Answer :
Sure! Let's solve each part of the problem step by step.
### (1) Sum of the first 15 terms of the AP: 11, 5, -1, -7
This sequence is an arithmetic progression (AP) where:
- The first term (a) is 11.
- The common difference (d) is 5 - 11 = -6.
To find the sum of the first n terms of an AP, we use the formula:
[tex]\[ S_n = \frac{n}{2} \times \left(2a + (n - 1)d\right) \][/tex]
For the first 15 terms (n = 15):
[tex]\[ S_{15} = \frac{15}{2} \times \left( 2 \times 11 + (15 - 1) \times (-6) \right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \times (22 - 84) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \times (-62) \][/tex]
[tex]\[ S_{15} = -465 \][/tex]
### (a) Sum of the first 5 terms of the AP: 0, 0.5, 0.75, 1
Here, the first term (a) is 0, and the common difference (d) is 0.5 - 0 = 0.5.
For the first 5 terms:
[tex]\[ S_{5} = \frac{5}{2} \times \left(2 \times 0 + (5 - 1) \times 0.5\right) \][/tex]
[tex]\[ S_{5} = \frac{5}{2} \times (0 + 2) \][/tex]
[tex]\[ S_{5} = \frac{5}{2} \times 2 \][/tex]
[tex]\[ S_{5} = 5 \][/tex]
### (c) Sum of the first 13 terms of the AP: 2, 2.25, 2.5
First term (a) is 2, and the common difference (d) is 2.25 - 2 = 0.25.
For the first 13 terms:
[tex]\[ S_{13} = \frac{13}{2} \times \left(2 \times 2 + (13 - 1) \times 0.25\right) \][/tex]
[tex]\[ S_{13} = \frac{13}{2} \times (4 + 3) \][/tex]
[tex]\[ S_{13} = \frac{13}{2} \times 7 \][/tex]
[tex]\[ S_{13} = 45.5 \][/tex]
### (d) Sum of the first 20 terms of the AP: -81, -78, -75
The first term (a) is -81 and the common difference (d) is -78 - (-81) = 3.
For the first 20 terms:
[tex]\[ S_{20} = \frac{20}{2} \times \left(2 \times -81 + (20 - 1) \times 3\right) \][/tex]
[tex]\[ S_{20} = 10 \times (-162 + 57) \][/tex]
[tex]\[ S_{20} = 10 \times (-105) \][/tex]
[tex]\[ S_{20} = -1050 \][/tex]
### (i) Sum of the first 25 terms of the AP: 1, 15, 13
The first term (a) is 1, and the common difference (d) is 15 - 1 = 14.
For the first 25 terms:
[tex]\[ S_{25} = \frac{25}{2} \times \left(2 \times 1 + (25 - 1) \times 14\right) \][/tex]
[tex]\[ S_{25} = \frac{25}{2} \times (2 + 336) \][/tex]
[tex]\[ S_{25} = \frac{25}{2} \times 338 \][/tex]
[tex]\[ S_{25} = 4225 \][/tex]
### (2) The first term of an AP is 126, and the last term is 387. Find [tex]\( S_{30} \)[/tex]
Here, the first term (a) is 126 and the last term (l) is 387.
Using the sum formula where the last term is known:
[tex]\[ S_n = \frac{n}{2} \times (a + l) \][/tex]
For 30 terms:
[tex]\[ S_{30} = \frac{30}{2} \times (126 + 387) \][/tex]
[tex]\[ S_{30} = 15 \times 513 \][/tex]
[tex]\[ S_{30} = 7695 \][/tex]
### (2) Find the 9th term where [tex]\( S_8 \)[/tex] of an AP is 116 and [tex]\( S_9 \)[/tex] is 117
The 9th term is the difference between the sums of the first 9 and 8 terms.
[tex]\[ a_9 = S_9 - S_8 \][/tex]
[tex]\[ a_9 = 117 - 116 \][/tex]
[tex]\[ a_9 = 1 \][/tex]
These detailed calculations explain the numerical results obtained for each question part.
### (1) Sum of the first 15 terms of the AP: 11, 5, -1, -7
This sequence is an arithmetic progression (AP) where:
- The first term (a) is 11.
- The common difference (d) is 5 - 11 = -6.
To find the sum of the first n terms of an AP, we use the formula:
[tex]\[ S_n = \frac{n}{2} \times \left(2a + (n - 1)d\right) \][/tex]
For the first 15 terms (n = 15):
[tex]\[ S_{15} = \frac{15}{2} \times \left( 2 \times 11 + (15 - 1) \times (-6) \right) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \times (22 - 84) \][/tex]
[tex]\[ S_{15} = \frac{15}{2} \times (-62) \][/tex]
[tex]\[ S_{15} = -465 \][/tex]
### (a) Sum of the first 5 terms of the AP: 0, 0.5, 0.75, 1
Here, the first term (a) is 0, and the common difference (d) is 0.5 - 0 = 0.5.
For the first 5 terms:
[tex]\[ S_{5} = \frac{5}{2} \times \left(2 \times 0 + (5 - 1) \times 0.5\right) \][/tex]
[tex]\[ S_{5} = \frac{5}{2} \times (0 + 2) \][/tex]
[tex]\[ S_{5} = \frac{5}{2} \times 2 \][/tex]
[tex]\[ S_{5} = 5 \][/tex]
### (c) Sum of the first 13 terms of the AP: 2, 2.25, 2.5
First term (a) is 2, and the common difference (d) is 2.25 - 2 = 0.25.
For the first 13 terms:
[tex]\[ S_{13} = \frac{13}{2} \times \left(2 \times 2 + (13 - 1) \times 0.25\right) \][/tex]
[tex]\[ S_{13} = \frac{13}{2} \times (4 + 3) \][/tex]
[tex]\[ S_{13} = \frac{13}{2} \times 7 \][/tex]
[tex]\[ S_{13} = 45.5 \][/tex]
### (d) Sum of the first 20 terms of the AP: -81, -78, -75
The first term (a) is -81 and the common difference (d) is -78 - (-81) = 3.
For the first 20 terms:
[tex]\[ S_{20} = \frac{20}{2} \times \left(2 \times -81 + (20 - 1) \times 3\right) \][/tex]
[tex]\[ S_{20} = 10 \times (-162 + 57) \][/tex]
[tex]\[ S_{20} = 10 \times (-105) \][/tex]
[tex]\[ S_{20} = -1050 \][/tex]
### (i) Sum of the first 25 terms of the AP: 1, 15, 13
The first term (a) is 1, and the common difference (d) is 15 - 1 = 14.
For the first 25 terms:
[tex]\[ S_{25} = \frac{25}{2} \times \left(2 \times 1 + (25 - 1) \times 14\right) \][/tex]
[tex]\[ S_{25} = \frac{25}{2} \times (2 + 336) \][/tex]
[tex]\[ S_{25} = \frac{25}{2} \times 338 \][/tex]
[tex]\[ S_{25} = 4225 \][/tex]
### (2) The first term of an AP is 126, and the last term is 387. Find [tex]\( S_{30} \)[/tex]
Here, the first term (a) is 126 and the last term (l) is 387.
Using the sum formula where the last term is known:
[tex]\[ S_n = \frac{n}{2} \times (a + l) \][/tex]
For 30 terms:
[tex]\[ S_{30} = \frac{30}{2} \times (126 + 387) \][/tex]
[tex]\[ S_{30} = 15 \times 513 \][/tex]
[tex]\[ S_{30} = 7695 \][/tex]
### (2) Find the 9th term where [tex]\( S_8 \)[/tex] of an AP is 116 and [tex]\( S_9 \)[/tex] is 117
The 9th term is the difference between the sums of the first 9 and 8 terms.
[tex]\[ a_9 = S_9 - S_8 \][/tex]
[tex]\[ a_9 = 117 - 116 \][/tex]
[tex]\[ a_9 = 1 \][/tex]
These detailed calculations explain the numerical results obtained for each question part.
