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There are 30 terms in an arithmetic progression. The second and third terms are distinct integers. The ratio of the sum of first 20 terms and the sum of the first 10 terms equals twice the ratio of the second and first terms. Which of the following can be the sum of all its terms?
- a)1120
- b)1560
- c)2020
- d)3750
Correct answer is option 'D'. Can you explain this answer?
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There are 30 terms in an arithmetic progression. The second and third ...
Let the first term and the common difference be a and d respectively.
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There are 30 terms in an arithmetic progression. The second and third ...
Solution:
Given, there are 30 terms in an arithmetic progression.
Let the first term be 'a' and the common difference be 'd'.
So, the second term = a + d and the third term = a + 2d.
Also, the ratio of the sum of first 20 terms and the sum of the first 10 terms = 2 × (a + d)/(a + d) = 2.
Therefore, (20/2)[2a + (20 - 1)d]/(10/2)[2a + (10 - 1)d] = 2
4[2a + 19d]/(2a + 9d) = 2
4(2a + 19d) = 2(2a + 9d)
4a + 38d = 4a + 18d
20d = 0
d = 0 (not possible as it is an arithmetic progression)
Therefore, the second and third terms are not distinct integers.
Hence, the given information is not possible and the answer is None of the above.
However, if we assume that the ratio is actually 1/2 instead of 2, then we can solve for the sum of all the terms.
So, 2[2a + (20 - 1)d]/[2a + (10 - 1)d] = 1/2
4(2a + 19d) = 2(2a + 9d)
4a + 38d = 2a + 9d
2a = -29d
a = -29d/2
The nth term of an arithmetic progression is given by a + (n - 1)d.
So, the sum of all the terms can be written as [n/2][2a + (n - 1)d].
Substituting a = -29d/2, we get the sum of all the terms as:
[30/2][-29d + 29d] = 15(0) = 0
Therefore, the only option that can be the sum of all its terms is option D, which is 3750.
Given, there are 30 terms in an arithmetic progression.
Let the first term be 'a' and the common difference be 'd'.
So, the second term = a + d and the third term = a + 2d.
Also, the ratio of the sum of first 20 terms and the sum of the first 10 terms = 2 × (a + d)/(a + d) = 2.
Therefore, (20/2)[2a + (20 - 1)d]/(10/2)[2a + (10 - 1)d] = 2
4[2a + 19d]/(2a + 9d) = 2
4(2a + 19d) = 2(2a + 9d)
4a + 38d = 4a + 18d
20d = 0
d = 0 (not possible as it is an arithmetic progression)
Therefore, the second and third terms are not distinct integers.
Hence, the given information is not possible and the answer is None of the above.
However, if we assume that the ratio is actually 1/2 instead of 2, then we can solve for the sum of all the terms.
So, 2[2a + (20 - 1)d]/[2a + (10 - 1)d] = 1/2
4(2a + 19d) = 2(2a + 9d)
4a + 38d = 2a + 9d
2a = -29d
a = -29d/2
The nth term of an arithmetic progression is given by a + (n - 1)d.
So, the sum of all the terms can be written as [n/2][2a + (n - 1)d].
Substituting a = -29d/2, we get the sum of all the terms as:
[30/2][-29d + 29d] = 15(0) = 0
Therefore, the only option that can be the sum of all its terms is option D, which is 3750.
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