The next number of this sequence is 365.

## How can we find the next number?

The actual sequence is “2 5 14 41 122”. Look, this sequence starts with the original number 2, then adds 3 with the original number then we got the second number 5. The next number is the 3 times difference of the past two numbers and add it to the previous number, which is (5 – 2) = 3 and 3 times 3 = 9. So, 9 + 5 = 14. With the same procedure, we can find the next number 41.

14 – 5 = 9

3 times 9 = 27

So, 27+14 = 41.

Let’s find the 6th number of this sequence:

5th number and 4th number is = 122, 41

Difference of this two is: 122 – 41 = 81

So, the 6th number is: 122 + (81*3) = 122 + 243 = 365

### Solution: 02

There is another some solution for this “2 5 14 41 122” sequence:

We can see that in this sequence there is an exponential growth of base 3 which firstly adds with 2. Like:

1th term: 2

2nd term: 2 + (3^1= 3) = 5

3rd term: 5 + (3^2= 9) = 14

4th term: 14 +(3^3= 27) = 41

5th term: 41 + (3^4 = 81) = 122

6th term: 122 (3^5= 243) = 365

And it goes on like this….

### Solution: 03

(x*3) -1

Let’s check it:

1st term: 2

2nd term: (2*3) – 1= 5

3rd term: (5*3) – 1 = 14

4th term: (14*3) -1 = 41

5th term: (41*3) – 1 = 122

6th term: (122*3) -1 = 365

**You can also do it like this:**

Take the first two terms (2 and 5). Then always sum all elements, multiple it by 2 then for nth term, subtract value by (n-3). I think you get it.

1st & 2nd term is 2, 5

3rd term: ((2 + 5) * 2) – 0 = 14

4th term: ((2 + 5 + 14)*2) – 1 = 41

5th term: ((2 + 5 + 14 + 41) * 2) – 2 = 122

6th term: ((2 + 5 + 14 + 41 + 122)×2) – 3 = 365

I think you got it.

Now some may wonder to **Find the sum of the series 2+5+14+41+122+…. up to nth terms.**

Here is simple formula for that:

Sn = ½[ ((3(3^n – 1)) / 2) + n]

Here is a simple C program for finding the nth term number and finding the sum of nth term number for 2 5 14 122… sequence.

#include<stdio.h

#include<math.h>

int main()

{

int a = 2, sum = 2, i, n;

scanf("%d", &n);

for(i=1; i<n; i++)

{

a = a + pow(3, i);

sum += a;

}

printf("%dth term number is: %d ", n, a);

printf("%dth term summation is: %d ", n, sum);

return 0;

}

Sample input:

6

Sample Output:

6th term number is: 365

6th term summation is: 549