# How do you write the nth term rule for the arithmetic sequence with a_7=34 and a_18=122?

Jul 31, 2016

${n}^{t h}$ term of the arithmetic sequence is $8 n - 22$.

#### Explanation:

${n}^{t h}$ term of an arithmetic sequence whose first term is ${a}_{1}$ and common difference is $d$ is ${a}_{1} + \left(n - 1\right) d$.

Hence ${a}_{7} = {a}_{1} + \left(7 - 1\right) \times d = 34$ i.e. ${a}_{1} + 6 d = 34$

and ${a}_{18} = {a}_{1} + \left(18 - 1\right) \times d = 122$ i.e. ${a}_{1} + 17 d = 122$

Subtracting firt equation from second equation, we get

$11 d = 122 - 34 = 88$ or $d = \frac{88}{11} = 8$

Hence ${a}_{1} + 6 \times 8 = 34$ or ${a}_{1} = 34 - 48 = - 14$

Hence ${n}^{t h}$ term of the arithmetic sequence is $- 14 + \left(n - 1\right) \times 8$ or $- 14 + 8 n - 8 = 8 n - 22$.

$\textcolor{b l u e}{{a}_{n} = 8 n - 22}$

#### Explanation:

The given data are

${a}_{7} = 34$ and ${a}_{18} = 122$

We can set up 2 equations

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

${a}_{7} = {a}_{1} + \left(7 - 1\right) \cdot d$

$34 = {a}_{1} + 6 \cdot d \text{ }$first equation
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
${a}_{n} = {a}_{1} + \left(n - 1\right) \cdot d$

${a}_{18} = {a}_{1} + \left(18 - 1\right) \cdot d$

$122 = {a}_{1} + 17 \cdot d \text{ }$second equation

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

By method of elimination using subtraction, let us use first and second equations

$34 = {a}_{1} + 6 \cdot d \text{ }$first equation

$122 = {a}_{1} + 17 \cdot d \text{ }$second equation

By subtraction, we have the result

$88 = 0 + 11 d$

$d = \frac{88}{11} = 8$

Solving now for ${a}_{1}$ using the first equation and $d = 8$

$34 = {a}_{1} + 6 \cdot d \text{ }$first equation

$34 = {a}_{1} + 6 \cdot 8 \text{ }$

$34 = {a}_{1} + 48$

${a}_{1} = - 14$

We can write the $n t h$ term rule now

#a_n=-14+8*(n-1)

${a}_{n} = - 14 - 8 + 8 n$

$\textcolor{b l u e}{{a}_{n} = 8 n - 22}$

God bless....I hope the explanation is useful.