# How do you write a rule for the nth term of the arithmetic sequence and then find a_10 for d=5, a_5=33?

Apr 23, 2017

${a}_{n} = 13 + 5 \cdot \left(n - 1\right)$
${a}_{10} = 58$

#### Explanation:

The general form of ${n}^{t h}$ term of an AP is given as:-

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

In this case, $d = 5$
$\therefore {a}_{n} = {a}_{1} + 5 \cdot \left(n - 1\right)$

${a}_{5} = {a}_{1} + 5 \cdot \left(5 - 1\right) = {a}_{1} + 5 \cdot 4 = {a}_{1} + 20$

But, ${a}_{5} = 33$

$\therefore 33 = {a}_{1} + 20$
$\implies {a}_{1} = 33 - 20 = 13$

Hence ${a}_{n} = 13 + 5 \cdot \left(n - 1\right)$

${a}_{10} = 13 + 5 \cdot \left(10 - 1\right) = 13 + 5 \cdot 9 = 13 + 45 = 58$