Question

The fifth term of a G.P. is x, eighth term of
a G.P. is y and eleventh term of a G.P. is z
verify whether y2 = x z.​

Answer 1

Step-by-step explanation:

$n ^{th} term \: of \: a \: geometric \: sequence \\ \: is \: given \: as \colon \: \: a_n = ar^{n - 1} \\ we \: are \: given \: that \colon \\ a_5 = x \\ \implies \: ar^{5 - 1} = x \\ \implies \: a {r}^{4} = x.....(1)\\ \\ a_8 = y \\ \implies \: ar^{8 - 1} = y\\ \implies \: a {r}^{7} = y.....(2)\\ \\ a_{11} = z \\ \implies \: ar^{11 - 1} = z\\ \implies \: a {r}^{10} = z.....(3) \\ let \: us \: now \: multiply \: equations \\ \: (1) \: and \: (3) \: we \: find \colon \: \\ xz \: = a {r}^{4} \: \times a {r}^{10} \\ \implies \: xz \: = {a}^{2} {r}^{14} = (a {r}^{7} )^{2} \\ \implies \: xz \: = y ^{2} \\ (\because \: y = a {r}^{7}) \\hence \: proved \colon \:$

Answer 2

Answer:

nth term of GP = $a_n=ar^{n-1}$

We are given that The fifth term of a G.P. is x

Substitute n = 5

$a_5=ar^{5-1}$

$x=ar^{4}$

Now we are given that  eighth term of  a G.P. is y

Substitute  n = 8

$a_8=ar^{8-1}$

$y=ar^{7}$

Now we are given that eleventh term of a G.P. is z

Substitute  n = 11

$a_{11}=ar^{11-1}$

$z=ar^{10}$

Now we are given that $y^2=xz$

So,  $(ar^{7})^2=ar^{4} \times ar^{10}$

$a^2r^{14}=a^2r^{14}$

Hence Verified  $y^2=xz$