Question

Identify the 42nd term of an arithmetic sequence where a1 = -12 and a27 = 66.

70
72
111
114

Answer 1

\(\large\color{black}{ a_1=-12 }\) and \(\large\color{black}{ a_{27}=-12 }\), So we can say that

\(\large\color{black}{ a_{27}=a_1+d(26) }\) Plug in your given values of a1, and a27.

\(\large\color{black}{ 66=-12+d(26) }\)

can you solve for d ?

Answer 2

yes ok thanks can you help me with another one? i'll post it

Answer 3

Hold on, not yet... you need 42nd term.

Answer 4

oh yeah that's right...

Answer 5

First, tell me what ` d ` equals.

Answer 6

3

Answer 7

Yes

Answer 8

So after you know that the difference is 3, and a1 is (-12) you can find a42.

You can use a1,
\(\large\color{black}{ a_{42}=a_{1}+d(42-1) }\) (knowing that a1=-12)


OR, you can use a27,
\(\large\color{black}{ a_{42}=a_{27}+d(42-27) }\) (knwoing that a27=66)

Answer 9

ok so I can use either one?

Answer 10

Oh yeah :)

Answer 11

thxx! ;)

Answer 12

tell me what \(\large\color{red}{ a_{42} }\) equals and we are good for this question.

Answer 13

give meh just a sec..

Answer 14

111

Answer 15

Using \(\large\color{red}{ a_{1} }\)

\(\large\color{black}{ a_{42}=a_{1}+d(42-1) }\)
\(\large\color{black}{ a_{42}=-12+3(41) }\)
\(\large\color{black}{ a_{42}=~~\color\red{??} }\)

──────────────────────────

Using \(\large\color{green}{ a_{27} }\)

\(\large\color{black}{ a_{42}=a_{27}+d(42-27) }\)
\(\large\color{black}{ a_{42}=66+d(15) }\)
\(\large\color{black}{ a_{42}=~~\color\green{??} }\)

Answer 16

Yes !

Answer 17

Awww thanks for writing all that out, such a help!!! :). Give me like 4 min then I'll post my other ? :)

Answer 18

Sure :)