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…………….

nth term of the 8th term(n8th) = a+(n-1)d

Common difference= 2nd term-first term

d= 4-1; = 3

= 1+(8-1)3

= 1+(7)3

= 1+21

n8th = 22

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…………….