What is the 8th term in the sequence?

Note: Use the explicit rule.

an=25−3n

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a8=

Which answer is the explicit rule for the sequence: 13, 10.5, 8, 5.5, 3, 0.5, ...

Aan=15.5−2.5n

Ban=15+2.5n

Can=15−2.5n

Dan=15.5+2.5n

A coin collector has 1044 coins. Each month, the coin collector purchases 9 more coins.

How many coins will the coin collector have after 34 months?

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What is the recursive rule for an=3n+5

Aa1=8;an=an−1+3

Ba1=5;an=an−1+3

Ca1=8;an=an−1+5

Da1=3;an=an−1+5

The asnwer to the following set of questions are as follows:

1. The 8th term of the sequence would be 1.

2. A. an=15.5−2.5n

3. The coin collector will have 1350 coins after 34 months

4. The recursive rule for an=3n+5 would be a1=8;an=an−1+3

I hope my answer has come to your help. God bless and have a nice day ahead!

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flightbath flightbath · Answer 2 · 2018-12-06T02:16:05-05:00

Answer:

Step-by-step explanation:

(1) Using the rule \(a_n=25-3n\)

put n=8 in the above formula we get

\(a_8=25-3(8)=1\)

Hence, \(a_8=1\)

(2) The given sequence has common difference -2.5

common difference that is \(d=a_2-a_1\)

We will use the formula of AP which is

\(a_n=a+(n-1)d\)

Here, a= 13, d=-2.5 and n=n on substituting the values in the formula we get

\(a_n=13+(n-1)(-2.5)\)

\(a_n=15.5-2.5n\)

Hence, Option A

(3)From the given information we can conclude

\(a=1044,d=9,n=34\)

We have to find \(a_n\)

On substituting the values we get

\(a_n=1044+(34-1)(9)\)

\(a_n=1044+297\)

\(a_n=1341\) coins the coin collector will have.

(4) Recursive rule for \(a_n=3n+5\) will be

Put n=1 in the given formula we get

\(a_1=3(1)+5=8\)

When n=2 we get

\(a_2=3(2)+5=11\)

When n=3 we get

\(a_3=3(3)+5=14\)

Hence, option A that is \(a_1=8;a_n=a_{n-1}+3\)