Ratio of mth and nth terms of an A. P. with given ratio of sums
Given that the ratio to sum of first m and n terms of an A.P. with first term ‘a’ and commond difference ‘d’ is m^2:n^2. The task is to find the ratio of mth and nth term of this A.P.
Examples:
Input: m = 3, n = 2
Output: 1.6667
Input: m = 5, n = 3
Output: 1.8
Approach:
Let the Sum of first m and n terms be denoted by Sm and Sn respectively.
Also, let the mth and nth term be denoted by tm and tn respectively.
Sm = (m * [ 2*a + (m-1)*d ])/2
Sn = (n * [ 2*a + (n-1)*d ])/2
Given: Sm / Sn = m^2 / n^2
Hence, ((m * [ 2*a + (m-1)*d ])/2) / ((n * [ 2*a + (n-1)*d ])/2) = m^2 / n^2
=> (2*a + (m-1)*d) / (2*a + (n-1)*d) = m / n
on cross multiplying and solving, we get
d = 2 * a
Hence, the mth and nth terms can be written as:
mth term = tm = a +(m-1)*d = a + (m-1)*(2*a)
nth term = tn = a +(n-1)*d = a + (n-1)*(2*a)
Hence the ratio will be:
tm / tn = (a + (m-1)*(2*a)) / (a + (n-1)*(2*a))
tm / tn = (2*m – 1) / (2*n – 1)
Below is the required implementation:
C++
#include <bits/stdc++.h>
using namespace std;
float CalculateRatio( float m, float n)
{
return (2 * m - 1) / (2 * n - 1);
}
int main()
{
float m = 6, n = 2;
cout << CalculateRatio(m, n);
return 0;
}
|
Java
import java.io.*;
class Nth {
static float CalculateRatio( float m, float n)
{
return ( 2 * m - 1 ) / ( 2 * n - 1 );
}
}
class GFG {
public static void main (String[] args) {
float m = 6 , n = 2 ;
Nth a= new Nth();
System.out.println(a.CalculateRatio(m, n));
}
}
|
Python3
def CalculateRatio(m, n):
return ( 2 * m - 1 ) / ( 2 * n - 1 );
if __name__ = = '__main__' :
m = 6 ;
n = 2 ;
print ( float (CalculateRatio(m, n)));
|
C#
using System;
class Nth {
float CalculateRatio( float m, float n)
{
return (2 * m - 1) / (2 * n - 1);
}
public static void Main () {
float m = 6, n = 2;
Nth a= new Nth();
Console.WriteLine(a.CalculateRatio(m, n));
}
}
|
PHP
<?php
function CalculateRatio( $m , $n )
{
return (2 * $m - 1) / (2 * $n - 1);
}
$m = 6; $n = 2;
echo CalculateRatio( $m , $n );
?>
|
Javascript
<script>
function CalculateRatio(m, n)
{
return (2 * m - 1) / (2 * n - 1);
}
let m = 6, n = 2;
document.write(CalculateRatio(m, n));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
Last Updated :
07 Jul, 2022
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