High School

8. In an Arithmetic Progression (AP), the third term is -40, and the thirteenth term is 10.

a. Find the first term and the common difference.
b. Find the [tex]30^{\text{th}}[/tex] term of the AP.
c. Which term of the AP is 35?

Answer :

We are given an arithmetic progression (AP) where the third term is given by

[tex]$$
a + 2d = -40
$$[/tex]

and the thirteenth term is given by

[tex]$$
a + 12d = 10,
$$[/tex]

where [tex]$a$[/tex] is the first term and [tex]$d$[/tex] is the common difference.

–––––––––––––––––––––––
Step 1. Find the common difference [tex]$d$[/tex].

Subtract the equation for the third term from that for the thirteenth term:

[tex]$$
\begin{align*}
(a + 12d) - (a + 2d) &= 10 - (-40) \\
12d - 2d &= 10 + 40 \\
10d &= 50 \\
d &= \frac{50}{10} = 5.
\end{align*}
$$[/tex]

–––––––––––––––––––––––
Step 2. Find the first term [tex]$a$[/tex].

Using the equation for the third term:

[tex]$$
\begin{align*}
a + 2d &= -40 \\
a + 2(5) &= -40 \\
a + 10 &= -40 \\
a &= -40 - 10 = -50.
\end{align*}
$$[/tex]

So, the first term is [tex]$a = -50$[/tex] and the common difference is [tex]$d = 5$[/tex].

–––––––––––––––––––––––
Step 3. Find the 30th term of the AP.

The [tex]$n^{\text{th}}$[/tex] term of an AP is given by

[tex]$$
a_n = a + (n - 1)d.
$$[/tex]

For the 30th term:

[tex]$$
\begin{align*}
a_{30} &= a + 29d \\
&= -50 + 29(5) \\
&= -50 + 145 \\
&= 95.
\end{align*}
$$[/tex]

–––––––––––––––––––––––
Step 4. Determine which term of the AP is [tex]$35$[/tex].

We set up the equation:

[tex]$$
a + (n - 1)d = 35.
$$[/tex]

Substitute [tex]$a = -50$[/tex] and [tex]$d = 5$[/tex]:

[tex]$$
\begin{align*}
-50 + (n - 1)5 &= 35 \\
(n - 1)5 &= 35 + 50 \\
(n - 1)5 &= 85 \\
n - 1 &= \frac{85}{5} = 17 \\
n &= 17 + 1 = 18.
\end{align*}
$$[/tex]

So, [tex]$35$[/tex] is the [tex]$18^\text{th}$[/tex] term of the AP.

–––––––––––––––––––––––
Final Answers:

a. The first term is [tex]$-50$[/tex] and the common difference is [tex]$5$[/tex].
b. The [tex]$30^{\text{th}}$[/tex] term is [tex]$95$[/tex].
c. The term equal to [tex]$35$[/tex] is the [tex]$18^{\text{th}}$[/tex] term.