diff --git a/C++/Program-21/README.md b/C++/Program-21/README.md
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+# Last digit of sum of partial Fibonacci Series
+
+## Problem statement:
+Given two non-negative integers m and n, where m <= n, find the last digit of the sum *Fm + Fm+1 + .... + Fn*
+
+### Input:
+Two non negative intergers m and n, such that m <= n.
+
+### Output: 
+The last digit of the sum of Fibonacci series starting from *Fm* to *Fn*.
+
+### Example:
+
+*Input:*
+```
+3 7
+```
+*Output:*
+```
+1
+```
+*Explanation:*
+```
+𝐹3 + 𝐹4 + 𝐹5 + 𝐹6 + 𝐹7 = 2 + 3 + 5 + 8 + 13 = 31
+Last digit of 31 = 1
+```
diff --git a/C++/Program-21/program.cpp b/C++/Program-21/program.cpp
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+/* Compute the last digit of the sum of partial Fibonacci series. (Given two non-negative integers m and n, where m <= n, find the last digit of the sum Fm + Fm+1 + .... + Fn) */
+
+#include<bits/stdc++.h>
+using namespace std;
+typedef long long int ll;
+
+void solve()
+{
+   ll n = 0, m = 0;
+   int b[60] = {0, 1, 2, 4, 7, 2, 0, 3, 4, 8, 3, 2, 6, 9, 6, 6, 3, 0, 4, 5, 0, 6, 7, 4, 2, 7, 0, 8, 9, 8, 8, 7, 6, 4, 1, 6, 8, 5, 4, 0, 5, 6, 2, 9, 2, 2, 5, 8, 4, 3, 8, 2, 1, 4, 6, 1, 8, 0, 9, 0};
+   int i = 0, j = 0, sum = 0;
+
+   cin >> n >> m;
+
+   i = (n-1) % 60;
+   j = m % 60;
+
+   if(b[j] >= b[i])
+   {
+      cout << b[j] - b[i] << "\n";
+   }
+   else
+   {
+      b[j] += 10;
+      cout << b[j] - b[i] << "\n";
+   }
+}
+
+int main()
+{
+   //for fast input output
+   ios_base::sync_with_stdio(false);cin.tie(NULL);
+
+   solve();
+ 
+   return 0;
+}
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