If we see the given series, 3, 4, 6, 9, x, 18, 24…

We find that 4 – 3 = 1, 6 – 4 = 2, so we see that it is not an AP as the difference between consecutive terms is not the same.

It is also not a GP as 4/3 is not equal to 6/4.

Now, if we have a closer look at the numbers we see that the difference between consecutive terms of the series: 4 – 3 = 1, 6 – 4 = 2, 9 – 6 = 3, starts from 1 and it increases by 1. So, we can equate x to 9 + 4 = 13. We see that 13 – 9 = 4, 18 – 13 = 5, 24 – 18 = 6 …

**Therefore x is equal to 13.**

The sequence of terms = 3, 4 , 6, 9, x, 18, 24.

We notice that the succesive increment is not constant but increasing one.

So the nth term an is must have a 2nd degree in n.

Let an = an^2+bn+c.

We now determine a, b and c.

a1 = a+b+c = 3.

a2 = 4a+2b+c = 4.

Therefore a2-a1 =3a+b = 4-3 = 1, or

3a+b = 1.....(1)

Similarly a3 = 9a+3b+c= 6.

So a3-a2 = 5a+ b = 6-4 = 2, or

5a+b = 2................(2).

(2)-(1): 5a-3a = 2-1 = 1.

=> 2a = 1. Or a= 1/2.

We substitute a = 1/2 in (1) : 3(1/2)+b = 1. So b= 1-3/2 = -1/2.

So we substitute a= 1/2 , b= -1/2 in a1 = a+b+c = 3.

=> (1/2-1/2+c = 3. So c= 3.

Therefore an = an^2+bn+c = (1/2)(n^2-n)+3.

So a1 = (1/2)(6^1-1)+3 = 3.

a2 (1/2)(2^2-2)+3 = 4.

a3 = (1/2)(3^3-3)+3 = 6.

a4(=(1/2)(4^2-4)+3 = 9.

a5 = (1/2)(5^2-5) +3 = 13.

a6 = 91/2)(6^2-6)+4 = 18.

a7 = (1/2)(7^2-7) +3 = 24.

Therefore** x = a5 = 13 **in the sequence generated by {an } = {(1/2)n^2-n)+3}**.**