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Question
If pth, qth, rth and sth terms of an A.P. be in G.P., then prove that p − q, q − r, r − s are in G.P.
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Solution
\[\text { Here, } a_p = a + \left( p - 1 \right)d\]
\[ a_q = a + \left( q - 1 \right)d\]
\[ a_r = a + \left( r - 1 \right)d\]
\[ a_s = a + \left( s - 1 \right)d\]
\[\text { It is given that }a_p , a_q , a_r \text { and } a_s\text { are in G . P } . \]
\[ \therefore \frac{a_q}{a_p} = \frac{a_r}{a_q} = \frac{a_q - a_r}{a_p - a_q} = \frac{q - r}{p - q} . . . . . . . (i)\]
\[\text { Similarly }, \frac{a_r}{a_q} = \frac{a_s}{a_r} = \frac{a_r - a_s}{a_q - a_r} = \frac{r - s}{q - r} . . . . . . . (ii)\]
\[\text { Using }\left( i \right) \text { and }\left( ii \right): \]
\[\frac{q - r}{p - q} = \frac{r - s}{q - r}, \]
\[\text { Therefore }, p - q, q - r\text { and } r - s \text { are in G . P } .\]
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