Question

NO LINKS!! Find the 10 terms of the sequence.

a1 = x, d = 3x

a1=
a2=
a3=
a4=
a5=
a6=
a7=
a8=
a9=
a10=

Answer 1

`x, \; 4x, \; 7x, \; 10x, \; 13x, \; 16x, \; 19x, \; 22x, \; 25x, \; 28x, ...`

Explanation:

`\boxed{\begin{minipage}{8 cm}\underline{General form of an arithmetic sequence}\\\\$a_n=a+(n-1)d$\\\\where:\\\phantom{ww}$\bullet$ $a_n$ is the nth term. \\ \phantom{ww}$\bullet$ $a$ is the first term.\\\phantom{ww}$\bullet$ $d$ is the common difference between terms.\\\phantom{ww}$\bullet$ $n$ is the position of the term.\\\end{minipage}}`

Given:

• a = x
• d = 3x

Substitute the given values of a and d into the formula to create an equation for the nth term:

`\implies a_n=x+(n-1)3x`

`\implies a_n=x+3nx-3x`

`\implies a_n=3nx-2x`

To find the first 10 terms of the given arithmetic sequence, substitute n = 1 through 10 into the equation.

`\begin{aligned}\implies a_1&=3(1)x-2x\\&=3x-2x\\&=x\end{aligned}`

`\begin{aligned}\implies a_2&=3(2)x-2x\\&=6x-2x\\&=4x\end{aligned}`

`\begin{aligned}\implies a_3&=3(3)x-2x\\&=9x-2x\\&=7x\end{aligned}`

`\begin{aligned}\implies a_4&=3(4)x-2x\\&=12x-2x\\&=10x\end{aligned}`

`\begin{aligned}\implies a_5&=3(5)x-2x\\&=15x-2x\\&=13x\end{aligned}`

`\begin{aligned}\implies a_6&=3(6)x-2x\\&=18x-2x\\&=16x\end{aligned}`

`\begin{aligned}\implies a_7&=3(7)x-2x\\&=21x-2x\\&=19x\end{aligned}`

`\begin{aligned}\implies a_8&=3(8)x-2x\\&=24x-2x\\&=22x\end{aligned}`

`\begin{aligned}\implies a_9&=3(9)x-2x\\&=27x-2x\\&=25x\end{aligned}`

`\begin{aligned}\implies a_(10)&=3(10)x-2x\\&=30x-2x\\&=28x\end{aligned}`

Therefore, the first 10 terms of the given arithmetic sequence are:

• 
  `x, \; 4x, \; 7x, \; 10x, \; 13x, \; 16x, \; 19x, \; 22x, \; 25x, \; 28x, ...`