Question

Write the first five terms of the sequence defined by the recursive formula

Answer 1

Answer:

The correct answer option is: $S_9=\frac{9}{2} (2+26)$

Step-by-step explanation:

We know that,

the sum of the first $n$ terms of an Arithmetic Sequence is given by:

$S_9=\frac{n(a_1+a_n)}{2}$

where $n$ is the number of terms,

$a_1$ is the first term of the sequence; and

$a_n$ is the first term of the sequence.

So for $a_n=3n-1$,

$a_1=3(1)-1=2$

and

$a_9=3(9)-1=26$

Putting these values in the formula to get:

$S_9=\frac{9(a_1+a_9)}{2}$

$S_9=\frac{9(2+26)}{2} \\\\S_9=\frac{9}{2} (2+26)$

First five terms:

$a_1=3(1)-1=2$

$S_1=\frac{1(2+2)}{2}$=2

$a_2=3(2)-1=5$

$S_2=\frac{2(2+5)}{2}$=7

$a_3=3(3)-1=8$

$S_2=\frac{3(2+8)}{2}$=15

$a_4=3(4)-1=11$

$S_4=\frac{4(2+11)}{2}$=26

$a_5=3(5)-1=14$

$S_5=\frac{5(2+14)}{2}$=40

Answer 2

Answer: the corrrect one is A s9=9/2(2+26)