


Discuss a problem in which the ratio of the sum of m and n terms of A.P is given. We need to find the ratio of the mth term to the nth term.
Input: m = 8, n = 4 Output: 2.142 Input: m = 3, n = 2 Output: 1.666 Input: m = 7, n = 3 Output: 2.6
Solution method
To find the ratio of the mth term to the nth term using code, we need to simplify the formula. Let Sm be the sum of the first m terms and Sn be the sum of the first n terms of A.P.
a - the first item,
d - the tolerance,
is given, Sm / Sn = m2 / n2
#S formula, Sm = (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 ]
Use 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
2an - 2am = nd 2a
The formula for the mitem is:
Tm = a (m-1)d
The ratio of the mth item to the nth item is,
Tm / Tn = a (m -1)d / a (n-1)d
Replace 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 -1So now we have a simple formula for finding the ratio of the m
thth term to the nthth term. Let's take a look at the C code. Example
C code for the above method#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 of finding the ratio of the m-th term to the n-th term under a given sum ratio. We solved this problem by simplifying the formula for the sum of m terms and the formula for the m-th semester. We also discussed a C program to solve this problem and we can implement it using programming languages like C, Java, Python etc. We hope you found this tutorial helpful.
The above is the detailed content of C++ Given the ratio of the sum of an arithmetic sequence, calculate the ratio of the Mth term and the Nth term. For more information, please follow other related articles on the PHP Chinese website!

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