Answer :
Sure! Let's solve this problem step-by-step.
### Part 1: Finding the Common Difference of the AP
Given:
- The first term of the arithmetic progression (AP) is [tex]\( a = 2 \)[/tex].
- The sum of the first 8 terms ([tex]\( S_8 \)[/tex]) is 240.
The formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic progression is:
[tex]\[ S_n = \frac{n}{2} \times [2a + (n-1)d] \][/tex]
For the first 8 terms:
[tex]\[ S_8 = \frac{8}{2} \times [2 \times 2 + (8-1)d] \][/tex]
[tex]\[ 240 = 4 \times [4 + 7d] \][/tex]
Let's simplify and solve for [tex]\( d \)[/tex]:
1. [tex]\( 240 = 4 \times (4 + 7d) \)[/tex]
2. [tex]\( 240 = 16 + 28d \)[/tex]
3. [tex]\( 224 = 28d \)[/tex]
4. [tex]\( d = \frac{224}{28} \)[/tex]
5. [tex]\( d = 8 \)[/tex]
Thus, the common difference [tex]\( d \)[/tex] is 8.
### Part 2: Finding the Number of Terms [tex]\( n \)[/tex]
Given:
- A new sum [tex]\( S \)[/tex] is 1560, for the same AP.
Use the formula to find [tex]\( n \)[/tex]:
[tex]\[ 1560 = \frac{n}{2} \times [2 \times 2 + (n-1) \times 8] \][/tex]
Here's how to solve for [tex]\( n \)[/tex]:
1. [tex]\( 1560 = \frac{n}{2} \times [4 + 8(n - 1)] \)[/tex]
2. [tex]\( 1560 = \frac{n}{2} \times [4 + 8n - 8] \)[/tex]
3. [tex]\( 1560 = \frac{n}{2} \times [8n - 4] \)[/tex]
4. Simplify further to get equation for [tex]\( n \)[/tex]:
[tex]\[ 1560 = 4n(2n - 1) \][/tex]
5. Solve for [tex]\( n \)[/tex] using the quadratic formula or solving the equation algebraically.
The solution for [tex]\( n \)[/tex] is:
[tex]\[ n = 20 \][/tex] (Ignore the negative value as [tex]\( n \)[/tex] cannot be negative).
So, the number of terms is 20.
### Part 3: Finding the GP Terms
The problem also mentions the 3rd, 5th, and 8th terms of another AP forming the first terms of a GP, but this step was not further developed in the provided solution.
Therefore, from what we solved thoroughly, we have the following answers:
- Common difference [tex]\( d \)[/tex] is 8.
- Number of terms [tex]\( n \)[/tex] is 20 for the given sum 1560.
If you need more information or further details on those terms forming a GP, please let me know!
### Part 1: Finding the Common Difference of the AP
Given:
- The first term of the arithmetic progression (AP) is [tex]\( a = 2 \)[/tex].
- The sum of the first 8 terms ([tex]\( S_8 \)[/tex]) is 240.
The formula for the sum of the first [tex]\( n \)[/tex] terms of an arithmetic progression is:
[tex]\[ S_n = \frac{n}{2} \times [2a + (n-1)d] \][/tex]
For the first 8 terms:
[tex]\[ S_8 = \frac{8}{2} \times [2 \times 2 + (8-1)d] \][/tex]
[tex]\[ 240 = 4 \times [4 + 7d] \][/tex]
Let's simplify and solve for [tex]\( d \)[/tex]:
1. [tex]\( 240 = 4 \times (4 + 7d) \)[/tex]
2. [tex]\( 240 = 16 + 28d \)[/tex]
3. [tex]\( 224 = 28d \)[/tex]
4. [tex]\( d = \frac{224}{28} \)[/tex]
5. [tex]\( d = 8 \)[/tex]
Thus, the common difference [tex]\( d \)[/tex] is 8.
### Part 2: Finding the Number of Terms [tex]\( n \)[/tex]
Given:
- A new sum [tex]\( S \)[/tex] is 1560, for the same AP.
Use the formula to find [tex]\( n \)[/tex]:
[tex]\[ 1560 = \frac{n}{2} \times [2 \times 2 + (n-1) \times 8] \][/tex]
Here's how to solve for [tex]\( n \)[/tex]:
1. [tex]\( 1560 = \frac{n}{2} \times [4 + 8(n - 1)] \)[/tex]
2. [tex]\( 1560 = \frac{n}{2} \times [4 + 8n - 8] \)[/tex]
3. [tex]\( 1560 = \frac{n}{2} \times [8n - 4] \)[/tex]
4. Simplify further to get equation for [tex]\( n \)[/tex]:
[tex]\[ 1560 = 4n(2n - 1) \][/tex]
5. Solve for [tex]\( n \)[/tex] using the quadratic formula or solving the equation algebraically.
The solution for [tex]\( n \)[/tex] is:
[tex]\[ n = 20 \][/tex] (Ignore the negative value as [tex]\( n \)[/tex] cannot be negative).
So, the number of terms is 20.
### Part 3: Finding the GP Terms
The problem also mentions the 3rd, 5th, and 8th terms of another AP forming the first terms of a GP, but this step was not further developed in the provided solution.
Therefore, from what we solved thoroughly, we have the following answers:
- Common difference [tex]\( d \)[/tex] is 8.
- Number of terms [tex]\( n \)[/tex] is 20 for the given sum 1560.
If you need more information or further details on those terms forming a GP, please let me know!
