diff --git a/Questions/Find_second_largest_Element_in_array.md b/Questions/Find_second_largest_Element_in_array.md
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+# Find Second largest element in an array
+
+Given an array of integers, our task is to write a program that efficiently finds the second largest element present in the array.
+
+## Input Format
+
+First line contains T that denotes the number of test cases. This is followed by T lines, each containing an integer, N.
+
+
+## Output Format
+
+Print the required answer for each test case.
+
+
+### Sample Input and Output
+
+```
+Input : arr[] = {12, 35, 1, 10, 34, 1}
+Output : The second largest element is 34.
+
+Input : arr[] = {10, 5, 10}
+Output : The second largest element is 5.
+
+Input : arr[] = {10, 10, 10}
+Output : The second largest does not exist.
+
+```
diff --git a/Solutions/Multiple_of_3_and_5.py b/Solutions/Multiple_of_3_and_5.py
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+++ b/Solutions/Multiple_of_3_and_5.py
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+def sumOfMultiple():
+
+    multiplyList = [];
+    for i in range(1000):
+        if i%3 == 0 or i%5 == 0:
+            multiplyList.append(i)
+
+    return sum(multiplyList)
+
+print(sumOfMultiple())
diff --git a/Solutions/Sum_square_difference.cpp b/Solutions/Sum_square_difference.cpp
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+++ b/Solutions/Sum_square_difference.cpp
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+#include <iostream>
+
+using namespace std;
+
+int main()
+{
+    int number;
+    long sumOfSqure, squreOfSum, difference;
+
+    cout<<"Enter Natural Number --> ";
+    std::cin >> number;
+
+    squreOfSum = (number*(number + 1)) / 2;         //formula to find sum of first n natural number.
+    squreOfSum = squreOfSum * squreOfSum;
+
+    sumOfSqure = (number*(number + 1)*(2*number + 1)) / 6;          //formula to find sum of squre of first n natural numbers.
+
+    difference = squreOfSum - sumOfSqure;
+
+    cout<<"Difference of the sum squre is -->  "<<difference;
+
+}
diff --git a/Solutions/Sum_squre_difference.py b/Solutions/Sum_squre_difference.py
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+++ b/Solutions/Sum_squre_difference.py
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+import math
+
+number = int(input("Enter Natural Number --> "))
+
+squreOfSum = math.pow((number*(number + 1)) // 2, 2)
+sumOfSqure = (number*(number + 1)*(2*number + 1)) // 6
+
+print("Difference of the sum squre is --> ", squreOfSum - sumOfSqure)