Arithmetic sequence.

Question

Consider the numbers $37,44,51,...,177$, which form an arithmetic sequence. A number $n$ is the sum of five distinct numbers from this sequence. How many possible values of $n$ are there?

My attempt,

I assume that the sequence is $1,2,3,4,5,...,21.$ So $1+2+3+4+5=15 $ which is the smallest and $17+18+19+20+21=95$ which is the largest.

So $n=15$~$21$

So there are $81$ possible numbers.

I've checked with the solution, and apparently $81$ is the correct answer. The proposed solution which is the totally not same with mine. Can anyone check with my solution? Thanks in advance.

Answer 1

This is correct, because every term of your sequence can be bijectively mapped using the rule $S=30+7n$ onto the other sequence.

Answer 2

The proposed solution by the author.

The common difference of the given sequence is $7$, and there are $21$ terms in the sequence, since $\frac{177-37}{7}=20$. Let $n$ be a sum of five terms in the sequence; that is $$n=185+7(a+b+c+d+e)$$ where $a,b,c,d$ and $e$ are distinct elements of the set $[0,1,2,...,19,20]$. Let $X=a+b+c+d+e$. Since we want to count the number of different possible values of $n$, it suffices to count the number of possible values of $X$. The least possible value of $X=0+1+2+3+4=10$ and the largest possible value of $X$ is $16+17+18+19+20=90$

Thus, every integer from $10$ to $90$ is a possible value of $X$. SInce $10$ and $90$ are the smallest and largest possible values of $X$, respectively, there are $90-10+1=81$ possible values of $X$ and so $81$ possible value of $n$