Answer :
To solve the problem, we need to determine the general structure of the arithmetic progression (AP) and the specific term where the value is 28.
Let's denote:
- The 3rd term of the AP as [tex]\( a_3 = 40 \)[/tex] (assuming "Fo" corresponds to 40).
- The 13th term of the AP as [tex]\( a_{13} = 0 \)[/tex].
The general formula for the [tex]\(n\)[/tex]-th term of an AP is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
where
- [tex]\(a\)[/tex] is the first term,
- [tex]\(d\)[/tex] is the common difference.
First, write down the equations for the 3rd and 13th terms using the general formula:
[tex]\[ a_3 = a + 2d = 40 \][/tex]
[tex]\[ a_{13} = a + 12d = 0 \][/tex]
Now, let's solve these two equations simultaneously to find [tex]\(a\)[/tex] and [tex]\(d\)[/tex].
### Step 1: Subtract the first equation from the second
[tex]\[ (a + 12d) - (a + 2d) = 0 - 40 \][/tex]
[tex]\[ a + 12d - a - 2d = -40 \][/tex]
[tex]\[ 10d = -40 \][/tex]
[tex]\[ d = -4 \][/tex]
### Step 2: Substitute [tex]\(d\)[/tex] back into one of the original equations to find [tex]\(a\)[/tex]
Using the equation [tex]\(a + 2d = 40\)[/tex]:
[tex]\[ a + 2(-4) = 40 \][/tex]
[tex]\[ a - 8 = 40 \][/tex]
[tex]\[ a = 48 \][/tex]
Now we know the first term [tex]\(a = 48\)[/tex] and the common difference [tex]\(d = -4\)[/tex].
### Step 3: Determine the term number where the term value is 28
We need to find [tex]\(n\)[/tex] such that [tex]\(a_n = 28\)[/tex]:
[tex]\[ 28 = a + (n-1)d \][/tex]
Substitute the values of [tex]\(a\)[/tex] and [tex]\(d\)[/tex]:
[tex]\[ 28 = 48 + (n-1)(-4) \][/tex]
[tex]\[ 28 = 48 - 4(n-1) \][/tex]
[tex]\[ 28 - 48 = -4(n-1) \][/tex]
[tex]\[ -20 = -4(n-1) \][/tex]
[tex]\[ 5 = n-1 \][/tex]
[tex]\[ n = 6 \][/tex]
Therefore, the term of the AP that is 28 is the 6th term.
Let's denote:
- The 3rd term of the AP as [tex]\( a_3 = 40 \)[/tex] (assuming "Fo" corresponds to 40).
- The 13th term of the AP as [tex]\( a_{13} = 0 \)[/tex].
The general formula for the [tex]\(n\)[/tex]-th term of an AP is given by:
[tex]\[ a_n = a + (n-1)d \][/tex]
where
- [tex]\(a\)[/tex] is the first term,
- [tex]\(d\)[/tex] is the common difference.
First, write down the equations for the 3rd and 13th terms using the general formula:
[tex]\[ a_3 = a + 2d = 40 \][/tex]
[tex]\[ a_{13} = a + 12d = 0 \][/tex]
Now, let's solve these two equations simultaneously to find [tex]\(a\)[/tex] and [tex]\(d\)[/tex].
### Step 1: Subtract the first equation from the second
[tex]\[ (a + 12d) - (a + 2d) = 0 - 40 \][/tex]
[tex]\[ a + 12d - a - 2d = -40 \][/tex]
[tex]\[ 10d = -40 \][/tex]
[tex]\[ d = -4 \][/tex]
### Step 2: Substitute [tex]\(d\)[/tex] back into one of the original equations to find [tex]\(a\)[/tex]
Using the equation [tex]\(a + 2d = 40\)[/tex]:
[tex]\[ a + 2(-4) = 40 \][/tex]
[tex]\[ a - 8 = 40 \][/tex]
[tex]\[ a = 48 \][/tex]
Now we know the first term [tex]\(a = 48\)[/tex] and the common difference [tex]\(d = -4\)[/tex].
### Step 3: Determine the term number where the term value is 28
We need to find [tex]\(n\)[/tex] such that [tex]\(a_n = 28\)[/tex]:
[tex]\[ 28 = a + (n-1)d \][/tex]
Substitute the values of [tex]\(a\)[/tex] and [tex]\(d\)[/tex]:
[tex]\[ 28 = 48 + (n-1)(-4) \][/tex]
[tex]\[ 28 = 48 - 4(n-1) \][/tex]
[tex]\[ 28 - 48 = -4(n-1) \][/tex]
[tex]\[ -20 = -4(n-1) \][/tex]
[tex]\[ 5 = n-1 \][/tex]
[tex]\[ n = 6 \][/tex]
Therefore, the term of the AP that is 28 is the 6th term.
