In a GP, if the (m + n)thterm is P and the (m – n)th term is q, then its mth term is

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Solution

The correct option is C
$T_{m + n} = p, T_{m - n} = q$
$\Rightarrow a r^{m + n - 1} = p a n d a r^{m - n - 1} = q m u l t i p l y i n g E g s . (i) a n d (i i), a^{2} r^{2 m - 2} = p q \Rightarrow (a r m - 1)^{2} = p q \Rightarrow a r^{m - 1} = \sqrt{p q} \Rightarrow T_{m} = \sqrt{p q}$

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Similar questions

In a G.P. if the $(m + n)^{t h}$ term is p and $(m - n)^{t h}$ term is q, then its $m^{t h}$ term is

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The (m + n)th and the (m - n)th terms of a GP are p and q respectively. Show that the mth and the nth terms of the GP are $\sqrt{p q}$ and ${(\frac{q}{p})}^{(\frac{m}{2 n})}$

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(m + n) t h

term is

P

and the

(m - n) t h

term is

q,

then its

m t h

term is:

In a G.P. if the ${(m + n)}^{t h}$ term is p and ${(m - n)}^{t h}$ term is q , then its $m^{t h}$ term is:

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In a GP, if the (m + n)thterm is P and the (m – n)th term is q, then its mth term is

The correct option is C Tm+n=p,Tm−n=q ⇒arm+n−1=pand arm−n−1=qmultiplying Egs. (i)and(ii),a2r2m−2=pq⇒(arm−1)2=pq⇒arm−1=√pq⇒Tm=√pq

The correct option is C
$T_{m + n} = p, T_{m - n} = q$
$\Rightarrow a r^{m + n - 1} = p a n d a r^{m - n - 1} = q m u l t i p l y i n g E g s . (i) a n d (i i), a^{2} r^{2 m - 2} = p q \Rightarrow (a r m - 1)^{2} = p q \Rightarrow a r^{m - 1} = \sqrt{p q} \Rightarrow T_{m} = \sqrt{p q}$