Answer :
To solve the problem, we need to find the terms of the arithmetic progression (AP) and the sum of all its terms, given some conditions.
Step-by-step Solution:
Understanding the Given Information:
- The sum of the first two terms: [tex]a_1 + a_2 = -3[/tex]
- Common difference [tex]d = 7[/tex]
- Last term [tex]a_n = 135[/tex]
First Two Terms:
- The first term [tex]a_1[/tex] of the AP.
- The second term [tex]a_2[/tex] can be expressed as [tex]a_1 + d[/tex] since the definition of an AP is that each term after the first is the sum of the previous term and the common difference.
Therefore, [tex]a_2 = a_1 + 7[/tex].
Plug these in the sum equation for the first two terms:
[
a_1 + (a_1 + 7) = -3]
[tex]2a_1 + 7 = -3[/tex]
[tex]2a_1 = -10[/tex]
[tex]a_1 = -5[/tex]
Hence, [tex]a_2 = a_1 + 7 = -5 + 7 = 2[/tex].
Finding the Number of Terms (n):
The general formula for the [tex]n[/tex]-th term in an AP is:
[
a_n = a_1 + (n-1) \times d]
Plug in the values:
[tex]135 = -5 + (n-1) \times 7[/tex]
Solve this equation:
[tex]135 + 5 = (n-1) \times 7[/tex]
[tex]140 = (n-1) \times 7[/tex]
[tex](n-1) = \frac{140}{7} = 20[/tex]
[tex]n = 21[/tex]
Thus, there are 21 terms in the AP.
Sum of All Terms:
The formula for the sum [tex]S_n[/tex] of the first [tex]n[/tex] terms of an AP is:
[tex]S_n = \frac{n}{2} \times (a_1 + a_n)[/tex]Substitute the known values:
[tex]S_{21} = \frac{21}{2} \times (-5 + 135)[/tex]
[tex]S_{21} = \frac{21}{2} \times 130[/tex]
[tex]S_{21} = 21 \times 65[/tex]
[tex]S_{21} = 1365[/tex]
Conclusion:
The arithmetic progression is: [tex]-5, 2, 9, ..., 135[/tex] with a total of 21 terms, and the sum of all these terms is 1365.
