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What is the fourth term of an AP of n terms whose sum is n(n + 1)?
- a)6
- b)8
- c)12
- d)20
Correct answer is option 'B'. Can you explain this answer?
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What is the fourth term of an AP of n terms whose sum is n(n + 1)?a)6b...
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What is the fourth term of an AP of n terms whose sum is n(n + 1)?a)6b...
Solution:
Given:
An Arithmetic Progression (AP) of n terms with sum n(n + 1)
Formula for sum of n terms of an AP:
Sum of n terms of an AP = (n/2)[2a + (n-1)d] where a is the first term and d is the common difference.
Given sum of n terms:
n(n + 1) = (n/2)[2a + (n-1)d]
Simplify:
n(n + 1) = n[2a + (n-1)d]
n + 1 = 2a + nd - d
n + 1 = 2a + d(n - 1) ---(i)
Using the formula for sum of n terms:
The sum of n terms can also be written as a + (a + d) + (a + 2d) + ... + [a + (n-1)d]
This is a sum of n terms in an AP.
Given sum is n(n + 1):
a + (a + d) + (a + 2d) + ... + [a + (n-1)d] = n(n + 1)
Substitute values:
a + (a + d) + (a + 2d) + ... + [a + (n-1)d] = n(n + 1)
n(2a + (n-1)d) = n(n + 1)
2a + (n-1)d = n + 1
2a + d(n-1) = n + 1 ---(ii)
Equating equations (i) and (ii):
From (i) and (ii), we have:
n + 1 = 2a + d(n - 1)
n + 1 = n + 1
d(n - 1) = 2a
d = 2a / (n - 1)
Find the fourth term (4th term):
The fourth term can be found using the formula: a + 3d (since the first term is a, and the common difference is d).
Substitute d = 2a / (n - 1):
4th term = a + 3d
4th term = a + 3 * (2a / (n - 1))
4th term = a + 6a / (n - 1)
Given sum of n terms is n(n + 1):
a + (a + d) + (a + 2d) + (a + 3d) + ... + [a + (n-1)d] = n(n + 1)
Comparing with the sum to find the 4th term:
a + (a + d) + (a + 2d) + (a + 3d) = n(n + 1)
Therefore, the 4th term of the AP is:
4th term = a + 6a / (n - 1)
4th term = 7a / (n - 1)
Therefore,
Given:
An Arithmetic Progression (AP) of n terms with sum n(n + 1)
Formula for sum of n terms of an AP:
Sum of n terms of an AP = (n/2)[2a + (n-1)d] where a is the first term and d is the common difference.
Given sum of n terms:
n(n + 1) = (n/2)[2a + (n-1)d]
Simplify:
n(n + 1) = n[2a + (n-1)d]
n + 1 = 2a + nd - d
n + 1 = 2a + d(n - 1) ---(i)
Using the formula for sum of n terms:
The sum of n terms can also be written as a + (a + d) + (a + 2d) + ... + [a + (n-1)d]
This is a sum of n terms in an AP.
Given sum is n(n + 1):
a + (a + d) + (a + 2d) + ... + [a + (n-1)d] = n(n + 1)
Substitute values:
a + (a + d) + (a + 2d) + ... + [a + (n-1)d] = n(n + 1)
n(2a + (n-1)d) = n(n + 1)
2a + (n-1)d = n + 1
2a + d(n-1) = n + 1 ---(ii)
Equating equations (i) and (ii):
From (i) and (ii), we have:
n + 1 = 2a + d(n - 1)
n + 1 = n + 1
d(n - 1) = 2a
d = 2a / (n - 1)
Find the fourth term (4th term):
The fourth term can be found using the formula: a + 3d (since the first term is a, and the common difference is d).
Substitute d = 2a / (n - 1):
4th term = a + 3d
4th term = a + 3 * (2a / (n - 1))
4th term = a + 6a / (n - 1)
Given sum of n terms is n(n + 1):
a + (a + d) + (a + 2d) + (a + 3d) + ... + [a + (n-1)d] = n(n + 1)
Comparing with the sum to find the 4th term:
a + (a + d) + (a + 2d) + (a + 3d) = n(n + 1)
Therefore, the 4th term of the AP is:
4th term = a + 6a / (n - 1)
4th term = 7a / (n - 1)
Therefore,
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What is the fourth term of an AP of n terms whose sum is n(n + 1)?a)6b)8c)12d)20Correct answer is option 'B'. Can you explain this answer?
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