# Simple arithmetic progression question

• Dec 23rd 2012, 06:27 AM
kinhew93
Simple arithmetic progression question
I have the answer to this but I'm not sure that I entirely understand it.

Find the number of terms in the AP: 7 + 9 +...+ (2n+1)

My solution:

The ath term will be 2a + 5 so 2a + 5 = 2n + 1 therefore a = n - 2

This gives me the right answer but what I don't understand is what the difference between n and a is? Initially I thought that they were the same value but of course that is not possible. I would normally think of the 'ath term' as the 'nth term' so is (2n+1) just an expression that happens to use n as a variable just to be awkward?

I would really appreciate it if somebody could clarify this for me.

Merry Christmas!
• Dec 23rd 2012, 08:04 AM
HallsofIvy
Re: Simple arithmetic progression question
"n" is a specific integer value while you "a" varies from 1 up to n- 2 and so counts the number of terms.

That is, if n is, say 10, then 7, 9, 11, ..., 2n+ 1 is 7, 9, 11, 13, 15, 17, 19, 21 while a is 1, 2, 3, 5, 5, 7, 8.
• Dec 23rd 2012, 10:48 AM
kinhew93
Re: Simple arithmetic progression question
Ok well I normally think of the 'nth' term as the term that is found in a particular place in the sequence/series ie when n=10 the 10th term in the series is 25.

In this case is n not related to this and just a fixed value? Sorry but I'm still a bit confused...
• Dec 23rd 2012, 11:06 AM
Plato
Re: Simple arithmetic progression question
Quote:

Originally Posted by kinhew93
Ok well I normally think of the 'nth' term as the term that is found in a particular place in the sequence/series ie when n=10 the 10th term in the series is 25. In this case is n not related to this and just a fixed value? Sorry but I'm still a bit confused...

I think that the confusion comes from the fact you are using \$\displaystyle n\$ in two different ways. Let's look at the OP.
Quote:

Originally Posted by kinhew93
I have the answer to this but I'm not sure that I entirely understand it. Find the number of terms in the AP: 7 + 9 +...+ (2n+1)

You are to find "the number of terms" as a function of \$\displaystyle n\$ and that is not the same as the \$\displaystyle n^{th}\$ term.

Actually you did the problem correctly. There are \$\displaystyle n-2\$ terms in that A.P.

There are \$\displaystyle 8=10-2\$ terms in \$\displaystyle 7+9+\cdots+(21=2[10]+1)\$.
• Dec 23rd 2012, 11:15 AM
emakarov
Re: Simple arithmetic progression question
Would it be more natural if the problem asked, "Find the number \$\displaystyle n\$ of terms in the AP: 7 + 9 +...+ (2\$\displaystyle a\$+1) "? Possibly, but this formulation is exactly equivalent to "Find the number \$\displaystyle a\$ of terms in the AP: 7 + 9 +...+ (2\$\displaystyle n\$+1)." Both variants say, "Fix some number and call it \$\displaystyle a\$ in the first variant and \$\displaystyle n\$ in the second variant. Consider the arithmetic progression 7 + 9 +...+ (2\$\displaystyle a\$+1) or, respectively, 7 + 9 +...+ (2\$\displaystyle n\$+1) and find the number of terms as a function of \$\displaystyle a\$, or, respectively, \$\displaystyle n\$."

The variable \$\displaystyle n\$ does not have a monopoly on denoting the index of a sequence. Any variable can be used for this. Similarly, even in a text talking about a sequence of numbers, \$\displaystyle n\$ can denote many different things. In this case, it denotes some number such that \$\displaystyle 2n + 1\$ is the last element of the sequence.
• Dec 23rd 2012, 10:12 PM
ibdutt
Re: Simple arithmetic progression question
The AP is: 7,9,11,…… ( 2n+1)
Ok Let us presume that the AP has ‘m’ terms thus we have for the given AP
First term a1 = 7 common difference d = 2, and last term i.e., mth term = 2n + 1
But am = a1 + ( m-1) d
= 7 + ( m-1) x 2 = 7 + 2m – 2 = 2m + 5
It is given that the last term am = 2n + 1
Thus we have 2m + 5 = 2n + 1
That give m = n -2
Hence the AP has n-2 terms.