MHB Divisibility of Terms in an Arithmetic Series

[mathdad]

Arithmetic Series?
Given the arithmetic series 5+14+23+...(to 241 terms), how many terms in the series are divisible by 5?

I need a good explanation and a good start.

 

[I like Serena]

Arithmetic Series?
Given the arithmetic series 5+14+23+...(to 241 terms), how many terms in the series are divisible by 5?

I need a good explanation and a good start.

241 terms is a lot of terms. Let's simplify the problem to, say, 1 term. How many are divisible by 5?
How about if we have 2 terms?
Or 3, 4, 5, 6, 7?
Can we discern a pattern? (Wondering)

 

[skeeter]

Arithmetic Series?
Given the arithmetic series 5+14+23+...(to 241 terms), how many terms in the series are divisible by 5?

I need a good explanation and a good start.

the nth term of an arithmetic series is $a_n = a_1+(n-1) \cdot d$, where $a_1$ is the 1st term and $d$ is the common difference between each consecutive term.

for the given series, $a_n = 5+(n-1) \cdot 9$

if $a_n$ is divisible by $5$, what does that say about the value of $(n-1)$ ?

 

[mathdad]

the nth term of an arithmetic series is $a_n = a_1+(n-1) \cdot d$, where $a_1$ is the 1st term and $d$ is the common difference between each consecutive term.

for the given series, $a_n = 5+(n-1) \cdot 9$

if $a_n$ is divisible by $5$, what does that say about the value of $(n-1)$ ?

I do not understand your question.

 

[Greg]

Each term in the series can be represented by $5+9n,\,0\le n\le240$. In order for a term to be divisible by $5$, $n$ must be divisible by $5$. Hence the number of terms divisible by $5$ must be $\frac{240}{5}+1=49$.

 

[mathdad]

Thank you everyone. Sorry that I could not show much work in this reply.