If pth ,qth ,rth and sth terms of an A.P. are in G.P. then show that (p-q), (q-r), (r-s) are also in G.P.
Hi!
Here is the answer to your question.
Let a and d be the first term and common difference of the A.P.
ap, aq,ar, as are in G.P.
⇒ a + (q – 1)d = k[a + (p – 1)d], a + (r – 1)d = k [a + (q – 1)d], a + (s – 1)d = k [a + (r – 1)d]
⇒ [a + (q – 1)d] – [a + (p – 1)d] = k[a + (p – 1)d] – [a + (p – 1)d],
[a + (r – 1)d] – [a + (q – 1)d] = k[a + (q – 1)d] – [a + (q – 1)d] and
[a + (s – 1)d] – [a + (r – 1)d] = k[a + (r – 1)d] – [a + (r – 1)d]
⇒ –d (p – q) = (k – 1) [a + (p – 1)d],
–d(q – r) = (k – 1) [a + (q – 1)d],
–d(r – s) = (k – 1) [a + (r – 1)d]
This shows that p – q, q – r and r – s are in G.P.
Cheers!