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The sum of first 6th term of an arithmetic progression is 42 the ratio of 10th term to its 30 term is 1:3 calculate the first and 13 term of AP?
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The sum of first 6th term of an arithmetic progression is 42 the ratio...
Given information:
- Sum of the first 6 terms = 42
- Ratio of 10th term to its 30th term = 1:3
Finding the common difference (d):
- Let the first term of the AP be 'a' and the common difference be 'd'.
- The sum of the first 6 terms of an AP is given by: S6 = 6/2 [2a + (6-1)d] = 42
- Simplifying the equation, we get: 3[2a + 5d] = 42
- 2a + 5d = 14
Finding the 10th term (T10) and 30th term (T30):
- Let the 10th term be T10 and the 30th term be T30.
- According to the given ratio, T10/T30 = 1/3
- T10 = a + 9d
- T30 = a + 29d
- Substituting these values in the ratio equation, we get: (a + 9d)/(a + 29d) = 1/3
- Solving this equation, we get: 3a + 27d = a + 29d
- 2a - 2d = 0
- a = d
Calculating the first term (a) and the 13th term (T13):
- Substituting the value of a = d in 2a + 5d = 14, we get: 2d + 5d = 14
- Solving this equation, we get: d = 2
- Substituting d = 2 in a = d, we get: a = 2
- The first term of the AP is a = 2
- The 13th term of the AP is T13 = a + 12d = 2 + 12(2) = 26
Therefore, the first term of the arithmetic progression is 2 and the 13th term of the arithmetic progression is 26.
- Sum of the first 6 terms = 42
- Ratio of 10th term to its 30th term = 1:3
Finding the common difference (d):
- Let the first term of the AP be 'a' and the common difference be 'd'.
- The sum of the first 6 terms of an AP is given by: S6 = 6/2 [2a + (6-1)d] = 42
- Simplifying the equation, we get: 3[2a + 5d] = 42
- 2a + 5d = 14
Finding the 10th term (T10) and 30th term (T30):
- Let the 10th term be T10 and the 30th term be T30.
- According to the given ratio, T10/T30 = 1/3
- T10 = a + 9d
- T30 = a + 29d
- Substituting these values in the ratio equation, we get: (a + 9d)/(a + 29d) = 1/3
- Solving this equation, we get: 3a + 27d = a + 29d
- 2a - 2d = 0
- a = d
Calculating the first term (a) and the 13th term (T13):
- Substituting the value of a = d in 2a + 5d = 14, we get: 2d + 5d = 14
- Solving this equation, we get: d = 2
- Substituting d = 2 in a = d, we get: a = 2
- The first term of the AP is a = 2
- The 13th term of the AP is T13 = a + 12d = 2 + 12(2) = 26
Therefore, the first term of the arithmetic progression is 2 and the 13th term of the arithmetic progression is 26.
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The sum of first 6th term of an arithmetic progression is 42 the ratio of 10th term to its 30 term is 1:3 calculate the first and 13 term of AP?
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The sum of first 6th term of an arithmetic progression is 42 the ratio of 10th term to its 30 term is 1:3 calculate the first and 13 term of AP? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The sum of first 6th term of an arithmetic progression is 42 the ratio of 10th term to its 30 term is 1:3 calculate the first and 13 term of AP? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of first 6th term of an arithmetic progression is 42 the ratio of 10th term to its 30 term is 1:3 calculate the first and 13 term of AP?.
The sum of first 6th term of an arithmetic progression is 42 the ratio of 10th term to its 30 term is 1:3 calculate the first and 13 term of AP? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The sum of first 6th term of an arithmetic progression is 42 the ratio of 10th term to its 30 term is 1:3 calculate the first and 13 term of AP? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of first 6th term of an arithmetic progression is 42 the ratio of 10th term to its 30 term is 1:3 calculate the first and 13 term of AP?.
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