# How do I find the nth term of the arithmetic sequence whose initial term and common difference d are given below? (A) a_1=4, d=3 (B) a_1= -8, d=5

Jun 27, 2018

(A)$\text{ } {n}^{t h} = 3 n + 1$

(B)$\text{ } {n}^{t h} = 5 n - 13$

#### Explanation:

the ${n}^{t h}$term of an Ap is given by

${n}^{t h} = a + \left(n - 1\right) d$

$\left(A\right)$

${n}^{t h} = 4 + \left(n - 1\right) 3$

${n}^{t h} = 4 + 3 n - 3$

$\therefore {n}^{t h} = 3 n + 1$

(B)

${n}^{t h} = - 8 + \left(n - 1\right) 5$

${n}^{t h} = - 8 + 5 n - 5$

$\therefore {n}^{t h} = 5 n - 13$

Jun 27, 2018

a) $\textcolor{w h i t e}{\text{d}} {a}_{n} = + 4 + 3 \left(n - 1\right)$

b) $\textcolor{w h i t e}{\text{d}} {a}_{n} = - 8 + 5 \left(n - 1\right)$

#### Explanation:

$\textcolor{b l u e}{\text{Solution to part A ( in detail )}}$

Given: ${a}_{1} = 4 \mathmr{and} d = 3$

I am doing it this way to get you used to the way sequences are written. I will be building up the the ${n}^{\text{th}}$ term

Let the position count in the sequence be $i$

i=1->a_i->color(white)("dd")a_1=4+ 0d larr" First term"#
$i = 2 \to {a}_{i} \to \textcolor{w h i t e}{\text{dd}} {a}_{2} = 4 + 1 d$
$i = 3 \to {a}_{i} \to \textcolor{w h i t e}{\text{dd}} {a}_{3} = 4 + 2 d$
$i = 4 \to {a}_{i} \to \textcolor{w h i t e}{\text{dd}} {a}_{4} = 4 + 3 d$
$i = 5 \to {a}_{i} \to \textcolor{w h i t e}{\text{dd}} {a}_{5} = 4 + 4 d$

Looking at the behaviour we have:

Set $d = 3 \mathmr{and} i = n$ giving:

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

${a}_{n} = 4 + \left(n - 1\right) 3$

Writing this as required by convention:

${a}_{n} = 4 + 3 \left(n - 1\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Solution to part B - using the above}}$

Set ${a}_{1} = - 8 , d = 5 \mathmr{and} i = n$ giving:

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

${a}_{n} = - 8 + \left(n - 1\right) 5$

${a}_{n} = - 8 + 5 \left(n - 1\right)$