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The [tex]$3^{\text{rd}}$[/tex] and [tex]$13^{\text{th}}$[/tex] terms of an A.P. are 40 and 0, respectively. Which term of the A.P. is 28?

Sagot :

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.
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