# if pth, qth, rthand sth terms of an A.P. are in G.P.,show that (p-q),(q-r) and (r-s) are also in G.P.

• -1

pth term of an AP = a + (p-1)d

qth term of the AP = a + (q-1)d

rth term of the AP = a + (r-1)d

sth term of the AP = a + (s-1)d

.

Since they are in GP, let x be the first term of GP and y be the common ratio.

.

Therefore,

a+(p-1)d = x .................... 1

a+(q-1)d = xy .................. 2

a+(r-1)d = xy2 ................. 3

a+(s-1)d = xy3................. 4

.

Hence,

Subtracting 1 from 2:-

[a+(q-1)d] - [a+(p-1)d] = xy - x

• -18

Srry clicked post answer by mistake..... here is the answer:-

.

pth term of an AP = a + (p-1)d

qth term of the AP = a + (q-1)d

rth term of the AP = a + (r-1)d

sth term of the AP = a + (s-1)d

.

Since they are in GP, let x be the first term of GP and y be the common ratio.

.

Therefore,

a+(p-1)d = x .................... 1

a+(q-1)d = xy .................. 2

a+(r-1)d = xy2................. 3

a+(s-1)d = xy3................. 4

.

Hence,

.

Subtracting 1 from 2:-

[a+(q-1)d] - {a+(p-1)d] = xy - x

(q-p)d = x(y-1) .................... 5

.

Similarily,

By subtracting,

2 from 3 :- (r-q)d = xy(y-1) ............... 6

3 from 4:- (s-r)d = xy2 (y-1) ............. 7

.

Hence

Dividing 6 by 5:-

(r-q) /(q-p)= xy/x = y ............... 8

.

Similarily,

Dividing 7 by 6:-

(s-r) / (r-q) = y ............. 9

.

From 8 and 9 :-

(r-q) / (q-p) = (s-r) / (r-q)

multiplying by (-1) throughout:-

(q-r) / (p-q) = (r-s) / (q-r)

.

Hence (p-q) , (q-r) and (r-s) are in GP.

• 65
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