3 years ago

Find the 30th term of an arithmetic sequence whose first term is 2 and common difference is 5.

Mathematics

2 answers:

Veronika [31]3 years ago

3 0

2 + 29 times 5 = 147

Alina [70]3 years ago

3 0

First, create a formula for the arithmetic sequence.
a(n) = a(1) + d(n - 1)
a(n) = 2 + 5(n - 1)
a(30) = 2 + 5(29)
a(30) = 2 + 145
a(30) = 147

Hope that helps!

You might be interested in

What is the length of segment A (2,1) B (6,6)

stepladder [879]

4.7 because if you plug it in- (2,1) and (6,6) it is 4.7 units

4 0

3 years ago

Read 2 more answers

Two angles are supplementary. If 5 times of one angle is 10 times of the other angle, find the two angles.

Luba_88 [7]

5x + 10x = 180
15x = 180
X = 12
5 x 12 = 60
10 x 12 = 120

The angles are 60° and 120°

7 0

3 years ago

Daniel baked 12 pies in 6 days Rates

bekas [8.4K]

That’s too much Pie and it’s waste a lot of money yooo and why would he do that dang wow :(

4 0

3 years ago

Find f(-3) if f(x) =x^2

Gnom [1K]

The answer is 9.
Hope I helped

8 0

3 years ago

Read 2 more answers

An arithmetic sequence is defined by the recursive formula t1 = 11, tn = tn - 1 - 13, where n ∈N and n > 1. Which of these is

galben [10]

coolio

$t_1=11$

$t_n=t_{n-1}-13$

so each term is ound by subtracting 13 from the previous term

an aritmetic sequence can be written as

$t_n=t_1+d(n-1)$ were

$t_n$ is the nth term

$t_1$ is the first term

d is common difference, which can also be found by doing $t_n-t_{n-1}=d$

n=wich term

we know that $t_1=11$ and we can find d

$t_n=t_{n-1}-13$ , $t_n-t_{n-1}=-13=d$

so te general term is $t_n=11-13(n-1)$ which can also be expanded and written as $t_n=-13n+24$

6 0

4 years ago