The common difference of a arithmetric progression are (+3) and (-2) respectively.

E.g => if the first term of this sequence is 4, the terms would be:
4, 7, 5, 8, 6, 9, 7
i.e 4, (+3), 7, (-2), 5, (+3), 8, (-2), 6, ...

The task here is to come up with a formulae to get the nth term of such sequence.

Solution....
First, this isn't a geometric progression but more like an arithmetic progression, it has two sequences therefore it must have two initial starting points....

Let the sequence appear in this structure,

a0,a1,a2,a3,a4,....(a)(n+2) or (a)(n) for all n(0 to infinity)................. Statement 1...

From statement one, we can derive two conditions,
a0= 4, and a1 =7....

Therefore, statement 1 appears like this;
4,7,a2,a3,a4,....(a)(n+2) or (a)(n).... Statement 2...

The nth term for both sequences therefore will be,

a(n+2) = a(n) + 1 ( if a0 = 4 and if and only if n = [0,2,4,6,..., 2(n)] ••••••• Equation 1...

Similarly,
a(n+2) = a(n) + 1 ( if a1 = 7 and if and only if n = [1,3,5,7,..., 2(n)+1] ••••••• Equation 2...

These formulas enables us to determine the nth term in an implicit approach, it could also be determined in an explicit manner but would require traversing equ 1 and 2....

So for Example,
If our solution appears like this

4,7,5,8,6,9,7,10,8,11,9
a0 = 4
a1=7

Let's find a7, of cause a7 =10....
Solution
Since n= 7, therefore let's apply equation
2 which means a1 =7
Therefore,
a7 = a ( n+2) means n =5...
So,
a(5 +2)= [a5 +1]....
a7 = [a5 + 1]
= [ (a3 + 1) + 1 ]
= [ ((( a1 + 1)+1)+1)]
Remember, a1 = 7
Therefore,
a7 = [((( 7+1) + 1)+1)]
= [((( + 1) +1)]
= [((8+1) + 1)]
=[((9)+1)]
= 10.... Answer

So you see, it works but this method can appear tedious when working with large terms, so I recommend using a programming language such as Java or C++ and applying my algorithm... To find more complex terms....

Thanks for reading!!!

Let n = 0.

then the sequence is 0,3,1,4,2,5,3,6,4

looking at that sequence, we can see than for all odd sequences, it is 0,1,2,3,4 and for all even sequences it is 3,4,5,6

Given a starting number v at index n to find the pattern
if n is odd, then the nth sequence is
((n - 1) / 2) + v

if n is even, then it is
((n / 2) + 2) + v

So given a sequence that begins with 4, to find the 7th sequence
given that 7 is odd, ((n - 1) / 2) + v = ((7 - 1) / 2) + 4 = 6/2 + 4 = 3+4 = 7

If 6th sequence
given that 6 is even ((n / 2) + 2) + v = (6/2) + 2 + 4 = 3 + 2 + 4 = 9

Given a sequence that begins with 17 to find the 12,345,678 - 12,345,680 numbers

12,345,678 sequence (12345678/2) + 2 + 17 = 6172858
12,345,679 sequence ((12345679 - 1 )/2) + 17 = 6172856
12,345,680 sequence (12345680/2) + 2 + 17 = 6172859

//language = java

public class Test
{
public static void main(final String... args)
throws Exception
{



// printing out first 7 terms
int ft=4;
int p1=3; int p2=-2;
System.out.println();
for(int x=1;x<=7;x++)
{
int res = nthTerm(ft,x,p1,p2);
System.out.print(" "+res);
}
}

//ft=first term, n = nth term, p1=first difference , p2 = second difference


public static int nthTerm(final int ft,final int n,final int p1,final int p2)
{
if(n<=1)
return ft;
else
{
int currentTerm = (n%2==0)?p1:p2;
if(n==2)
return ft+currentTerm;
return nthTerm(ft,n-1,p1,p2)+currentTerm;
}
}
}

using c++

======================================================================
#include <iostream>

using std::cin;
using std::cout;

int main(){

cout << "A program to find the nth term of a sequence with the difference (+3) and (-2)\n\n";

int first;
int nth_term;

cout << "Enter the first term of the sequence: \t";
cin >> first;
cout << std::endl;

cout << "Enter the nth term you want to find: \t";
cin >> nth_term;
cout << std::endl;

int answer = first;

for (int i = 0; i < nth_term; i++){

cout << answer << " ";

if ((i%2)!= 0){
answer = (answer + (-2));
}

else {

answer = (answer + 3);
}
}

cout << "\n\n";
cout << "The " << nth_term << " of the sequence is: " << answer;

return 0;
}

Math Tutor

Purpose
This assignment is designed to help students in using C++ to perform basic calculations. Students will also gain experience with generating randon numbers.

Problem Statement
Write a program that can be used as a math tutor for a young student. The program should display two random numbers between 1 and 9 to be added, subtracted, multimplied, and divided, such as 3 * 9 =
Help me with this math problem using C++




After the student has entered an answer and pressed the [Enter] key, the program should display the correct answer so the student can see if his or her answer is correct.

Sample Output
3 + 9 = 12
That's Correct
3 - 9 = 6
Sorry, That's not correct, the correct answer is -6
3 X 9 = 27
That's Correct
3 / 9 = 0.333
Sorry, That's not correct, the correct answer is 0