Certainly! Let's tackle this problem step-by-step.
Given:
- 3rd term ([tex]\(a_3\)[/tex]) = 40
- 13th term ([tex]\(a_{13}\)[/tex]) = 0
In an arithmetic progression (A.P.), the nth term can be expressed as:
[tex]\[a_n = a + (n - 1)d\][/tex]
where:
- [tex]\(a\)[/tex] is the first term,
- [tex]\(d\)[/tex] is the common difference.
From the given information, we can set up the following equations:
1. For the 3rd term:
[tex]\[a + 2d = 40\][/tex]
2. For the 13th term:
[tex]\[a + 12d = 0\][/tex]
To find the common difference ([tex]\(d\)[/tex]), we can subtract the first equation from the second equation:
[tex]\[
(a + 12d) - (a + 2d) = 0 - 40
\][/tex]
This simplifies to:
[tex]\[
10d = -40
\][/tex]
Solving for [tex]\(d\)[/tex]:
[tex]\[
d = \frac{-40}{10} = -4
\][/tex]
Now, we substitute [tex]\(d = -4\)[/tex] back into the first equation to find the first term ([tex]\(a\)[/tex]):
[tex]\[
a + 2(-4) = 40
\][/tex]
[tex]\[
a - 8 = 40
\][/tex]
[tex]\[
a = 48
\][/tex]
With [tex]\(a = 48\)[/tex] and [tex]\(d = -4\)[/tex], we want to find which term of the A.P. is 28. Let this term be the [tex]\(n\)[/tex]-th term.
Using the nth term formula again:
[tex]\[
a_n = a + (n - 1)d
\][/tex]
and substituting [tex]\(a_n = 28\)[/tex]:
[tex]\[
28 = 48 + (n - 1)(-4)
\][/tex]
Simplifying the equation:
[tex]\[
28 = 48 - 4n + 4
\][/tex]
[tex]\[
28 = 52 - 4n
\][/tex]
Rearranging to solve for [tex]\(n\)[/tex]:
[tex]\[
28 - 52 = -4n
\][/tex]
[tex]\[
-24 = -4n
\][/tex]
[tex]\[
n = \frac{-24}{-4} = 6
\][/tex]
So, the term of the arithmetic progression that equals 28 is the 6th term.
Final answer:
The 6th term of the arithmetic progression is 28.
