Questions about arithmetic and geometric sequence and series?

Question

1) Show that the sequence formed by squaring each term of the sequence a1,a1r,a1r^2,a1r^3, ..... is also a geometric sequence.


2) The numbers a,12 and b form an arithmetic sequence and b>a>0. If a+1,16,b+43, is a geometric sequence find the values of a and b

3) The sum of 3 numbers which form an arithmetic sequence is 36. If 1, 4 and 43 are added to each term respectively the resulting sequence is geometric. Find three numbers

4) If x, y, z form an arithmetic sequence show that 2^x, 2^y, 2^z form a geometric sequence.

Answer 1

I'll show you how to set 'em up, but *you* do the arithmetic

(1)
Clearly a1 is also a1*r^0 so we have

a1r^0, a1r^1, a1r^2, ... and the ratio of each term is r. Now, squaring each term gives

a1^2*r^0, a1^2*r^2, a1^2*r^4 and the ratio between terms will be r² so it's still a geometric sequence.

(2)
21-a = b-12 and 16/(a+1) = (b+43)/16. Do a bit of juggling and substitution.

(3)
Let the numbers be a, b, and c. Then
a+k = b and b+k = c (by definition of an arithmetic sequence. Also 3a+2k = 36 and
(b+4)/(a+1) = (c+43)/(b+4)
Again, juggle, substitute, and solve.

(4)
Use the laws of exponents. In particular
(x^a)/(x^b) = x^(b-a) and remember the definitions of arithmetic and geometric sequences


Doug

Answer 2

Dough_dounaghue’s solution is correct. Solution that I have given below which I think is easily comprehendible to the questions 2) & 3)
Equation to the sum of a series of AP is Sn=n/2[a+l] ;where n is the number of terms, a and l are the first and last terms respectively.
Also when n is an odd number we have Sn = n* (Mid -term of the series)
Consequently in the given problem, 36= 3*b (since b is the mid-term of the sequence)
Hence the mid term b= 12
Therefore sequence can be written as, (12-d) , 12, (12+d); and this sequence has no unique solution. ie. d is a whole number > = 0 , usually < 12
Given that when 1, 4 , 43 are added respectively to the sequence it forms a GMS.
Ie. (13-d), 16, (55+d) is a GM.
Consequently, 16/ (13-d) = (55+d)/16
Even by trial and error method we can find that d=9
If X, Y, Z, are in GMS
Then, X= (13-d)=13-9=4 ; & Y=16 , Z=(55+9) =64 Which is in GMS.
The three numbers are 4, 16,64