Which term of following GP is 256.
\[2\],\[2\sqrt 2 \],\[4\]…………..
Last updated date: 26th Mar 2023
•
Total views: 306.9k
•
Views today: 4.83k
Answer
306.9k+ views
Hint: In the questions we have to find the common ratio and then using the formula of the nth term of GP we can find which position the term 256 holds by equating it.
Complete step-by-step answer:
In the given series let's name the terms first i.e. \[{a_1}\] = 2, \[{a_2}\] =\[2\sqrt 2 \], \[{a_3}\]= 4
Let 256 be the \[{a_n}\]th term
Now, the common ratio \[r\] is given by
\[r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = .......... = \dfrac{{{a_n}}}{{{a_{n - 1}}}}\]
∴ \[r\]=\[\dfrac{{2\sqrt 2 }}{2}\]=\[\sqrt 2 \]
Using the formula of nth term
\[{a_n} = a{r^{n - 1}}\]
\[ \Rightarrow 256 = 2{\left( {\sqrt 2 } \right)^{n - 1}}\]
\[ \Rightarrow 128 = {\left( {\sqrt 2 } \right)^{n - 1}}\]
Converting 128 in terms of power of 2
\[ \Rightarrow {2^7} = {2^{\dfrac{{n - 1}}{2}}}\]
Since, bases are equal, therefore powers can also be equated.
\[ \Rightarrow 7 = \dfrac{{n - 1}}{2}\]
\[ \Rightarrow 14 = n - 1\]
\[\therefore n = 15\]
∴ 256 is the 15th term.
Note: Geometric progression is a sequence of numbers where each new term after the first is obtained by multiplying the preceding term by a constant r called common ratio. The common ratio is given by the formula \[r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = .......... = \dfrac{{{a_n}}}{{{a_{n - 1}}}}\].
Complete step-by-step answer:
In the given series let's name the terms first i.e. \[{a_1}\] = 2, \[{a_2}\] =\[2\sqrt 2 \], \[{a_3}\]= 4
Let 256 be the \[{a_n}\]th term
Now, the common ratio \[r\] is given by
\[r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = .......... = \dfrac{{{a_n}}}{{{a_{n - 1}}}}\]
∴ \[r\]=\[\dfrac{{2\sqrt 2 }}{2}\]=\[\sqrt 2 \]
Using the formula of nth term
\[{a_n} = a{r^{n - 1}}\]
\[ \Rightarrow 256 = 2{\left( {\sqrt 2 } \right)^{n - 1}}\]
\[ \Rightarrow 128 = {\left( {\sqrt 2 } \right)^{n - 1}}\]
Converting 128 in terms of power of 2
\[ \Rightarrow {2^7} = {2^{\dfrac{{n - 1}}{2}}}\]
Since, bases are equal, therefore powers can also be equated.
\[ \Rightarrow 7 = \dfrac{{n - 1}}{2}\]
\[ \Rightarrow 14 = n - 1\]
\[\therefore n = 15\]
∴ 256 is the 15th term.
Note: Geometric progression is a sequence of numbers where each new term after the first is obtained by multiplying the preceding term by a constant r called common ratio. The common ratio is given by the formula \[r = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{{a_3}}}{{{a_2}}} = .......... = \dfrac{{{a_n}}}{{{a_{n - 1}}}}\].
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
