
How do you graph \[{a_n} = 2{\left( 3 \right)^{n - 1}}\] ?
Answer
545.7k+ views
Hint: Here in this given equation given the nth term of geometric series when we give the n values we get the geometric series now give the values to the n like 0, 1, 2, 3, … simultaneously we get the values of \[{a_0}\] , \[{a_1}\] , \[{a_2}\] … Now we construct the coordinates of a given series i.e., (x,y) by using the coordinates to construct the graph. We assign the value of n values as x and the determined value \[{a_n}\] as y and we plot the graph.
Complete step-by-step answer:
Consider the equation \[{a_n} = 2{\left( 3 \right)^{n - 1}}\]
Which is a geometric sequence, The sequence cannot be represented by a continuous graph as \[{a_n}\] is only defined \[\forall n \in N\]
By giving the n values …-3,-2,-1, 0, 1, 2, 3, … simultaneously we get the values of \[{a_n}\]
Put n=-3
Then \[ \Rightarrow \,{a_{ - 3}} = 2{\left( 3 \right)^{ - 3 - 1}}\]
\[ \Rightarrow \,\,\,{a_{ - 3}} = \dfrac{2}{{{3^4}}}\]
\[\therefore \,\,\,{a_{ - 3}} = 0.02469\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( { - 3,0.02469} \right)\]
Put n=-2
\[ \Rightarrow \,{a_{ - 2}} = 2{\left( 3 \right)^{ - 2 - 1}}\]
\[ \Rightarrow \,\,\,{a_{ - 2}} = \dfrac{2}{{{3^3}}}\]
\[\therefore \,\,\,{a_{ - 2}} = 0.07407\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( { - 2,0.07407} \right)\]
Put n=-1
\[ \Rightarrow \,{a_{ - 1}} = 2{\left( 3 \right)^{ - 1 - 1}}\]
\[ \Rightarrow \,\,\,{a_{ - 1}} = \dfrac{2}{{{3^2}}}\]
\[\therefore \,\,\,{a_{ - 1}} = 0.2222\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( { - 1,0.2222} \right)\]
Put n=0
\[ \Rightarrow \,{a_0} = 2{\left( 3 \right)^{0 - 1}}\]
\[ \Rightarrow \,\,\,{a_0} = \dfrac{2}{3}\]
\[\therefore \,\,\,{a_0} = 0.6666\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( {0,0.6666} \right)\]
Put n=1
\[ \Rightarrow \,{a_1} = 2{\left( 3 \right)^{1 - 1}}\]
\[ \Rightarrow \,{a_1} = 2{\left( 3 \right)^0}\]
\[\therefore \,\,\,{a_1} = 2\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( {1,0} \right)\]
Put n=2
\[ \Rightarrow \,{a_2} = 2{\left( 3 \right)^{2 - 1}}\]
\[ \Rightarrow \,{a_2} = 2{\left( 3 \right)^1}\]
\[\therefore \,\,\,{a_2} = 6\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( {2,6} \right)\]
Put n=3
\[ \Rightarrow \,{a_3} = 2{\left( 3 \right)^{3 - 1}}\]
\[ \Rightarrow \,{a_3} = 2{\left( 3 \right)^2}\]
\[\therefore \,\,\,{a_2} = 18\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( {3,18} \right)\]
And so on
Now, the graph of the of the given geometric progression series \[{a_n} = 2{\left( 3 \right)^{n - 1}}\] is given by
Note: First we simplify the given equation and we plot the graph for the points. The graph is plotted x-axis versus y axis. The graph is two dimensional. By the equation of a graph, we can plot the graph by assuming the value of x. we can’t assume the value of y. because the value of y depends on the value of x. Hence, we have plotted the graph.
Complete step-by-step answer:
Consider the equation \[{a_n} = 2{\left( 3 \right)^{n - 1}}\]
Which is a geometric sequence, The sequence cannot be represented by a continuous graph as \[{a_n}\] is only defined \[\forall n \in N\]
By giving the n values …-3,-2,-1, 0, 1, 2, 3, … simultaneously we get the values of \[{a_n}\]
Put n=-3
Then \[ \Rightarrow \,{a_{ - 3}} = 2{\left( 3 \right)^{ - 3 - 1}}\]
\[ \Rightarrow \,\,\,{a_{ - 3}} = \dfrac{2}{{{3^4}}}\]
\[\therefore \,\,\,{a_{ - 3}} = 0.02469\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( { - 3,0.02469} \right)\]
Put n=-2
\[ \Rightarrow \,{a_{ - 2}} = 2{\left( 3 \right)^{ - 2 - 1}}\]
\[ \Rightarrow \,\,\,{a_{ - 2}} = \dfrac{2}{{{3^3}}}\]
\[\therefore \,\,\,{a_{ - 2}} = 0.07407\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( { - 2,0.07407} \right)\]
Put n=-1
\[ \Rightarrow \,{a_{ - 1}} = 2{\left( 3 \right)^{ - 1 - 1}}\]
\[ \Rightarrow \,\,\,{a_{ - 1}} = \dfrac{2}{{{3^2}}}\]
\[\therefore \,\,\,{a_{ - 1}} = 0.2222\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( { - 1,0.2222} \right)\]
Put n=0
\[ \Rightarrow \,{a_0} = 2{\left( 3 \right)^{0 - 1}}\]
\[ \Rightarrow \,\,\,{a_0} = \dfrac{2}{3}\]
\[\therefore \,\,\,{a_0} = 0.6666\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( {0,0.6666} \right)\]
Put n=1
\[ \Rightarrow \,{a_1} = 2{\left( 3 \right)^{1 - 1}}\]
\[ \Rightarrow \,{a_1} = 2{\left( 3 \right)^0}\]
\[\therefore \,\,\,{a_1} = 2\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( {1,0} \right)\]
Put n=2
\[ \Rightarrow \,{a_2} = 2{\left( 3 \right)^{2 - 1}}\]
\[ \Rightarrow \,{a_2} = 2{\left( 3 \right)^1}\]
\[\therefore \,\,\,{a_2} = 6\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( {2,6} \right)\]
Put n=3
\[ \Rightarrow \,{a_3} = 2{\left( 3 \right)^{3 - 1}}\]
\[ \Rightarrow \,{a_3} = 2{\left( 3 \right)^2}\]
\[\therefore \,\,\,{a_2} = 18\]
Therefore, co-ordinate \[\left( {x,y} \right) = \left( {3,18} \right)\]
And so on
Now, the graph of the of the given geometric progression series \[{a_n} = 2{\left( 3 \right)^{n - 1}}\] is given by
Note: First we simplify the given equation and we plot the graph for the points. The graph is plotted x-axis versus y axis. The graph is two dimensional. By the equation of a graph, we can plot the graph by assuming the value of x. we can’t assume the value of y. because the value of y depends on the value of x. Hence, we have plotted the graph.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

The draft of the Preamble of the Indian Constitution class 10 social science CBSE

Who gave "Inqilab Zindabad" slogan?

Why is there a time difference of about 5 hours between class 10 social science CBSE

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

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

