The sum of the first n terms of an arithmetic series is given by: Sn=5/2n^2 + 7/2n.
Determine: 3.2.1 The common difference.
3.2.2 The tenth term
the first term would be S1
= (5/2)(1) + (7/2)(1) = 6
the sum of the first 2 terms is S2
= (5/2)(4) + (7/2)(2) = 17
so the 2nd term = S2 - S1 = 17-6 = 11
which means the common difference must be 5
(just checking:
the first 4 terms would be : 6, 11, 16, 21
and the sum of the first 4 terms = 54
by your formula:
S4 = (5/2)(16) + (7/2)(4) = 54)
so term(10) = a+9d
= 6 + 9(5)
= 51
The first and last term of an A.p are -3 and 145 respectively.if the common difference is 4 .find 12th
To determine the common difference of an arithmetic series, we need to examine the difference between consecutive terms.
In the given arithmetic series, the sum of the first n terms is expressed as Sn = (5/2)n^2 + (7/2)n.
3.2.1 The common difference:
To find the common difference, we compare the general term of the series, Tn, to the general term of the previous term, Tn-1.
The general term of an arithmetic series can be found using the formula:
Tn = a + (n-1)d
Here, a represents the first term of the arithmetic series, and d represents the common difference.
From the given formula for the sum of the first n terms (Sn = (5/2)n^2 + (7/2)n), we can rewrite it to find the general term:
Tn = (Sn - Sn-1) + Tn-1
Let's find T10 using the given formula for Sn and Sn-1:
T10 = [(5/2)(10^2) + (7/2)10] - [(5/2)(9^2) + (7/2)9]
Simplifying both equations:
T10 = [(5/2)(100) + (7/2)(10)] - [(5/2)(81) + (7/2)(9)]
T10 = [250 + 70] - [202.5 + 31.5]
T10 = 320 - 234
T10 = 86
Therefore, the tenth term of the arithmetic series is 86.
3.2.2 The tenth term:
We have already found the tenth term in the previous step, which is 86.