Question

5th 8th and 11th term of GP are p,q and s respectively prove that q^2=ps

Answer 1

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a subscript 5 = a r^[5–1 ]= a r^4 = p … (1)

a subscript 8 = a r^[8–1 ]= a r^7 = q … (2)

a subscript 11 = a r^[11–1 ]= a r^10 = s … (3)

Dividing equation (2) by (1), we obtain

[a r^7 ]/[a r^4=  q/p

r^3 =q/p.....................................................(4)

Dividing equation (3) by (2), we obtain

[a r^10 ]/[a r^7]=  s/q

r^3 =s/q....................................................(5)

Equating the values of r^3 obtained in (4) and (5), we obtain

q/p=s/q

q^2 =ps

Thus, the given result is proved.

Answer 2

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