How do you write a rule for the nth term of the arithmetic sequence given $a_{20} = 240, a_{15} = 170$ $a_20=240, a_15=170$ ?

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Answer 1 · 2017-07-20T06:43:36

$n^{t h}$ $n^(th)$ term $a_{n} = 14 n - 40$ $a_n=14n-40$

$n^{t h}$ $n^(th)$ term $a_{n}$ $a_n$ of an arithmetic sequence, whose first term is $a_{1}$ $a_1$ and common difference is $d$ $d$ is given by

$a_{n} = a_{1} + (n - 1) d$ $a_n=a_1+(n-1)d$

Hence $a_{20} = a_{1} + 19 d = 240$ $a_20=a_1+19d=240$ ..........(A)

and $a_{15} = a_{1} + 14 d = 170$ $a_15=a_1+14d=170$ ..........(B)

Subtracting (B) from (A), $5 d = 70$ $5d=70$ i.e. $d = \frac{70}{5} = 14$ $d=70/5=14$

and hence putting this in (B), we get

$a_{1} + 14 \times 14 = 170$ $a_1+14xx14=170$

or $a_{1} = 170 - 14 \times 14 = 170 - 196 = - 26$ $a_1=170-14xx14=170-196=-26$

Hence $a_{n} = - 26 + (n - 1) \times 14$ $a_n=-26+(n-1)xx14$

$= - 26 + 14 n - 14 = 14 n - 40$ $=-26+14n-14=14n-40$

Answer 2 · 2017-07-20T06:45:07

$a_{n} = - 26 + 14 (n - 1)$ $a_"n"=-26+14(n-1)$

$ω=(a_20-a_15)/5=(240-170)/5=70/5=14$

$a_1=a_15-14ω=170-196=-26$

So for the $n^{th}$ term we know :

$a_"n"=a_1+(n-1)ω=-26+14(n-1)$

To check out answer let's calculate $a_20$

$a_20=-26+14*19=-26+266=240$

Which is true.

How do you write a rule for the nth term of the arithmetic sequence given a_20=240, a_15=170a20=240,a15=170a_20=240, a_15=170?