SEQSUM - Editorial

PROBLEM LINK: Sequence Sum

Author: Rishabh Rathi
Tester: Rishabh Rathi

DIFFICULTY:

CAKEWALK

PREREQUISITES:

Maths

PROBLEM:

You are given a sequence - 11, 17, 23, 29, ……

You need to find the total number of sweets needed to distribute among the first N children (sum of first N elements).

NOTE : Since answer can be large, print answer modulo 10^9+7

EXPLANATION:

The sequence given is an Arithmetic Progression. Here, first term (a) is 11 and common difference (d) is 6.

Sum of N terms of arithmetic progression, \sum_{i = 1}^{N} a_i

= \frac{N [2a + (n – 1)d]}{2}

=\frac{N [2*11 + (N – 1)*6]}{2}

=\frac{N [22 + 6N - 6]}{2}

=\frac{N [16 + 6N]}{2}

=N [8 + 3N]

SOLUTIONS:

Setter's Solution (Python)
import sys
def input():
	return sys.stdin.readline().strip()

mod = 10**9 + 7

t = int(input())
for _ in range(t):
	n = int(input())
	ans = (n*(8 + 3*n))%mod
	print(ans)
Tester's Solution (CPP)
#include <bits/stdc++.h>  
using namespace std;

const int mod = 1e9 + 7;

int main() {
	ios_base::sync_with_stdio(0);  
	cin.tie(0);

	long long int n, ans;
	int t; cin>>t;

	while(t--) {
		cin>>n;
		ans = ((n%mod) * (8 + 3*n)%mod)%mod;
		cout<<ans<<"\n";
	}
	return 0;
}

Feel free to share your approach. In case of any doubt or anything is unclear, please ask it in the comments section. Any suggestions are welcome :smile: