Can someone help me on a few questions which I got stuck on and show me how you got the answer?
1. Evaluate: Sigma sign n on the top, 1 underneath then in the middle 2r-7.
2. The first term of an aritmetic series is 14. The common difference and the sum of the first n terms are both -6. Find n.
3. The first three terms of an arithmetic series are 2x+1, 3x+2 and 5x-1. Find x.
Thanks in advance
Hi there...Originally Posted by Confuzzled?
For Qn 1, for typing easeness, i will denote the Sigma sign n on the top, 1 underneath as (S1).
(S1) (2r-7)
= (S1) (2r) - (s1) (7)
= 2 (S1)(r) -7n
=2 (1/2)(n)(n+1) - 7n
=n2 + n - 7n
=n2 - 6n
=n(n-6)
For Qn 2, 1st term = a = 14 ; Sn = d = -6
Sn = n/2(2a + (n-1)d) => eq 1
Substitute the following values on top into eq 1 and solve for n.
For Qn 3,
1st Term = 2x + 1
2nd Term = 3x + 2
3rd Term = 5x - 1
As they are in arithmetic series, they will have a common difference.
Thus,
(3x + 2) - (2x + 1) = (5x - 1) - (3x + 2)
Solve the above equation for x.
Hope these can be of some help to u.
Arithmetic series:
an = a1 +(n-1)d
Sn = (n/2)(a1 +an)
------------------
1. Evaluate: Sigma sign n on the top, 1 underneath then in the middle 2r-7.
Your "Sigma sign, n on top, 1 underneath" means Summation of the terms from r=1 to r=n. It is the value/sum of the series from r=1 to r=n.
term = 2r -7
a1 = 2(1) -7 = -5
a2 = 2(2) -7 = -3
a3 = 2(3) -7 = -1
So, d = 2
And, an = a1 +(n-1)d = -5 +(n-1)(2) = -5 +2n -2 = 2n -7
Hence, "Sigma sign, n on top, 1 underneath"[2r -7] = Sn
= (n/2)[a1 +an]
= (n/2)[-5 +2n -7]
= (n/2)[2n -12]
= n(n-6) ----------------answer.
-------------
2. The first term of an aritmetic series is 14. The common difference and the sum of the first n terms are both -6. Find n.
a1 = 14
d = -6
Sn = -6
Sn = (n/2)[a1 +an] = (n/2)[a1 +(a1 +(n-1)d)] = (n/2)[2a1 +(n-1)d]
Substitutions,
-6 = (n/2)[2(14) +(n-1)(-6)]
-6 = (n/2)[28 -6n +6]
-6 = (n/2)[34 -6n]
-6 = n(17 -3n)
-6 = 17n -3n^2
3n^2 -17n -6 = 0
Use the Quadratic Formula to solve for n, or use factoring if possible,
(3n +1)(n -6) = 0
3n +1 = 0
3n = -1
n = -1/3 -----reject, because n must be a positive integer.
n -6 = 0
n = 6 ----------answer.
-----------------------------
3. The first three terms of an arithmetic series are 2x+1, 3x+2 and 5x-1. Find x.
a1 = 2x +1
a2 = 3x +2
a3 = 5x -1
an = a1 +(n-1)d
a2 = a1 +(2-1)d
a2 = a1 +d
3x +2 = (2x +1) +d
3x +2 -2x -1 = d
x+1 = d -------------***
an = a1 +(n-1)d
a3 = a1 +(3-1)d
a3 = a1 +2d
5x -1 = (2x +1) +2(x+1) <----d = x+1
5x -1 = 2x +1 +2x +2
5x -1 = 4x +3
5x -4x = 3 +1
x = 4 ------------answer.
OKie.. no problem... Solving this question requires the same concept as question 1.
Now u r dealing wif summation of the 1st 6 terms for the sequence (4 - r/2)
There's two methods u can adopt.
First method will be by substitution.
r=1 ; 4 -1/2 = 3 1/2
r=2 ; 4 -2/2 = 3
r=3 ; 4 -3/2 = 2 1/2
r=4 ; 4 -4/2 = 2
r=5 ; 4 -5/2 = 1 1/2
r=6 ; 4 -6/2 = 1
By summation, 1. Sigma (n on the top, 6 on the bottom) in the middle: (4-r/2)
= 3.5+3+2.5+2+1.5+1
=13.5
Method 2:
S1 (4 -r/2)
=S1(4) - S1(r/2)
=4(6) - 1/2 S1(r)
=24 - 1/2(n/2)(n+1)
=24 - 1/4(n)(n+1)
Considering n=6,
24 - 1/4(6)(6+1)
=24 - 1.5(7)
=13.5
Solving this type of problem will require you to have a basic knowledge on summation of series. So i suggest u read up abit on it before u attempt. Get the basic concepts rite first.
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