Question

2. Find the missing term in the series given below. 2, 10, 24, ?, 70, 102​

Answer 1

Step-by-step explanation:

Consider the arithmetic progression 2, 8, 14, 20, 26. The series provided is the sum of this progression; thus, one could look up the formula or derive it; the n’th term of the arithmetic progression is 2+(n-1)6; there are n/2 pairs; thus, noting that one can pair 1st and n’th, 2nd and n-1’th, …, the sum of arithmetic progression is

n/2 (4+6(n-1)) = n/2(6n-2) = 3n^2 - n

I then show this generates the first five terms

For n = 1, 3*1^2 - 1 = 3 - 1 = 2

For n = 2, 3*2^2 - 2 = 3*4 - 2 = 12 - 2 = 10

For n = 3, 3*3^2 - 3 = 3*9 - 3 = 27 - 3 = 24

For n = 4, 3*4^2 - 4 = 3*16 - 4 = 48 - 4 = 44

For n = 5, 3*5^2 - 5 = 3*25 - 5 = 75 - 5 = 70

The formula 3 n^2 - n has successfully generated the first 5 terms.

Answer 2

Answer:

The missing term in the series $2, 10, 24, ?, 70, 102$ is 44.

Step-by-step explanation:

Let us consider the arithmetic progression $2, 8, 14, 20, 26$. The series provided is the sum of this progression. Now we could look up the formula or derive it; the $n^{th}$ term of the arithmetic progression is $2+(n-1)6$; there are $n/2$ pairs; thus, noting that one can pair $1st$ and $n^{th}$, $2nd$ and $(n-1)^{th$, …

The sum of an arithmetic progression is"

$\frac{n}{2} (4+6(n-1)) = \frac{n}{2} (6n-2) = 3n^2 - n$

Now we will show this generates the first five terms:

For $n = 1, 3\times1^2 - 1 = 3 - 1 = 2$

For $n = 2, 3\times2^2 - 2 = 3\times4 - 2 = 12 - 2 = 10$

For $n = 3, 3\times3^2 - 3 = 3\times9 - 3 = 27 - 3 = 24$

For $\bold{n = 4, 3\times4^2 - 4 = 3\times16 - 4 = 48 - 4 = 44}$

For $n = 5, 3\times5^2 - 5 = 3\times25 - 5 = 75 - 5 = 70$

Therefore, the missing number is 44.