Question

If Pth

, q

th and rth terms of an A.P. are in G.P. then the common ratio of this G.P. is​

Answer 1

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

ANSWER

In AP, first term be a

Difference be d

∴pthterm ⇒a+(p−1)d

q th term a+(q−1)d

r thterm a+(r−1)d

Since they are in GP

and let first term in G.P be A

and common ration be r

∴ As given

a+(p−1)d=A...(1)

a+(q−1)d=Ar...(2)

a+(r−1d)=Ar ²...(3)

subtracting (1) from (2)

[(q−1)−(p−1)]d=A[r−1]

(q−p)d=A(r−1)...(4)

subtracting (2) form (3)

(r−q)d=Ar(r−1)...(5)

Dividing (5), we get

$= \dfrac{ar(r - 1)}{a(r - 1)} = \dfrac{(r - q)d}{(q - p)d}$

∴$r = \dfrac{q - r}{p - q}$

Hence proved.

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Answer 2

Given :

• If pth, qth and rth  terms of an A.P. are in G.P.

To Find :

• Common ratio of GP

Explanation :

p , q and r are in AP.

=> q-p = r-q

=> 2q = p + r ... (1)

Now they are also in GP.

=> q = pc , r = pc²

where , c is common ratio

Now sub. these in (1) , we get ,

2(pc)=p+pc²

=> c²-2c+1 = 0

=> (c-1)² = 0

=> c = 1

So from(1),

(p-q) / (q-r)=1=c

and (q-r)/(p-q) = 1