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Answer :
Sure! Let's solve the problem step-by-step.
### Given Arithmetic Progression (AP)
The terms of the arithmetic progression are: 600, 650, 700, 750. This means:
- The first term [tex]\( a_1 \)[/tex] is 600.
- The common difference [tex]\( d \)[/tex] can be found by subtracting the first term from the second term: [tex]\( d = 650 - 600 = 50 \)[/tex].
### (a) Find the 30th term of the AP
To find the [tex]\( n \)[/tex]-th term of an arithmetic progression, we use the formula:
[tex]\[ a_n = a_1 + (n - 1) \cdot d \][/tex]
For the 30th term ([tex]\( n = 30 \)[/tex]):
[tex]\[ a_{30} = 600 + (30 - 1) \cdot 50 \][/tex]
[tex]\[ a_{30} = 600 + 29 \cdot 50 \][/tex]
[tex]\[ a_{30} = 600 + 1450 \][/tex]
[tex]\[ a_{30} = 2050 \][/tex]
Thus, the 30th term of the AP is 2050.
### (b) Find the sum of the first 30 terms of the AP
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an arithmetic progression is given by the formula:
[tex]\[ S_n = \frac{n}{2} \cdot (2a_1 + (n - 1) \cdot d) \][/tex]
For the first 30 terms ([tex]\( n = 30 \)[/tex]):
[tex]\[ S_{30} = \frac{30}{2} \cdot (2 \cdot 600 + (30 - 1) \cdot 50) \][/tex]
[tex]\[ S_{30} = 15 \cdot (1200 + 1450) \][/tex]
[tex]\[ S_{30} = 15 \cdot 2650 \][/tex]
[tex]\[ S_{30} = 39750 \][/tex]
Thus, the sum of the first 30 terms of the AP is 39,750.
I hope this helps you understand how to solve the problem! If you have any further questions, feel free to ask.
### Given Arithmetic Progression (AP)
The terms of the arithmetic progression are: 600, 650, 700, 750. This means:
- The first term [tex]\( a_1 \)[/tex] is 600.
- The common difference [tex]\( d \)[/tex] can be found by subtracting the first term from the second term: [tex]\( d = 650 - 600 = 50 \)[/tex].
### (a) Find the 30th term of the AP
To find the [tex]\( n \)[/tex]-th term of an arithmetic progression, we use the formula:
[tex]\[ a_n = a_1 + (n - 1) \cdot d \][/tex]
For the 30th term ([tex]\( n = 30 \)[/tex]):
[tex]\[ a_{30} = 600 + (30 - 1) \cdot 50 \][/tex]
[tex]\[ a_{30} = 600 + 29 \cdot 50 \][/tex]
[tex]\[ a_{30} = 600 + 1450 \][/tex]
[tex]\[ a_{30} = 2050 \][/tex]
Thus, the 30th term of the AP is 2050.
### (b) Find the sum of the first 30 terms of the AP
The sum [tex]\( S_n \)[/tex] of the first [tex]\( n \)[/tex] terms of an arithmetic progression is given by the formula:
[tex]\[ S_n = \frac{n}{2} \cdot (2a_1 + (n - 1) \cdot d) \][/tex]
For the first 30 terms ([tex]\( n = 30 \)[/tex]):
[tex]\[ S_{30} = \frac{30}{2} \cdot (2 \cdot 600 + (30 - 1) \cdot 50) \][/tex]
[tex]\[ S_{30} = 15 \cdot (1200 + 1450) \][/tex]
[tex]\[ S_{30} = 15 \cdot 2650 \][/tex]
[tex]\[ S_{30} = 39750 \][/tex]
Thus, the sum of the first 30 terms of the AP is 39,750.
I hope this helps you understand how to solve the problem! If you have any further questions, feel free to ask.
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