If p t h , q t h , r t h and s t h terms of an A.P. be in G.P., then (p - q),(q - r),(r - s) will be in Options: A . &nbsp G.P B . &nbsp A.P C . &nbsp H.P D . &nbsp None of these

Answer: Option A : A If a and d be the first term and common difference of the A.P. Then T p = a + (p - 1)d, T q = a + (q - 1)d and T r = a+(r - 1)d. If T p , T q , T r are in G.P. Then its common ratio R = T q T p = T r T q = T q − T r T p − T q = [ a + ( q − 1 ) d ] − [ a + ( r − 1 ) d ] [ a + ( p − 1 ) d ] − [ a + ( q − 1 ) d ] = q − r p − q Similarly, we can show that R = q − r p − q = r − s q − r Hence (p - q), (q - r),(r - s) be in G.P.

If pth,qth,rth and sth terms of an A.P. be in G.P., then (p - q),(q

Discussion Forum : Sequences And Series

Question -

If $p^{t h}$ , $q^{t h}$ , $r^{t h}$ and $s^{t h}$ terms of an A.P. be in G.P., then (p - q),(q - r),(r - s) will be in

Options:

A . G.P

B . A.P

C . H.P

D . None of these

Answer: Option A
:
A

If a and d be the first term and common difference of the A.P.

Then $T_{p}$ = a + (p - 1)d, $T_{q}$ = a + (q - 1)d and

$T_{r}$ = a+(r - 1)d.

If $T_{p}, T_{q}, T_{r}$ are in G.P.

Then its common ratio R = $\frac{T_{q}}{T_{p}}$ = $\frac{T_{r}}{T_{q}}$ = $\frac{T_{q} - T_{r}}{T_{p} - T_{q}}$

= $\frac{[a + (q - 1) d] - [a + (r - 1) d]}{[a + (p - 1) d] - [a + (q - 1) d]}$ = $\frac{q - r}{p - q}$

Similarly, we can show that R = $\frac{q - r}{p - q}$ = $\frac{r - s}{q - r}$

Hence (p - q), (q - r),(r - s) be in G.P.

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