, then $n^{th}$ term of an G.P will be,
$$t_{n}=ar^{\left( n-1\right) }$$.......(1)
Complete step-by-step solution:
So here it is given that, the $$\left( p+q\right)^{th} $$ term is m, so we can write,
$$t_{p+q}=m$$
$$\Rightarrow ar^{\left( p+q-1\right) }=m$$......(2) [using formula (1)]
And $$\left( p-q\right)^{th} $$ term is n, therefore,
$$t_{p-q}=n$$
$$\Rightarrow ar^{\left( p-q-1\right) }=n$$......(3)
Now by multiplying the equation (2) and (3), we get,
$$\left( ar^{p+q-1}\right) \left( ar^{p-q-1}\right) =m\times n$$
$$\Rightarrow a\times a\times r^{p+q-1}\times r^{p-q-1}=m\times n$$
$$\Rightarrow a^{2}\times r^{(p+q-1+p-q-1)}=mn$$ [$$\because a^{m}\cdot a^{n}=a^{m+n}$$]
$$\Rightarrow a^{2}\times r^{(2p-2)}=mn$$
$$\Rightarrow a^{2}\times r^{2(p-1)}=mn$$
$$\Rightarrow a^{2}\left( a^{p-1}\right)^{2} =mn$$ [$$\because a^{mn}=\left( a^{m}\right)^{n} $$]
$$\Rightarrow \left( ar^{p-1}\right)^{2} =mn$$ [$$\because a^{m}\cdot b^{m}=\left( ab\right)^{m} $$]
$$\Rightarrow ar^{p-1}=\sqrt{mn}$$
As we know that the expression of $p^{th}$ term, $$t_{p}=ar^{p-1}$$
Therefore we get, $$t_{p}=\sqrt{mn}$$
Hence, the correct option is option B.
Note: To solve this type of question you need to know that, whenever you divide or multiply two equations then you have to do it side wise, i.e, left side of one equation multiplied with left side of other equation and similarly applied for division.