The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

Asked by Abhisek | 1 year ago |  74

##### Solution :-

Let’s take a to be the first term and r to be the common ratio of the G.P.

Then according to the question, we have

a5 = a r5–1 = a r4 = p … (i)

a= a r8–1 = a r7 = q … (ii)

a11 = a r11–1 = a r10 = s … (iii)

Dividing equation (ii) by (i), we get

$$r^3=\dfrac{q}{p}$$ ...............(iv)

Dividing equation (iii) by (ii), we get

$$r^3=\dfrac{s}{q}$$ ...............(v)

Equating the values  of r3 obtained in (iv) and (v), we get

$$\dfrac{q}{p}=\dfrac{s}{q}$$

q2 = ps

Hence proved

Answered by Pragya Singh | 1 year ago

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