Act.al.4 What is the 50th term

[karush]

$\tiny{act.al.4}$
The 3rd and 4th terms of an arithmetic sequence are 13 and 18, respectively. What is the 50th term of the sequence?
$a.\ {248}\quad b.\ {250}\quad c.\ {253}\quad d.\ {258}\quad e.\ {263}$

ok according to Sullivan's textbook
$a_1=a\quad a_n=a_{n-1}+d$
so $d=5$ and $a_1=3$
and the nth term is
$a_n = a_1 + (n -1)d$
then
$a_50= 3+ (50 -1)5=$ 248

 

[HOI]

Yes, since two consecutive terms are 13 and 18, d, the "common difference", is 18- 13= 5. Calling the first term $a_1$, the second difference is $a_2= a_1+ 5$ and the third difference is $a_1+ 10= 13$ so that $a_1= 13- 10= 3$. Well done!

And then the 50th term is the first term, $a_1= 3$, plus d= 5, added 50- 1= 49 times. 49 times 5= 245. The 50th term is 3+ 245= 248, exactly what have! Excellent!

 

[karush]

mahalo,

I assume $a_1$ never goes negative

the hard part is trying to remember these formulas

 

[HOI]

DON'T memorize formulas, memorize definitions! Everything I wrote follows directly from the definition of "arithmetic sequence".