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For any natural number tt, suppose the sum of the first tt terms of an arithmetic progression is (n + 2n2). If the nth term of the progression is divisible by 9, then the smallest possible value of n is
- a)4
- b)7
- c)9
- d)8
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
For any natural number tt, suppose the sum of the first tt terms of an...
**Solution:**
To find the smallest possible value of n, we need to find the smallest value of tt such that the nth term of the arithmetic progression is divisible by 9.
Let's assume that the first term of the arithmetic progression is aa and the common difference is dd.
The sum of the first tt terms of an arithmetic progression can be given by the formula:
Sum = (tt/2) * (2a + (tt-1)d)
Given that the sum of the first tt terms is (n + 2n^2), we can write the equation as:
(n + 2n^2) = (tt/2) * (2a + (tt-1)d) ---(1)
We also know that the nth term of the arithmetic progression is given by the formula:
nth term = a + (n-1)d
Since the nth term is divisible by 9, we can write the equation as:
a + (n-1)d = 9k ---(2), where k is a positive integer.
**Simplifying the Equations:**
From equation (2), we can write:
a = 9k - (n-1)d
Substituting this value of a in equation (1), we get:
(n + 2n^2) = (tt/2) * (2(9k - (n-1)d) + (tt-1)d)
Simplifying this equation further, we get:
(n + 2n^2) = (tt/2) * (18k + (tt-1)d)
Dividing both sides of the equation by n, we get:
1 + 2n = (tt/2n) * (18k + (tt-1)d)
Since the left-hand side of the equation is an integer, the right-hand side must also be an integer.
Therefore, (tt/2n) must be an integer.
**Analyzing the Factors of tt and 2n:**
The factors of tt can be either odd or even.
If tt is odd, then 2n must also be odd, as 2n is a factor of tt.
If tt is even, then 2n must be even, as 2n is a factor of tt.
In either case, (tt/2n) will always be an integer.
**Conclusion:**
From the above analysis, we can conclude that (tt/2n) is always an integer, regardless of the values of tt and 2n.
Therefore, the smallest possible value of n can be any natural number.
Hence, the correct answer should be "None of the above" instead of option 'B'.
To find the smallest possible value of n, we need to find the smallest value of tt such that the nth term of the arithmetic progression is divisible by 9.
Let's assume that the first term of the arithmetic progression is aa and the common difference is dd.
The sum of the first tt terms of an arithmetic progression can be given by the formula:
Sum = (tt/2) * (2a + (tt-1)d)
Given that the sum of the first tt terms is (n + 2n^2), we can write the equation as:
(n + 2n^2) = (tt/2) * (2a + (tt-1)d) ---(1)
We also know that the nth term of the arithmetic progression is given by the formula:
nth term = a + (n-1)d
Since the nth term is divisible by 9, we can write the equation as:
a + (n-1)d = 9k ---(2), where k is a positive integer.
**Simplifying the Equations:**
From equation (2), we can write:
a = 9k - (n-1)d
Substituting this value of a in equation (1), we get:
(n + 2n^2) = (tt/2) * (2(9k - (n-1)d) + (tt-1)d)
Simplifying this equation further, we get:
(n + 2n^2) = (tt/2) * (18k + (tt-1)d)
Dividing both sides of the equation by n, we get:
1 + 2n = (tt/2n) * (18k + (tt-1)d)
Since the left-hand side of the equation is an integer, the right-hand side must also be an integer.
Therefore, (tt/2n) must be an integer.
**Analyzing the Factors of tt and 2n:**
The factors of tt can be either odd or even.
If tt is odd, then 2n must also be odd, as 2n is a factor of tt.
If tt is even, then 2n must be even, as 2n is a factor of tt.
In either case, (tt/2n) will always be an integer.
**Conclusion:**
From the above analysis, we can conclude that (tt/2n) is always an integer, regardless of the values of tt and 2n.
Therefore, the smallest possible value of n can be any natural number.
Hence, the correct answer should be "None of the above" instead of option 'B'.
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Community Answer
For any natural number tt, suppose the sum of the first tt terms of an...
It is given,
The terms are 3, 7, 11, 15, 19, 23, 27,......
The terms are 3, 7, 11, 15, 19, 23, 27,......
27 is the first term in the series divisible by 9.
27 is the 7th term.
Therefore, the least possible value of n is 7.
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