Series / Sequences: Complete the equation

We start with the right column in the exercises.

Between each number there is a jump of +3:$4+3=7$

$7+3=10$

Etcetera.

Now we move to the left column of the exercises.

Between each number there is a jump of +1:

$2+1=3$

$3+1=4$

Now we can figure out which exercise is missing:

The left digit will be:$2-1=1$

The right digit will be:$4-3=1$

And the missing exercise is:$1+1$

It is apparent, that for each successive number, a square is added in length and one in width.

Hence, the rule using the variable n is:

$a(n)=n^2$

Therefore, the eighth term will be:

$n^2=8\times8=16$

We calculate by replacing$n=11$

$2\times11+2=$

First we solve the multiplication exercise and then we add 2:

$22+2=24$

We know that the first term of the series is 15.

From here we can easily write the entire series, until we see if we reach 1.  

15, 13, 11, 9, 7, 5, 3, 1

The number 1 is indeed an element of the series!

The third term in the sequence is the term $a_3$meaning in the general term formula given:

$a_n= \frac{n}{2}$We need to substitute the position (of the requested term in the sequence):

$n=3$Let's do this:

$a_{\underline{n}}= \frac{\underline{n}}{2} \\ n=\underline{3}\\ \downarrow\\ a_{\underline{3}}=\frac{\underline{3}}{2}$When we substituted in place of n the position (of the requested term in the sequence): 3, the substitution is shown with an underline in the expression above,

Therefore, the correct answer is answer C.

The fourth and fifth terms in the sequence are the terms: $a_4,\hspace{4pt}a_5$meaning in the general term formula given:

$a_n=10n-9$we need to substitute the position (of the requested term in the sequence):

$n=4$$$for - $a_4$and-

$$$n=5$for-

$a_5$ Let's do this for the fourth term:

$a_{\underline{n}}= 10\underline{n}-9 \\ n=\underline{4}\\ \downarrow\\ a_{\underline{4}}= 10\cdot\underline{4}-9=40-9\\ a_4=31$when we substituted in place of n the position (of the requested term in the sequence): 4, substitution is shown with an underline in the expression above,

Similarly, for the fifth term, $a_5$we get:

$a_{\underline{5}}= 10\cdot\underline{5}-9=50-9\\ a_5=41$ which means that:

$a_4=31,\hspace{4pt}a_5=41$Therefore the correct answer is answer A.