C++ Ratio of Mth and Nth Terms of an A. P. with Given Ratio of Sums

Discuss a problem where we are given a ratio of sums of m and n terms of A.P. We need to find the ratio of mth and nth terms.

Input: m = 8, n = 4
Output: 2.142

Input: m = 3, n = 2
Output: 1.666

Input: m = 7, n = 3
Output: 2.6

Approach to Find the Solution

To find the ratio of m^th and n^th terms using code, we need to simplify the formula. Let S_m be the sum of first m terms and S_n be the sum of the first n terms of the A.P.

a - first term,

d - common difference,

Given, S_m / S_n = m² / n²

Formula for S, S_m = (m/2)[ 2*a + (m-1)*d ]

m² / n² = (m/2)[ 2*a + (m-1)*d ] / (n/2)[ 2*a + (n-1)*d ]

m / n = [ 2*a +(m-1) *d ] / [ 2*a +(m-1) *d ]

Using cross multiplication,

n[ 2*a + (m−1)*d ] = m[ 2*a + (n−1)*d ]

2an + mnd - nd = 2am + mnd - md

2an - 2am = nd - md

(n - m)2a = (n-m)d

d = 2a

The formula of m^th term is,

T_m = a + (m-1)d

Ratio of m^th and n^th term is,

T_m / T_n = a + (m-1)d / a + (n-1)d

Replacing d with 2a,

Tm / Tn = a + (m-1)*2a / a + (n-1)*2a

Tm / Tn = a( 1 + 2m − 2 ) / a( 1 + 2n − 2 )

Tm / Tn = 2m - 1 / 2n - 1

So now we have a simple formula to find the ratio of m^th and n^th terms. Let’s see C++ code for this.

Example

C++ Code for the Above Approach

#include <bits/stdc++.h>
using namespace std;
int main(){
    float m = 8, n = 4;
    // calculating ratio by applying formula.
    float result = (2 * m - 1) / (2 * n - 1);
    cout << "The Ratio of mth and nth term is: " << result;
    return 0;
}

Output

The ratio of mth and nth term is: 2.14286

Conclusion

In this tutorial, we discussed a problem to find the ratio of mth and nth term with a given ratio of sums which we solved by simplifying the formula of the sum of m terms and formula of mth term. We also discussed the C++ program for this problem which we can do with programming languages like C, Java, Python, etc. We hope you find this tutorial helpful.