Question

write an explicit formula for the sequence 7,2,-3,-8,-13,...then find a14

Answer 1

\[\Large a_n=a_1+(n-1)d\]

Answer 2

a(n)=nth term
a(1)=first term of the sequence
d=common difference

Answer 3

you are already given the 1st term which is .......?
you can find the common difference by subtracting the 2nd term from the 1st term OR by subtracting the 3rd term from the 2nd term.....
after you have found those, plug them in the general formula for arithmetic sequences i gave above...

Answer 4

$$7\to2\to-3\to-8\to-13\to\dots$$Notice each time it decreases by \(5\), i.e. \(7-5=2,2-5=-3,-3-5=-8,-8-5=-13\). So we can write a recurrence relation relating a term to the one right before it:$$a_n=a_{n-1}-5$$Now, what if we have \(a_{n-2}\) and want to find \(a_n\)? Simple -- we just subtract \(5\) *twice*. In fact, we can figure out any term \(a_n\) in our sequence by subtracting \(5\) a total of \(n-1\) times -- or, equivalently, subtracting \(5(n-1)\) -- from our first term \(a_1=7\).$$a_n=7-5(n-1)=7-5n+5=12-5n$$

Answer 5

its 5n - 12

Answer 6

how do I find a14? that's what im having trouble with @perurabo but thanks everyone !

Answer 7

just substitute

Answer 8

@teebabyy $$a_n=12-5n\\a_{14}=12-5(14)=12-70=-58$$

Answer 9

thanks for your help guys!