Question

In an AP the 6th term is half the 4th term and the 3rd term is 15.

a. Find the fist term and the common difference.

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

Answer 1

Arithmetic 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

Answer 2

"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
<b>a = 21 , b = -3 </b>

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

Answer 3

nth term of 9,12,15,18

Answer 4

If 3rd tarm of an AP is four times the first team and 6th term is 17 find the AP