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The pth term of an AP is (3p – 1)/6. The sum of the first n terms of the AP is
- a)n(3n + 1)
- b)n/12 (3n + 1)
- c)n/12 (3n – 1)
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
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The pth term of an AP is (3p – 1)/6. The sum of the first n term...
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The pth term of an AP is (3p – 1)/6. The sum of the first n term...
Pth term is 3p-1/6
Consider nth term simply by change in variable that is 3n-1/6
By putting n=1 u get 1/3 =a
Sn=n/2(a+ nth term) nth term being the last term
Solving u get Sn=n/12(3n+1)
Consider nth term simply by change in variable that is 3n-1/6
By putting n=1 u get 1/3 =a
Sn=n/2(a+ nth term) nth term being the last term
Solving u get Sn=n/12(3n+1)
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The pth term of an AP is (3p – 1)/6. The sum of the first n term...
Given: The pth term of an AP is (3p+1)/6.
To find: The sum of the first n terms of the AP.
Formula: The sum of n terms of an AP is given by Sn = n/2[2a + (n-1)d], where a is the first term, d is the common difference, and n is the number of terms.
Approach:
1. Find the first term and common difference of the AP using the given expression for the pth term.
2. Substitute the values in the formula for the sum of n terms of an AP to get the required expression.
Solution:
1. Finding the first term and common difference of the AP:
Using the expression for the pth term of an AP, we have:
(3p+1)/6 = a + (p-1)d
Multiplying both sides by 6, we get:
3p+1 = 6a + 6(p-1)d
Simplifying, we get:
6d = (3p+1-6a)/5
Since d is constant for all terms of the AP, we can equate the above expression to a constant k:
6d = k
Substituting this in the previous equation, we get:
3p+1 = 6a + 5kp
Solving for a, we get:
a = (3-5k)p + 1/6
Since a is the first term of the AP, we can write:
a = a1
Thus, we have:
a1 = (3-5k)p + 1/6
And:
d = k/6
2. Finding the sum of n terms of the AP:
Using the formula for the sum of n terms of an AP, we have:
Sn = n/2[2a1 + (n-1)d]
Substituting the values of a1 and d, we get:
Sn = n/2[2((3-5k)p + 1/6) + (n-1)(k/6)]
Simplifying, we get:
Sn = n/12[3n - 3kp + 2]
Comparing this with the options given, we see that the correct answer is (c) n/12(3n-1). Therefore, the sum of the first n terms of the AP is n/12(3n-1).
To find: The sum of the first n terms of the AP.
Formula: The sum of n terms of an AP is given by Sn = n/2[2a + (n-1)d], where a is the first term, d is the common difference, and n is the number of terms.
Approach:
1. Find the first term and common difference of the AP using the given expression for the pth term.
2. Substitute the values in the formula for the sum of n terms of an AP to get the required expression.
Solution:
1. Finding the first term and common difference of the AP:
Using the expression for the pth term of an AP, we have:
(3p+1)/6 = a + (p-1)d
Multiplying both sides by 6, we get:
3p+1 = 6a + 6(p-1)d
Simplifying, we get:
6d = (3p+1-6a)/5
Since d is constant for all terms of the AP, we can equate the above expression to a constant k:
6d = k
Substituting this in the previous equation, we get:
3p+1 = 6a + 5kp
Solving for a, we get:
a = (3-5k)p + 1/6
Since a is the first term of the AP, we can write:
a = a1
Thus, we have:
a1 = (3-5k)p + 1/6
And:
d = k/6
2. Finding the sum of n terms of the AP:
Using the formula for the sum of n terms of an AP, we have:
Sn = n/2[2a1 + (n-1)d]
Substituting the values of a1 and d, we get:
Sn = n/2[2((3-5k)p + 1/6) + (n-1)(k/6)]
Simplifying, we get:
Sn = n/12[3n - 3kp + 2]
Comparing this with the options given, we see that the correct answer is (c) n/12(3n-1). Therefore, the sum of the first n terms of the AP is n/12(3n-1).
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The pth term of an AP is (3p – 1)/6. The sum of the first n terms of the AP isa)n(3n + 1)b)n/12 (3n + 1)c)n/12 (3n – 1)d)none of theseCorrect answer is option 'B'. Can you explain this answer?
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