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If the sum of n terms of an Arithmetic Progression is given by Sn = 3n2 + 2n. Find its first term and common difference.
- a)2, 3
- b)4, 5
- c)5, 6
- d)5, 7
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If the sum of n terms of an Arithmetic Progression is given by Sn = 3n...
Given Sn = 3n2 + 2n
∴ S1 = 3 × 12 + 2 × 1 = 5
S2 = 3 × 22 + 2 × 2 = 16
The sequence is 5, 11, …..
a = 5 , d = 11 - 5 = 6
∴ S1 = 3 × 12 + 2 × 1 = 5
S2 = 3 × 22 + 2 × 2 = 16
The sequence is 5, 11, …..
a = 5 , d = 11 - 5 = 6
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If the sum of n terms of an Arithmetic Progression is given by Sn = 3n...
To find the first term and common difference of an arithmetic progression (AP), we will use the given sum of n terms (Sn) formula and compare it with the general formula for the sum of an AP.
Given:
Sn = 3n² - 2n
We know that the sum of n terms of an AP is given by:
Sn = (n/2)(2a + (n-1)d)
Where:
a = first term of the AP
d = common difference of the AP
n = number of terms
Comparing the given sum formula with the general sum formula, we can equate the two equations:
3n² - 2n = (n/2)(2a + (n-1)d)
Simplifying the equation:
3n² - 2n = an + (n² - n)d/2
Multiplying both sides by 2 to get rid of the fractions:
6n² - 4n = 2an + (n² - n)d
Now, let's compare the coefficients of n², n, and the constant term on both sides of the equation:
Coefficient of n²:
6 = a
Coefficient of n:
-4 = 2a + d
Constant term:
0 = (n² - n)d
From the constant term, we can see that either d = 0 or n² - n = 0. Since the number of terms cannot be zero, we have n² - n = 0.
Solving n² - n = 0:
n(n - 1) = 0
So, n = 0 or n - 1 = 0
Since n cannot be zero, we have n - 1 = 0
n = 1
Now, substitute n = 1 into the equation for the coefficient of n:
-4 = 2a + d
Substituting the value of a = 6 into the equation above, we get:
-4 = 2(6) + d
-4 = 12 + d
d = -16
So, the first term of the AP is a = 6 and the common difference is d = -16.
Therefore, the correct answer is option C) 5, 6.
Given:
Sn = 3n² - 2n
We know that the sum of n terms of an AP is given by:
Sn = (n/2)(2a + (n-1)d)
Where:
a = first term of the AP
d = common difference of the AP
n = number of terms
Comparing the given sum formula with the general sum formula, we can equate the two equations:
3n² - 2n = (n/2)(2a + (n-1)d)
Simplifying the equation:
3n² - 2n = an + (n² - n)d/2
Multiplying both sides by 2 to get rid of the fractions:
6n² - 4n = 2an + (n² - n)d
Now, let's compare the coefficients of n², n, and the constant term on both sides of the equation:
Coefficient of n²:
6 = a
Coefficient of n:
-4 = 2a + d
Constant term:
0 = (n² - n)d
From the constant term, we can see that either d = 0 or n² - n = 0. Since the number of terms cannot be zero, we have n² - n = 0.
Solving n² - n = 0:
n(n - 1) = 0
So, n = 0 or n - 1 = 0
Since n cannot be zero, we have n - 1 = 0
n = 1
Now, substitute n = 1 into the equation for the coefficient of n:
-4 = 2a + d
Substituting the value of a = 6 into the equation above, we get:
-4 = 2(6) + d
-4 = 12 + d
d = -16
So, the first term of the AP is a = 6 and the common difference is d = -16.
Therefore, the correct answer is option C) 5, 6.
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