Question

If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is

विकल्प

• (a) \[\frac{p - q}{q - r}\] 

• (b) \[\frac{q - r}{p - q}\] 

• (c) pqr

• (d) none of these

   

Answer 1

(b) \[\frac{q - r}{p - q}\] 

Let a be the first term and d be the common difference of the given A.P.
Then, we have:

\[p^{th} \text{ term }, a_p = a + \left( p - 1 \right)d\]
\[ q^{th} \text{ term }, a_q = a + \left( q - 1 \right)d\]
\[ r^{th} \text{ term }, a_r = a + \left( r - 1 \right)d\]
\[\text{ Now, according to the question the p^{th} , the q^{th} and the r^{th} terms are in G . P } . \]
\[ \therefore \left( a + \left( q - 1 \right)d \right)^2 = \left( a + \left( p - 1 \right)d \right) \times \left( a + \left( r - 1 \right)d \right)\]
\[ \Rightarrow a^2 + 2a \left( q - 1 \right)d + \left( \left( q - 1 \right)d \right)^2 = a^2 + ad\left( r - 1 + p - 1 \right) + \left( p - 1 \right) \left( r - 1 \right) d^2 \]
\[ \Rightarrow ad\left( 2q - 2 - r - p + 2 \right) + d^2 \left( q^2 - 2q + 1 - pr + p + r - 1 \right) = 0\]
\[ \Rightarrow a\left( 2q - r - p \right) + d\left( q^2 - 2q - pr + p + r \right) = 0 \left( \because d cannot be 0 \right)\]
\[ \Rightarrow a = - \frac{\left( q^2 - 2q - pr + p + r \right)d}{\left( 2q - r - p \right)}\]
\[ \therefore \text{ Common ratio }, r = \frac{a_q}{a_p}\]
\[ = \frac{a + \left( q - 1 \right)d}{a + \left( p - 1 \right)d}\]
\[ = \frac{\frac{\left( q^2 - 2q - pr + p + r \right)d}{\left( p + r - 2q \right)} + \left( q - 1 \right)d}{\frac{\left( q^2 - 2q - pr + p + r \right)d}{\left( p + r - 2q \right)} + \left( p - 1 \right)d}\]
\[ = \frac{q^2 - 2q - pr + p + r + pq + rq - 2 q^2 - p - r + 2q}{q^2 - 2q - pr + p + r + p^2 + pr - 2pq - p - r + 2q}\]
\[ = \frac{pq - pr - q^2 + qr}{p^2 + q^2 - 2pq}\]
\[ = \frac{p\left( q - r \right) - q\left( q - r \right)}{\left( p - q \right)^2}\]
\[ = \frac{\left( p - q \right)\left( q - r \right)}{\left( p - q \right)^2}\]
\[ = \frac{\left( q - r \right)}{\left( p - q \right)}\]
\[ \]
\[\]