AP2 - AP - Complete The Series (Easy) ediorial SPOJ - UCS - Unleash-Coding-Skills

## Tuesday, 1 May 2018

The link for the problem  AP2 - AP - Complete The Series (Easy) is:

http://www.spoj.com/problems/AP2/

The logic used in this problem  AP2 - AP - Complete The Series (Easy) is:

This problem is used AP formulas.We have A and A[N-3] and Sum of Series (S) in the given input.But we don't know about N and we need to calculate it.

We know the sum of Series of AP: ((A+A[N]/2)*N
Given A which is equal to
A=A+2*D

So A = A - 2*D

and A[N] is given by A[N] = A[N-2] + 2*DSo ((A+A[N]/2)*N  becomes :
S = ((A-2*D+A[N-2]+2*D)/2)*N

In this, we can know the value of N by:
N=2*S/(A+A[N-2])

For Finding D in these formulas we have to use the thing:

suppose consider the series for example:

3 6 9 12 15 18 21 24

Suppose we know A 9 and A[N-2] 18 So to find the D for this we must use the formula:

A[N-2]-A/(No of terms in between them +1)

The following is CPP solution for this  "AP2 - AP - Complete The Series (Easy)" is:

```#include<iostream>
#include<cstdio>
#define LL long long int
using namespace std;

int main()
{
int T;
cin>>T;
while(T--)
{
LL a3,an3,sum,d,n,a;
cin>>a3>>an3>>sum;
n = (2*sum)/(a3+an3);
d = (an3-a3)/(n-5);
a = a3-(2*d);
cout<<n<<endl;
for(int i=0;i<n;i++)
cout<<(a+(i*d))<<" ";
cout<<endl;
}
return 0;
}```

Happy Coding........