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Given the formula: $A_n = A_1+ (n-1)d$

I'm trying to look for $d$, if given the the second and $17^{\text{th}} $ terms, namely $37$ and $82$.

I can't seem to figure out where to start; if $A_1$ were at least given I might have a starting point, but here I'm completely lost now that both are missing.

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  • $\begingroup$ Find $A(17)-A(2)$ which should be a computable multiple of $d.$ $\endgroup$ – coffeemath Jan 27 at 13:51
  • $\begingroup$ How many terms are there from $A_2$ to $A_{17}$? $\endgroup$ – KM101 Jan 27 at 13:51
  • $\begingroup$ @KM101 Starting at $A(2),$ the common difference $d$ should be added $15$ times to get $A(17.) $\endgroup$ – coffeemath Jan 27 at 14:00
  • $\begingroup$ @coffeemath For $n$ terms, the difference is added $(n-1)$ times, I know. This is equivalent to finding $d$ given the first and sixteenth terms. $\endgroup$ – KM101 Jan 27 at 14:08
  • $\begingroup$ Then how many d added from second to seventeenth? $\endgroup$ – coffeemath Jan 27 at 14:12
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Hint: $$a_{17} - a_2 = 45 = 16d - d = 15d$$

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  • $\begingroup$ A(17) - A(2) = A(15) -> ( *A(17) - *A(2) ) / 15 = 3 = d. Giving an A(1) of 34. I got it solved shortly after posting. I think I got thrown off by online texts saying A starts at A(0) as opposed to my A(1) start. Thanks. $\endgroup$ – Arvayne Jan 27 at 14:15
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The $n$th term of an arithmetic sequence is

$$a+(n-1)d$$ where $a$ is the first term and $d$ is the common difference between each term.

You have:

$$a+16d=82\tag1$$ $$a+d=37\tag2$$ When we perform $(1)-(2)$, what do we get?

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