Friday

May 6, 2016
Posted by **Math** on Monday, September 17, 2012 at 2:49am.

a. Find the fist term and the common difference.

b. How many times are needed to give a sum that is less than 65?

- Arithmetic Progression -
**Bosnian**, Monday, September 17, 2012 at 4:45amArithmetic progression :

an = a1 + ( n - 1 ) * d

In this case :

a6 = a1 + ( 6 - 1 ) * d

a6 = a1 + 5 d

a4 = a1 + ( 4 - 1 ) *d

a4 = a1 + 3 d

a6 = ( 1 / 2 ) a4

a6 = ( 1 / 2 ) ( a1 + 3 d )

a6 = a1 / 2 + 3 d / 2

a6 = a6

a1 + 5 d = a1 / 2 + 3 d / 2 Substract 5 d to both sides

a1 + 5 d - 5 d = a1 / 2 + 3 d / 2 - 5 d

a1 = a1 / 2 + 3 d / 2 - 10 d / 2

a1 = a1 / 2 - 7 d / 2 Substract a1 / 2 to both sides

a1 - a1 / 2 = a1 / 2 - 7 d / 2 - a1 / 2

a1 / 2 = - 7 d / 2 Multiply both sides by 2

a1 = - 7 d

a3 = a1 + ( 3 - 1 ) * d

a3 = a1 + 2 d

a3 = 15

15 = a1 + 2 d

15 = - 7 d + 2 d

15 = - 5 d Divide both sides by - 5

15 / - 5 = d

- 3 = d

d = - 3

a1 = - 7 d

a1 = - 7 * ( - 3 ) = 21

a1 = 21

a2 = a1 + ( 2 - 1 ) * d = a1 + d = 21 + ( - 3 ) = 21 - 3 = 18

a3 = a1 + ( 3 - 1 ) * d = a1 + 2 d = 21 + 2 * ( - 3 ) = 21 + ( - 6 ) = 21 - 6 = 15

a4 = a1 + ( 4 - 1 ) * d = a1 + 3 d = 21 + 3 * ( - 3 ) = 21 + ( - 9 ) = 21 - 9 = 12

a5 = a1 + ( 5 - 1 ) * d = a1 + 4 d = 21 + 4 * ( - 3 ) = 21 + ( - 12 ) = 21 - 12 = 9

a6 = a1 + ( 6 - 1 ) * d = a1 + 5 d = 21 + 5 * ( - 3 ) = 21 + ( - 15 ) = 21 - 15 = 6

AP:

21 , 18 , 15 , 12 , 9 , 6

b.

The sum of the n members of a arithmetic progression :

Sn = ( n / 2 ) ( a1 + an )

In this case :

65 < ( n / 2 ) ( a1 + an )

65 < ( n / 2 ) * ( 21 + an ) Multiply both sides by 2

2 * 65 < 2 * ( n / 2 ) * ( 21 + an )

135 < n * ( 21 + an )

Sn = ( n / 2 ) ( a1 + an )

For n = 6

S6 = ( 6 / 2 ) ( 21 + 6 )

S6 = 3 * 27 = 81

81 > 65 That is not solution

For n = 5

S5 = ( 5 / 2 ) ( 21 + 9 )

S5 = ( 5 / 2 ) * 30 = 150 / 2 = 75

75 > 65 That is not solution

For n = 4

S4 = ( 4 / 2 ) ( 21 + 12 )

S4 = 2 * 33 = 66

66 > 65 That is not solution.

For n = 3

S3 = ( 3 / 2 ) ( 21 + 15 )

S3 = ( 3 / 2 ) * 36 = 108 / 2 = 54

54 < 65 That is solution.

a1 + a2 + a3 = 21 + 18 + 15 = 54 - Arithmetic Progression -
**Reiny**, Monday, September 17, 2012 at 8:18am"the 6th term is half the 4th term" ---> a+5d = (1/2)(a+3d)

2a + 10d = a + 3d

a + 7d = 0

"the 3rd term is 15. " ---> a + 2d = 15

subtract those two equations ...

5d = -15

d = -3

sub into

a+ 2d = 15 ---> a - 6 = 15

a = 21**a = 21 , b = -3**

let the number of terms be n

(n/2)[42 - (n-1)(-3) < 65

n [ 42 + 3n - 3 ] < 130

3n^2 + 39n -130 < 0

Consider 3n^2 + 39n - 130 = 0

n = 2.75 or a negative

but n must be a whole number

let n=3

21 + 18 + 15 < 65

3 terms are needed - Arithmetic Progression -
**Anonymous**, Monday, October 26, 2015 at 6:17pmnth term of 9,12,15,18