Question

For a sequence, if $\rm t_{n} =\frac{7^{n+3} }{5^{n+2}}$, show that the sequence is a GP., find the first term and the common ratio.

Answer 1

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Answer 2

Answer:

First term, $t_{1} = \frac{7^{4}}{5^{3}}$

Common ratio $= \frac{7}{5}$

Step-by-step explanation:

As per the given data, we have the nth term of G.P:

$t_{n}=\frac{7^{n+3}}{5^{n+2}}$

For the first term, put n = 1 in $t_{n}=\frac{7^{n+3}}{5^{n+2}}$

Therefore,

$t_{1} = \frac{7^{1+3}}{5^{1+2}}$

∴ First term, $t_{1} = \frac{7^{4}}{5^{3}}$

For the second term, put n = 2 in $t_{n} = \frac{7^{n+3}}{5^{n+2}}$

Therefore,

$t_{2} = \frac{7^{2+3}}{5^{2+2}}$

∴ Second term, $t_{2} = \frac{7^{5}}{5^{4}}$

For the third term, put n = 3 in $t_{n} = \frac{7^{n+3}}{5^{n+2}}$

Therefore,

$t_{3} = \frac{7^{3+3}}{5^{3+2}}$

∴ Third term, $t_{2} = \frac{7^{6}}{5^{5}}$

Common ratio $= \frac{Second\ term}{First\ term}$

Common ratio $= \frac{ \frac{7^{5}}{5^{4}}}{\frac{7^{4}}{5^{3}}}$

∴ Common ratio $= \frac{7}{5}$

Also,

Common ratio $= \frac{Third\ term}{Second\ term}$

Common ratio $= \frac{ \frac{7^{6}}{5^{5}}}{\frac{7^{5}}{5^{4}}}$

∴ Common ratio $= \frac{7}{5}$

Hence, the sequence is a G.P