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 - Maths - Sequences and Series

Question

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 !!

Ankush Jain · Accepted Answer

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

a_p, a_q,a_r, a_s are in G.P.

$∴ \frac{a_{q}}{a_{p}} = \frac{a_{r}}{a_{q}} = \frac{a_{s}}{a_{r}}$

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

a_q = a + (q-1)d

a_r = a + (r-1)d

a_s = a + (s-1)d

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

⇒ 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 (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]

On solving the respective equations, we get

This shows that (p – q), (q – r) and (r – s) are in G.P.

-14

This shows that (p â€“ q), (q â€“ r) and (r â€“ s) are in G P

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 !!

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 !!