# Which term of the G.P. 2, 8, 32,… is 131072?

Last updated date: 19th Mar 2023

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Answer

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Hint: Find the general term of the G.P. and equate it to 131072 to find ‘n’. This will give the position of the term 131072 in the G.P.

The given series 2, 8, 32,… is in G.P. (Geometric Progression)

We need to find ‘n’ for the ${n^{th}}$ term $Tn = 131072$ …(1)

In a G.P. The first term is denoted as ‘a’ and the common ratio is denoted as ‘r’. Common ratio is defined as the ratio between two consecutive terms in the G.P. It will be the same for any two consecutive terms in the G.P.

So, we will find the value of ‘a’ and ‘r’.

$

r = \dfrac{8}{2} = 4 \\

r = \dfrac{{32}}{8} = 4 \\

$

$a = 2,r = 4$ …(2)

The formula for general term of a G.P. is $Tn = a{r^{n - 1}}$ …(3)

Substitute (1), (2) in (3)

$

Tn = a{r^{n - 1}} \\

131072 = \left( 2 \right){\left( 4 \right)^{n - 1}} \\

{4^{n - 1}} = 65536 \\

$

Applying the law of indices, ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$ we get,

$

\dfrac{{{4^n}}}{{{4^1}}} = 65536 \\

{4^n} = 262144 \\

$

Expressing 262144 as exponents of 4, we will get

${4^n} = {4^9}$

We know that if the bases are equal, the exponents are also equal.

$n = 9$

Hence, 131072 is the 9th term in the given G.P.

Note: Expressing in terms of exponents can be done using trial and error. The law of indices must be well-known by the students to solve such problems in a jiffy.

The given series 2, 8, 32,… is in G.P. (Geometric Progression)

We need to find ‘n’ for the ${n^{th}}$ term $Tn = 131072$ …(1)

In a G.P. The first term is denoted as ‘a’ and the common ratio is denoted as ‘r’. Common ratio is defined as the ratio between two consecutive terms in the G.P. It will be the same for any two consecutive terms in the G.P.

So, we will find the value of ‘a’ and ‘r’.

$

r = \dfrac{8}{2} = 4 \\

r = \dfrac{{32}}{8} = 4 \\

$

$a = 2,r = 4$ …(2)

The formula for general term of a G.P. is $Tn = a{r^{n - 1}}$ …(3)

Substitute (1), (2) in (3)

$

Tn = a{r^{n - 1}} \\

131072 = \left( 2 \right){\left( 4 \right)^{n - 1}} \\

{4^{n - 1}} = 65536 \\

$

Applying the law of indices, ${a^{m - n}} = \dfrac{{{a^m}}}{{{a^n}}}$ we get,

$

\dfrac{{{4^n}}}{{{4^1}}} = 65536 \\

{4^n} = 262144 \\

$

Expressing 262144 as exponents of 4, we will get

${4^n} = {4^9}$

We know that if the bases are equal, the exponents are also equal.

$n = 9$

Hence, 131072 is the 9th term in the given G.P.

Note: Expressing in terms of exponents can be done using trial and error. The law of indices must be well-known by the students to solve such problems in a jiffy.

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