Arithmetic progressions-nth terms, formulas,sum of nth term (MTH 112)

MTH 112
For Example
1, 4, 7, 10, 13,…………21
We have some important terms we’ve to know. Some of them are common difference (c.d), first term (a), number of terms (n+1)
nth = a+(n-1)d
L = a+(n-1)d
To get common difference, you’ll subtract 2nd term from 1st term or 3rd term from 2nd term (2nd term- 1st term or 3rd term- 2nd term)
First term (a) = 1 ( i.e the first number)
9th = a + (n-1)d
= 1 + (9-1)3
= 1 + (8)3
= 1 + 24
= 25
a, a+d, a+2d, a+3d……..
Common difference = d
Sum of the nth term = n(a+L)/2
Substitute L as a + (n-1)d
S terms = n(a+[a+(n-1)d]/2
= n(a+a+(n-1)d)/2
=n(2a+(n-1)d)/2
=

(1) Find 8th terms of the series and 8th sum of the series.1,4,7,10,13…………….

Sum of the 8th term = n/2[ 2a+(n-1)d]
= 8/2[(2×1) +(8-1)3]
= 4[2+7•3]
= 4[2+21]
= 4×23 ; = 92

(2) Given an arithmetic sequence as -5,x,y,z,3………. obtain the value of x,y and z
SOLUTION

Nth term of 8 = a + (n-1)d

a = -5 ……………..1st term
a+d = x ……………2nd term
a+2d = y ………….3rd term
a+3d = 2 ………….4th term
a+4d = 3 ………….5th term
-5+4d = 3
4d = 3+5
4d = 8
d= 2
Therefore,
X = a+d
= -5+2
= -3
Y = a+2d
=-5+2(2)
=-5+4
=-1
Z = a+3d
= -5+3(2)
= -5+6
=1
The value of x,y and z = (-3,-1 and 1)

(3) How many terms of the following series may be taken so that their sum is 66 in -9, -6 , -3……..

Sn = n/2[2a+(n-1)d]
66 = n/2[2a+(n-1)dCommon difference = -6-(-9)
d= 3
66 = n/2[2(-9)+(n-1)3]
66 = n/2[-18+3n-3]
66 = n/2[3n-21]
66 = (3n²-21)/2
(3n²-21n)/2 =66
Cross multiply both sides together
3n²-21= 132
3n²-21-132=0
Divide through by 3
n²-7-44=0
Using factorization method
n²-7n-44=0
n²-11n+4n-44=0
(n²-11n)+(4n-44)=0
n(n-11)+4(n-11)=0
n+4=0 ; = -4
n-11=0 ; = 11
So therefore n=-4 or 11
The number of terms are -4 and 11

(4) if x = acosø and y = bsinø. Show that X²/a² + y²/b² = 1
SOLVE

x = acosø and y = bsinø
x² = a²cos²ø and y² = b²sin²ø
To find a²
x² = a²cos²ø(divide both sides by cos²ø)
X²/cos∅ = a²
a² =X²/cos∅
To find b²
y² = b²sin²ø
b²=y²/sin²∅
So let’s substitute them into the equation
X²/a² + y²/b²
a²cos²ø÷x²/cos²ø + b²sin²ø÷y²/sin²ø
a²cos²ø×cos²ø/x² + b²sin²ø×sin²ø/y²
Recall, y² = b²sin²ø, x² = a²cos²ø
a²(cosø)⁴/x² + b²(sinø)⁴/y²
a²cos⁴∅/a²cos²∅ + b²sin⁴∅/b²sin²∅ ( a² and b² will cancel out)
Cos⁴∅/cos²∅ + sin⁴∅/sin²∅
Cos²ø + sin²ø
Recall from trigonometry identities
Cos²ø + sin²ø = 1
So therefore, X²/a² + y²/b² = 1

Find 8th terms of the series and 8th sum of the series.1,4,7,10,13…………….