# 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.(example of ncert).plzzz explain it couldn't understand !!

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

ap, aq,ar, as are in G.P.

Now, using A.P. , we have ap = a + (p-1)d,

aq = a + (q-1)d

ar = a + (r-1)d

as = a + (s-1)d

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]

Taking, a + (q – 1)d = k[a + (p – 1)d]

On subtracting  a + (p – 1)d on both sides, we get

⇒ [a + (q – 1)d] – [a + (p – 1)d] = k[a + (p – 1)d] – [a + (p – 1)d],

Similarly,  on taking  a + (r – 1)d = k [a + (q – 1)d] and  a + (s – 1)d = k [a + (r – 1)d] and subtracting a + (q – 1)d and a + (r – 1)d on both sides respectively, we get

[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 (pq) = (k – 1) [a + (p – 1)d]
d(qr) = (k – 1) [a + (q – 1)d],

d(rs) = (k – 1) [a + (r – 1)d]

On solving the respective equations, we get

This shows that (pq), (qr) and (rs) are in G.P.

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