Sum of Arithmetic Geometric Sequence (original) (raw)

Last Updated : 7 Aug, 2022

In mathematics, an arithmetico–geometric sequence is the result of the term-by-term multiplication of a geometric progression with the corresponding terms of an arithmetic progression.

is an arithmetico–geometric sequence.
Given the value of a(First term of AP), n(Number of terms), d(Common Difference), b(First term of GP), r(Common ratio of GP). The task is find the sum of first n term of the AGP.
Examples:

Input : First term of AP, a = 1, Common difference of AP, d = 1, First term of GP, b = 2, Common ratio of GP r = 2, Number of terms, n = 3 Output : 34 Explanation Sum = 12 + 222 + 3*23 = 2 + 8 + 24 = 34

The nth term of an arithmetico–geometric sequence is the product of the n-th term of an arithmetic sequence and the nth term of a geometric one. Arithmetico–geometric sequences arise in various applications, such as the computation of expected values in probability theory. For example Counting Expected Number of Trials until Success.
n-th term of an AGP is denoted by: tn = [a + (n - 1) * d] * (b * rn-1)
Method 1: (Brute Force)
The idea is to find each term of the AGP and find the sum.
Below is the implementation of this approach:

C++ `

// CPP Program to find the sum of first n terms. #include<bits/stdc++.h> using namespace std;

// Return the sum of first n term of AGP int sumofNterm(int a, int d, int b, int r, int n) {
// finding the each term of AGP and adding // it to sum. int sum = 0; for (int i = 1; i <= n ; i++)
sum += ((a + (i -1) * d) * (b * pow(r, i - 1)));
return sum; }

// Driven Program int main() { int a = 1, d = 1, b = 2, r = 2, n = 3; cout << sumofNterm(a, d, b, r, n) << endl; return 0; }

Java

// Java Program to find the sum of first n terms. import java.io.*;

class GFG {

// Return the sum of first n term of AGP
static int sumofNterm(int a, int d, int b, int r, int n)
{      
    // finding the each term of AGP and adding
    // it to sum.
    int sum = 0;
    for (int i = 1; i <= n ; i++)    
        sum += ((a + (i -1) * d) * (b * Math.pow(r, i - 1)));  
    return sum;
}


// Driven Program
public static void main(String args[])
{
    int a = 1, d = 1, b = 2, r = 2, n = 3;
    System.out.println(sumofNterm(a, d, b, r, n));
    
}

}

// This code is contributed by Nikita Tiwari.

Python3

Python3 code to find the

sum of first n terms.

import math

Return the sum of first

n term of AGP

def sumofNterm( a , d , b , r , n ): # finding the each term # of AGP and adding it to sum. sum = 0 for i in range(1,n+1): sum += ((a + (i -1) * d) * (b * math.pow(r, i - 1))) return int(sum)

Driven Code

a = 1 d = 1 b = 2 r = 2 n = 3 print(sumofNterm(a, d, b, r, n))

This code is contributed by "Sharad_Bhardwaj".

C#

// C# Program to find the sum of first n terms. using System;

class GFG {

// Return the sum of first n term of AGP
static int sumofNterm(int a, int d, int b, int r, int n)
{ 
    // Finding the each term of AGP 
    // and adding it to sum.
    int sum = 0;
    for (int i = 1; i <= n ; i++) 
        sum += (int)((a + (i -1) * d) *
                     (b * Math.Pow(r, i - 1))); 
    return sum;
}


// Driver Code
public static void Main()
{
    int a = 1, d = 1, b = 2, r = 2, n = 3;
    Console.Write(sumofNterm(a, d, b, r, n));
}

}

// This code is contributed by vt_m.

PHP

d,d, d,b, r,r, r,n) { // finding the each term // of AGP and adding // it to sum. $sum = 0; for ($i = 1; i<=i <= i<=n ; $i++) sum+=((sum += ((sum+=((a + ($i -1) * $d) * ($b * pow($r, $i - 1))); return $sum; } // Driver Code a=1;a = 1; a=1;d = 1; b=2;b = 2; b=2;r = 2; $n = 3; echo(sumofNterm($a, d,d, d,b, r,r, r,n)); // This code is contributed by Ajit. ?>

JavaScript

Output:

Time Complexity: O(nlogn) since using a inbuilt pow function inside a loop

Auxiliary Space: O(1) as using constant variables

Method 2: (Using Formula)

Proof,

Series, Sn = ab + (a+d)br + (a+2d)br2 + ..... + (a + (n-1)d)brn-1

Multiplying Sn by r, rSn = abr + (a+d)br2 + (a+2d)br3 + ..... + (a + (n-1)d)brn

Subtract rSn from Sn, (1 - r)Sn = [a + (a + d)r + (a + 2d)r2 + ...... + [a + (n-1)d]rn-1] - [ar + (a + d)r2 + (a + 2d)r3 + ...... + [a + (n-1)d]rn] = b[a + d(r + r2 + r3 + ...... + rn-1) - [a + (n-1)d]rn] (Using sum of geometric series Sn = a(1 - rn-1)/(1-r)) = b[a + dr(1 - rn-1)/(1-r) - [a + (n-1)d]rn]

Below is the implementation of this approach:

CPP `

// CPP Program to find the sum of first n terms. #include<bits/stdc++.h> using namespace std;

// Return the sum of first n term of AGP int sumofNterm(int a, int d, int b, int r, int n) { int ans = 0;
ans += a;
ans += ((d * r * (1 - pow(r, n-1)))/(1-r));
ans -= (a + (n-1)*d)pow(r, n);
return (ansb)/(1-r); }

// Driven Program int main() { int a = 1, d = 1, b = 2, r = 2, n = 3; cout << sumofNterm(a, d, b, r, n) << endl; return 0; }

Java

// Java Program to find the sum of first n terms.

import java.io.; import java.math.;

class GFG {

// Return the sum of first n term of AGP
static int sumofNterm(int a, int d, int b, int r, int n)
{
    int ans = 0;    
    ans += a;    
    ans += ((d * r * (1 - (int)(Math.pow(r, n-1))))/(1-r));    
    ans -= (a + (n-1)*d)*(int)(Math.pow(r, n));    
    return (ans*b)/(1-r);
}
 

// Driven Program
public static void main(String args[])
{
    int a = 1, d = 1, b = 2, r = 2, n = 3;
    System.out.println(sumofNterm(a, d, b, r, n));
    
}

}

// This code is contributed by Nikita Tiwari.

Python3

Python3 code to find

the sum of first n terms.

import math

Return the sum of

first n term of AGP

def sumofNterm( a , d , b , r , n ):

ans = 0
ans += a

ans += ((d * r * (1 - math.pow(r, n-1))
                            )/(1-r))

ans -= (a + (n-1)*d)*math.pow(r, n)

return int((ans*b)/(1-r))

Driven Code

a = 1 d = 1 b = 2 r = 2 n = 3 print(sumofNterm(a, d, b, r, n) )

This code is contributed by "Sharad_Bhardwaj".

C#

// C# Program to find the sum of first n terms. using System;

class GFG {

// Return the sum of first n term of AGP
static int sumofNterm(int a, int d, int b, int r, int n)
{
    int ans = 0; 
    ans += a; 
    ans += ((d * r * (1 - (int)(Math.Pow(r, n-1))))
                                           / (1-r)); 
    ans -= (a + (n-1) * d) *
            (int)(Math.Pow(r, n)); 
    
    return (ans * b) / (1 - r);
}


// Driver Code
public static void Main()
{
    int a = 1, d = 1, b = 2, r = 2, n = 3;
    Console.Write(sumofNterm(a, d, b, r, n));
}

}

// This code is contributed by vt_m.

PHP

JavaScript

Output:

Time Complexity: O(logn)

Auxiliary Space: O(1)