Question

Check whether the following sequence are G. P. if so, write tn
1) 2,6,18,54

Answer 1

GIVEN :–

• A given sequence 2,6,18,54........

TO FIND :–

• nth term of G.P. = ?

SOLUTION :–

• Sequence –

$\\ \bf \implies 2,6,18,54........$

• The ratio of terms of Geometrical progression always be equal.

• So that –

$\\ \bf \implies ratio(r)= \dfrac{6}{2} = \dfrac{18}{6} =\dfrac{54}{18} = 3$

• Hence , It's a G.P. series.

__________________________________________

• We know that nth term of G.P. –

$\\ \large \implies {\boxed{\bf T_{n} = a {r}^{n - 1}}}$

• Here –

$\\ \bf \: \: \: {\huge{.}} \:\:\: a = 2$

$\\ \bf \: \: \: {\huge{.}} \:\:\: r = 3$

• Now put the values –

$\\ \bf \implies T_{n} = (2){(3)}^{n - 1}$

$\\ \bf \implies T_{n} = (2){(3)}^{n} \bigg( \dfrac{1}{3} \bigg)$

$\\ \large \implies { \boxed{ \bf T_{n} = \dfrac{2}{3}{(3)}^{n}}}$

Answer 2

Answer:

$\huge\fbox{ෆ╹Answer ╹ෆ}$

Given:-

Sequence of 2,6,18,54

Solution:-

So here

$= > a = 2$

$= > r_{1} = \frac{6}{2}$

$= 3$

And

$= > r_{2} = \frac{18}{6}$

$= 3$

$from \: equation \: (1) \: and \: (2)$

$= > r_{1} = r_{2}$

$hence \: common \: ratio \: is \: same \:$

$so \: the \: sequence \: are \: gp.$