Posted by Math on Monday, September 17, 2012 at 2:49am.
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
"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
nth term of 9,12,15,18