SEQUENCE AND SERIES A10

The 1st term of a geometric progression is 2 and the 5th term is 162. Find the common ratio.

Solution

G₁=2, G₅=162, n=5

G_n = G₁r^n-1;

G₅ = G₁r^5-1;

G₅ = G₁r⁴;

162 = 2 x r⁴;

162 = 2r⁴;

 2       2

81 = r⁴;  Finding the fourth root of 81,

r = 3

Hence the common ratio is 3.

REPEATING DECIMAL A7

MATRIX A8

REPEATING DECIMAL A6

SEQUENCE AND SERIES A9

The sum of the 1st five terms of a geometrical progression is 484. If the common ratio is 3 find the 1st term.

solution

S₅ = 484 G₁ = ?, r=3, n=5

S_n = G₁(rⁿ-1)/r-1

               

S₅ = G₁(r⁵-1)/r-1

               

484= G₁(3⁵-1)/3-1

               

484= G₁(243-1)

                  2

2 x 484= G₁(242)

                  

968= 242 G₁

⁴ 968   =  G₁

  242

G₁ = 4

Hence the 1st term is 4.

STATISTICS A3

The masses of 8 water melons in kilograms were recorded as follows:

8, 4, 3, 6, 7, 2, 1,  and 1. Calculate the median of these data.

Solution

Arrange the data in order of magnitude starting with the smallest.

1, 1, 2, 3, 4, 6, 7, 8

There are eight data so far. We get the two data in the middle which are 3 and 4.

Median can be calculated by finding the average of the two middle values.

so, median  = 3 + 4

                        2

                      = 3.5

Hence median is 3.5.

SIMULTANEOUS EQUATIONS A3

Solve the following equations by substitution method.

6x + y = 15

x +  3y = 11

solution

6x  + y = 15----------- (i)

x + 3y = 11---------- (ii)

from equation (i)

6x + y = 15

y = 15 - 6x ----------(iii)

Substitute equation (iii) in (ii) above,

x + 3(15 - 6x) = 11

x + 45 - 18x = 11

x - 18x = 11- 45

-17x  = -34

-17x  = -34

-17       -17

x = 2

From equation (iii)

y = 15 - 6x ----------(iii)

y = 15 – 6(2)

y = 15 – 12

y = 3

Hence x=2 and y=3.

RADIANS 5

SEQUENCE AND SERIES A8

Find the sum of the 1st eight terms of the geometrical progression 2+4+8+16+…….

solution

G₁ = 2, r=2, n=8

S_n = G₁(rⁿ-1)/r-1

               

S₈ = 2(2⁸-1)/2-1

             

S₈ = 2(2⁸-1)/1

              

S₈ = 2(2⁸-1)

S₈ = 2(256-1)

S₈ = 2(255)

S₈ = 510

Hence the sum of the 1st eight terms is 510.

POLYNOMIALS A1

When mx³+nx²-18x-45 is divided by x²-9 the remainder is zero. Calculate the values of m and n.

solution

x²-9 = 0

(x)²-(3)²=0

(x+3)(x-3) = 0

x+3= 0  OR  x-3 = 0

x= 3  OR  x= -3

When x=3,

 m(3)³+n(3)²-18(3)-45=0

27m + 9n -54 -45 = 0

27m + 9n -99 = 0

27m + 9n = 99  …………………..(1)

When x=-3,

 m(-3)³+n(-3)²-18(-3)-45=0

-27m + 9n + 54 - 45 = 0

-27m + 9n + 9 = 0

-27m + 9n = -9 ………………………………(2)

Solving 1 and 2 simultaneously,

  27m + 9n = 99  …………………..(1)

-27m + 9n = -9 ………………..……(2)

By using elimination method,

      27m + 9n = 99  

-  (-27m) + 9n = -9

    54m + 0    = 108

54m = 108

54         54

m = 2

from equation (1),

27(2) + 9n = 99  

54 + 9n = 99  

9n = 99 - 54

9n = 45

9n = 45

9       9

n=5

Hence m=2 and n=5.