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

Jul 20, 2017

${n}^{t h}$ term ${a}_{n} = 14 n - 40$

#### Explanation:

${n}^{t h}$ term ${a}_{n}$ of an arithmetic sequence, whose first term is ${a}_{1}$ and common difference is $d$ is given by

${a}_{n} = {a}_{1} + \left(n - 1\right) d$

Hence ${a}_{20} = {a}_{1} + 19 d = 240$ ..........(A)

and ${a}_{15} = {a}_{1} + 14 d = 170$ ..........(B)

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

and hence putting this in (B), we get

${a}_{1} + 14 \times 14 = 170$

or ${a}_{1} = 170 - 14 \times 14 = 170 - 196 = - 26$

Hence ${a}_{n} = - 26 + \left(n - 1\right) \times 14$

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

Jul 20, 2017

${a}_{\text{n}} = - 26 + 14 \left(n - 1\right)$

#### Explanation:

ω=(a_20-a_15)/5=(240-170)/5=70/5=14

a_1=a_15-14ω=170-196=-26

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

a_"n"=a_1+(n-1)ω=-26+14(n-1)

To check out answer let's calculate ${a}_{20}$

${a}_{20} = - 26 + 14 \cdot 19 = - 26 + 266 = 240$

Which is true.