Arithmetic Progression
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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?

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  (n1)(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