Sequences

The 1st, the 3rd, and the 7th terms of an arithmetic sequence withdistinct terms are in geometric sequence. The sum of these three termsis 28.

a) Find these three terms.

let T(1)=a, T(3)=a+2d, T(7)=a+6d ,

The 1st, the 3rd, and the 7th terms of an arithmetic sequence withdistinct terms are in geometric sequence
=>(a+2d)/(a)=(a+6d)/(a+2d)
=>(a+2d)^2=a(a+6d)--(1)

The sum of these three terms is 28
=> T(1)+T(3)+T(7)=28
=>3a+8d=28

solve (1)and(2)

we get a=4 ,d = 2

T(1)=a=4
T(3)=a+2d=4+2x4=8
T(7)=a+6d=4+2x6=16//




2008-06-09 15:58:51 補充：
b)Find the common difference of the arithmetic sequence.

common difference=d=2//

C) Find the common ratio of the geometric sequence.

the arithmetic sequence=T(7)/T(3) or T(3)/T(1)=16/8 or 8/4 = 2//

Let a be the first term d be the common difference
the sequence = a+(n-1)d

1st term = a
3rd term = a+2d
7th term = a+6d

a) , b)

a+a+2d+a+6d = 28
3a+8d=28 ----- 1

(a+2d)/a = (a+6d)/(a+2d)
(a+2d)(a+2d) = a(a+6d)
a^2+4ad+4d^2 = a^2 + 6ad
4d^2 - 2ad = 0
2d^2 - ad = 0
d(2d-a) = 0
2d-a = 0 or d=0 (rejected)
2d-a = 0
a = 2d ----- 2

sub 2 into 1

3(2d) + 8d = 28
14d = 28
d = 2

sub d=2 into 2
a = 2(2)
a = 4

therefore the 1st term is 4,
the third term is 4+2(2)=8,
the 7th term = 4+6(2) = 16

Common difference is 2

c)
Common ratio of the geometric sequence is 8/4 = 16/8 = 2