All categories

3 years ago

Find the 64th term of the following arithmetic sequence. 17, 26, 35, 44, ...

Mathematics

1 answer:

goblinko [34]3 years ago

8 0

take the difference

26-17=9

35-26=9

the first term is 17

and the nth term is 64

use the formula

tn=a+(n-1) d

let a be 17

let n be 64

let d be 9

then you substitute

t64=17+(64-1)9

=17+(63)9

=17+567

=584

so the 64th term is 584.

You might be interested in

Can anybody help me with my schoolwork right now?

denis23 [38]

Sure, what is the question.

6 0

2 years ago

What is the probability of p(2 of spades)

serious [3.7K]

Answer:

<h2><u><em>**Please rate brainliest and vote 5 and give thanks**</em></u></h2>

Step-by-step explanation:

● A standard deck of cards has:

● 52 Cards in 13 values and 4 suits

● Suits are Spades, Clubs, Diamonds

and Hearts

● Each suit has 13 card values:

2-10, 3 “face cards” Jack, Queen, King (J, Q, K)

and and Ace (A)

Basic Card Probabilities

● If you draw a card at random, what is the

probability you get:

● A Spade? P(Spade)=13/52

● A Face card? P(Face Card)=12/52 (or simply 3/13)

● A Red Ace? P(Red Ace) = 2/52

6 0

2 years ago

Jennifer is allowed to watch no more than

Vedmedyk [2.9K]

Answer:

Step-by-step explanation:

10-4=6

5 0

2 years ago

6. MONEY Len bought a $24.99 pair of pants

tatuchka [14]

Answer:

$2.06 tax

Step-by-step explanation:

$24.99 pants

$27.05 total cost with tax

24.99 + tax = 27.05

tax = 27.05 - 24.99

27.05 - 24.99 = 2.06

7 0

3 years ago

For each of the following sequences, • Give a formula for the nth term in the sequence, • Give a recursive definition for the se

Andrew [12]

Answer:

(a) $n^{th} = n$

$f(1) = 1$

$f(n) = f(n-1) + 1$

(b) $n^{th} = 2^{n-1}$

$f(1) = 1$

$f(n) = f(n-1) * 2$

$f(1) = 1$

$f(n) = f(n-1) * n$

Step-by-step explanation:

(a) This is a sequence of consecutive number

$n^{th} = n$

$f(1) = 1$

$f(n) = f(n-1) + 1$

(b) This is a sequence of 2 to the power of n - 1. The next number is twice time of this number

$n^{th} = 2^{n-1}$

$f(1) = 1$

$f(n) = f(n-1) * 2$

$n^{th} = n!$

$f(1) = 1$

$f(n) = f(n-1) * n$

5 0

2 years ago