Question

T4 of a G.P. in x, T10=y and T16 =z .Then
a) x^2 =yz
b)z^2 =xy
c)y^2=zx
d)none of these

Answer 1

Hey there!!

Here's your answer..

We know that the general term of a G.P is given Tn = $a r^{n-1}$

In the question, it is given that T4 is x

By substituting n = 4, we get T4 = x = $a r^{4-1} = a r^{3}$

Given that, T10 = y

⇒ n = 10 and T10 = y = $a r^{10-1} = a r^{9}$

Also, given that T16 = z

⇒ n = 16 and T16 = z = $a r^{16-1} = a r^{15}$

Let us now find the value of y²

y² = $(a r^{9}) ^{2}$ = $a^{2} r^{18}$

Now, let us find the value of xz

xz = $(a r^{3}).(a r^{15}) = a^{2} r^{18}$

So, y² = xz is the required answer,

Hope it helps!!

Answer 2

let first term of GP be a and common ratio be r

than 4th term is ar^3

10 th term is ar^9

16 the term is ar^15

ar^3,ar^9,ar^15

are in gp with common ratio r^6

hence proved