Question

If n arithmetic means are inserted between \[20\] and \[80\] , such that the ratio of the first mean to the last mean is \[1:3\] , then find the value of \[n\] .

Answer 1

Hint: We have to find the value of \[n\] i.e. the number of arithmetic means inserted between the given numbers \[20\] and \[80\] . We will solve this question using the concept of properties and various formulas of the arithmetic progression (A.P.) . First we would consider the total number of terms in the series formed after the insertion of n number of arithmetic means , then using the formula for \[{n^{th}}\] term of an arithmetic progression new would find a relation in the terms of the common difference \[\left( d \right)\] and the number of terms \[\left( n \right)\] of the series . And then using the given ratio and using the formula for the \[{n^{th}}\] term of the arithmetic progression we would obtain another relation between the common difference \[\left( d \right)\] and the number of terms \[\left( n \right)\] of the series and finally on eliminating the term of common difference \[\left( d \right)\] from both the relations by dividing first relation by the second and hence on simplifying the ratio formed , we would obtain the value of \[n\] .

Complete step-by-step solution:
Given :
\[n\] arithmetic means are inserted between \[20\] and \[80\] . The ratio of the first mean to the last mean is \[1:3\] .
We also know that the ratio can be written in form of fraction as :
\[a:b = \dfrac{a}{b}\]
Hence , the ratio of the first mean to the last mean is \[\dfrac{1}{3}\] .
After inserting \[n\] arithmetic means the series becomes as : \[20{\text{ }},{\text{ }}{a_1},{\text{ }}{a_2}{\text{ }},{\text{ }}{a_3}{\text{ }},{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \ldots .{\text{ }},{\text{ }}{a_n},{\text{ }}80\] .
Now , as we have inserted the “n” number of terms between \[20\] and \[80\] . The number of terms becomes as :
\[N = n + 2\]
[As we already have two terms \[20\] and \[80\] in the series and we have inserted \[n\] number of terms .]
Now , we have been given that :
\[\dfrac{{{a_1}}}{{{a_n}}} = \dfrac{1}{3}\]
We know that the formula for \[{n^{th}}\] term of an A.P. is given as :
\[{a_n} = a + \left( {n - 1} \right)d\]
Now , from the series we get the value of first terms \[\left( a \right)\] of the series as :
\[a = 20\]
Now , using the formula of the \[{n^{th}}\] term of an A.P. we get the expression for last term of the series as :
\[{a_N} = a + \left( {N - 1} \right)d\]
\[80 = 20 + \left( {\left( {n + 2} \right) - 1} \right)d\]
Simplifying the expression , we get the expression as :
\[60 = \left( {n + 1} \right)d - - - \left( 1 \right)\]
Now using the relation of the ratio , we can write the expression as :
\[\dfrac{{{a_1}}}{{{a_n}}} = \dfrac{1}{3}\]
We know that the term \[{a_1}\] and \[{a_n}\] of the series are actually the second term and the \[{\left( {n + 1} \right)^{th}}\] term . So , we can represent the terms of the ratio as :
\[{a_1} = a + d\]
Also ,
\[{a_n} = a + \left( {\left( {n + 1} \right) - 1} \right)d\]
\[{a_n} = a + nd\]
Substituting the two , we get the ratio as :
\[\dfrac{{a + d}}{{a + nd}} = \dfrac{1}{3}\]
Cross multiplying the terms , we get the ratio as :
\[3\left( {a + d} \right) = a + nd\]
\[3a + 3d = a + nd\]
On simplifying the terms , we can write the relation as :
\[2a = nd - 3d\]
\[2a = \left( {n - 3} \right)d\]
Putting the value of a , we get the expression as :
\[2 \times 20 = \left( {n - 3} \right)d\]
\[40 = \left( {n - 3} \right)d - - - \left( 2 \right)\]
Now , we will eliminate the value of \[d\] .
Dividing equation \[\left( 1 \right)\] by equation \[\left( 2 \right)\] , we get the expression as :
\[\dfrac{{\left( {n + 1} \right)d}}{{\left( {n - 3} \right)d}} = \dfrac{{60}}{{40}}\]
On , cancelling the terms we get the expression as :
\[\dfrac{{\left( {n + 1} \right)}}{{\left( {n - 3} \right)}} = \dfrac{3}{2}\]
Cross multiplying the terms , we get the expression as :
\[2\left( {n + 1} \right) = 3\left( {n - 3} \right)\]
\[2n + 2 = 3n - 9\]
On simplifying , we get the expression as :
\[n = 11\]
Hence , the number of arithmetic means inserted between \[20\] and \[80\] is \[11\] .

Note: We can also use the substitution method or any other method for solving the two relations to get the value of the number of terms inserted . Don’t get confused with the word written arithmetic mean in the question , it is only given for creating the sense of confusion . It only means that n numbers are inserted between \[20\] and \[80\] .
The formula of mean of the three progression is given as :
\[\left( 1 \right)\] A.P.
\[Arithmetic{\text{ }}mean = \dfrac{{\left( {a + b} \right)}}{2}\]
\[\left( 2 \right)\] G.P.
\[mean = \sqrt {bc} \]
\[\left( 3 \right)\] H.P.
Harmonic mean = \[mean = \dfrac{{2ab}}{{a + b}}\]