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The sum of n terms of an AP is 3n2 + 5n. A.P. is
- a)8, 14, 20, 26
- b)8, 22, 42, 68
- c)22, 68, 114, ....
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
The sum of n terms of an AP is 3n2 + 5n. A.P. isa)8, 14, 20, 26b)8, 22...
The sum of n terms of an A.P. = 3n2+5n
then sum of n-1 terms= 3(n-1)2+5(n-1)
So nth term of A.P will be Tn=3n2+5n- 3(n-1)2-5(n-1)= 3n2+5n-3n2-3+6n-5n+5=6n+2
in this way by putinng values of n= 1,2,3,4,5,6,7,_______n-1, n
we will get the AP as 8,14,20,26,32,38,44,____and the 16th term will be= 6*16+2=98
Most Upvoted Answer
The sum of n terms of an AP is 3n2 + 5n. A.P. isa)8, 14, 20, 26b)8, 22...
Solution:
Given: Sum of n terms of an AP = 3n^2 + 5n
Let 'a' be the first term and 'd' be the common difference of the AP.
Formula for sum of n terms of an AP is given by:
Sn = (n/2)[2a + (n - 1)d]
Substituting the given values, we get:
3n^2 + 5n = (n/2)[2a + (n - 1)d]
6n^2 + 10n = n[2a + (n - 1)d]
6n + 10 = 2a + (n - 1)d
Equation (1)
To find the AP, we need to use the formula for nth term of an AP:
an = a + (n - 1)d
Let's find the value of a using equation (1):
6(1) + 10 = 2a + (1 - 1)d
2a = 16
a = 8
So, the first term of the AP is 8.
Now, let's find the common difference 'd'. Substituting the values of 'a' and 'n' in equation (1), we get:
6n + 10 = 2(8) + (n - 1)d
6n + 10 = 16 + nd - d
5d = 6n - 6
d = (6n - 6)/5
Since 'd' should be a constant value for all terms of the AP, let's check if this formula for 'd' satisfies the second term of the AP:
a2 = a + d
a2 = 8 + (6 - 6)/5
a2 = 8
So, the second term of the AP is also 8.
Hence, the AP is: 8, 14, 20, 26, ...
This matches option (A).
Given: Sum of n terms of an AP = 3n^2 + 5n
Let 'a' be the first term and 'd' be the common difference of the AP.
Formula for sum of n terms of an AP is given by:
Sn = (n/2)[2a + (n - 1)d]
Substituting the given values, we get:
3n^2 + 5n = (n/2)[2a + (n - 1)d]
6n^2 + 10n = n[2a + (n - 1)d]
6n + 10 = 2a + (n - 1)d
Equation (1)
To find the AP, we need to use the formula for nth term of an AP:
an = a + (n - 1)d
Let's find the value of a using equation (1):
6(1) + 10 = 2a + (1 - 1)d
2a = 16
a = 8
So, the first term of the AP is 8.
Now, let's find the common difference 'd'. Substituting the values of 'a' and 'n' in equation (1), we get:
6n + 10 = 2(8) + (n - 1)d
6n + 10 = 16 + nd - d
5d = 6n - 6
d = (6n - 6)/5
Since 'd' should be a constant value for all terms of the AP, let's check if this formula for 'd' satisfies the second term of the AP:
a2 = a + d
a2 = 8 + (6 - 6)/5
a2 = 8
So, the second term of the AP is also 8.
Hence, the AP is: 8, 14, 20, 26, ...
This matches option (A).
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The sum of n terms of an AP is 3n2 + 5n. A.P. isa)8, 14, 20, 26b)8, 22...
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