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Two APs have the same common difference. The difference between their 100th terms is 100, then the difference between their 1000th terms is
- a)100
- b)1000
- c)10000
- d)10
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
Two APs have the same common difference. The difference between their ...
**Solution:**
Let us assume that the common difference of the two arithmetic progressions (APs) is 'd'.
The difference between the 100th terms of the two APs is given as 100. This means that the 100th term of the second AP is 100 greater than the 100th term of the first AP.
Let the 100th term of the first AP be 'a' and the 100th term of the second AP be 'a + 100'.
We know that the nth term of an AP is given by the formula:
\(T_n = a + (n-1)d\)
where \(T_n\) is the nth term, 'a' is the first term, 'n' is the position of the term, and 'd' is the common difference.
**Difference between the 100th terms:**
Using the formula for the nth term, we can write:
\(a + (100-1)d\) (for the first AP)
\(a + 100 + (100-1)d\) (for the second AP)
The difference between these two terms is given as 100:
\(a + (100-1)d - (a + 100 + (100-1)d) = 100\)
Simplifying this equation, we get:
\(a + 99d - a - 100 - 99d = 100\)
Canceling out the 'a' and 'd' terms, we are left with:
\(-100 = 100\)
This is not a true statement, which means that our assumption that the difference between the 100th terms is 100 is incorrect.
Therefore, the given information is not possible.
Hence, there is no correct answer for this question.
Let us assume that the common difference of the two arithmetic progressions (APs) is 'd'.
The difference between the 100th terms of the two APs is given as 100. This means that the 100th term of the second AP is 100 greater than the 100th term of the first AP.
Let the 100th term of the first AP be 'a' and the 100th term of the second AP be 'a + 100'.
We know that the nth term of an AP is given by the formula:
\(T_n = a + (n-1)d\)
where \(T_n\) is the nth term, 'a' is the first term, 'n' is the position of the term, and 'd' is the common difference.
**Difference between the 100th terms:**
Using the formula for the nth term, we can write:
\(a + (100-1)d\) (for the first AP)
\(a + 100 + (100-1)d\) (for the second AP)
The difference between these two terms is given as 100:
\(a + (100-1)d - (a + 100 + (100-1)d) = 100\)
Simplifying this equation, we get:
\(a + 99d - a - 100 - 99d = 100\)
Canceling out the 'a' and 'd' terms, we are left with:
\(-100 = 100\)
This is not a true statement, which means that our assumption that the difference between the 100th terms is 100 is incorrect.
Therefore, the given information is not possible.
Hence, there is no correct answer for this question.
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Community Answer
Two APs have the same common difference. The difference between their ...
The formula for nth term of an AP is aₙ = a + (n - 1) d
Here, aₙ is the nth term, a is the first term, d is the common difference and n is the number of terms.
For first A.P., a₁₀₀ = a₁ + (100 - 1) d
a₁₀₀ = a₁ + 99d ------ (1)
a₁₀₀₀ = a₁ + (1000 - 1)d
a₁₀₀₀ = a₁ + 999d ------ (2)
For second A.P.,
b₁₀₀ = b₁+ (100 - 1)d
b₁₀₀ = b₁ + 99d ------ (3)
b₁₀₀₀ = b₁ + (1000 - 1)d
b₁₀₀₀ = b₁ + 999d ------ (4)
Given that, difference between 100th term of these A.P.s = 100
Thus, from equations (1) and (3) we have
(a₁ + 99d ) - (b₁ + 99d ) = 100
a₁ - b₁ = 100 ...(5)
Difference between 1000th terms of these A.P.s
Thus, from equations (2) and (4) we have
(a₁ + 999d ) - (b₁ + 999d ) = a₁ - b₁
But a₁ - b₁ = 100 [From equation(5)]
Hence, the difference between the 1000th terms of these A.P. will be 100.
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