Question

Find an equation for the nth term of the arithmetic sequence.
a10 = 32, a12 = 106

an = -301 + 37(n - 1)
an = -301 + 37(n - 2)
an = -301 + 37(n + 1)
an = -301 - 37(n + 1)

Answer 1

Answer:

$a_n=-301 +37(n-1)$

Step-by-step explanation:

arithmetic sequence formula:  $a_n=a +(n-1)d$

where $a$ is the first term and $d$ is the common difference

Given:

  $a_{10}=32$

⇒ $a +(10-1)d=32$

⇒ $a+9d=32$

Given:

    $a_{12}=106$

⇒ $a +(12-1)d=106$

⇒ $a+11d=106$

Rearrange the first equation to make $a$ the subject:

a = 32 - 9d

Now substitute into the second equation and solve for $d$

    (32 - 9d) + 11d = 106

⇒ 32 + 2d = 106

⇒ 2d = 106 - 32 = 74

⇒ d = 74 ÷ 2 = 37

Substitute found value of $d$ into the first equation and solve for $a$:

a + (9 x 37) = 32

a + 333 = 32

a = 32 - 333 = -301

Therefore, the equation is: $a_n=-301 +37(n-1)$