Question

nth term of g.p 4,-3/2,9/16=​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The nth term is $(-1)^{n-1}\ (\frac{3^{n-1}}{2^{3n-5}} )$

Step-by-step explanation:

The general form of G.P is $a$, $ar$, $ar^2$, ... , $ar^{n-1}$,

where $a$ is the first term, $r$ is the common ratio, $n$ is the number of terms in the G.P and $ar^{n-1}$ is the $n^{th}$ term.

Then the formula for calculating is

$a_n=ar^{n-1}$

Common ratio = (any term)/preceding term

Consider the given G.P as follows:

$4, -3/2,\ 9/16$

Here, first term, $a=4$

Common ratio, $r=\frac{-3/2}{4}$

                             $=-\frac{3}{8}$

Then, nth term of G.P is given by

$a_n=ar^{n-1}$ ...... (1)

Substitute the values $4$ for $a$ and $-3/8$ for in the formula (1) as follows:

⇒ $a_n=(4)(-\frac{3}{8} )^{n-1}$

⇒ $a_n=2^2(-\frac{3}{2^3} )^{n-1}$ (Since $4=2^2$ and $8=2^3$)

Recall the laws of exponents,

(i) $(\frac{a}{b})^n = \frac{a^n}{b^n}$

(ii) $a^n= \frac{1}{a^{-n}}$

(iii) $(an)^n=a^nb^n$

(iv) $a^na^m=a^{m+n}$

Further, simplify as follows:

⇒ $a_n=(-1)^{n-1}\ 2^2(\frac{3^{n-1}}{2^{3(n-1)}} )$ (Using laws (i) and (iii))

⇒ $a_n=(-1)^{n-1}\ (\frac{1}{2^{-2}} )(\frac{3^{n-1}}{2^{3n-3}} )$ (Using law (ii))

⇒ $a_n=(-1)^{n-1}\ (\frac{3^{n-1}}{2^{3n-3-2}} )$ (Using law (iv))

⇒ $a_n=(-1)^{n-1}\ (\frac{3^{n-1}}{2^{3n-5}} )$

Thus, the nth term is $(-1)^{n-1}\ (\frac{3^{n-1}}{2^{3n-5}} )$.

#SPJ2