Sequence and Series question?

Question

(a) The first term of a geometric series is 7, its last term is 448 and its sum is 889. Find the common ratio.

(b) Given that 1/(y-x) , 1/(2y) , and 1/(y-z) are consecutive terms of an arithmetic sequence, prove that x, y, z are consecutive terms of a geometric sequence.

TY!

Answer 1

formula fr sum of the series is
sum=a(r^n-1)/r-1
wher r- the ratio
a-first term
n-number of terms
nth term=a.r^n-1
a=7
sum=889
nth term=448
using the values n the equatrion
r=2

(b)
2.1/2.y=1/y-x+1/y-z
1/y=y-x-z/(y^2+x.z-zy-xy)
thus,
2.y^2+xz-zy-xy=y^2-yx-yz
thus
y^2=xz
hence
x, y, z are consecutive terms of a geometric sequence

Answer 2

Part a
You could use the sequence given but since everything is a multiple of 7, you can divide by 7 and work on the sequence 1, r, r^2, r^3 . . . up to 64 with sum of 127. This could lead you to see what r is immediately.

Part b.
From the definition of geometric sequence, then y/x = z/y which leads to y^2 = xz so this is what you want to prove. From the definition of arithmetic sequence, then
1/(2y) - 1/(y - x) = 1/(y - z) - 1/(2y).
You multiply each side by 2y and then by both the other denominators to eliminate fractions. Some terms will then cancel out and leave what you are trying to prove.

Answer 3

448 = 7*r^n so r^n =64 .As n must be an int eger we can choose n=2 r=8 which would'n
give 889 or r= 2 and n=6
S= 448*2-7=889 OK.
2)1/2y-1(y-x) = (-y-x)/2y(y-x) = (y+z)/2y(y-z)==> (-y-x)/(y-x)= (y+z)/(y-z)

What we have to proof is that z/y =y/x

(-y/x -1)/(y/x+1)=(z/y+1)/(1-z/y)

If we call y/x=a and z/y= b

(-1-a)/(1+a)=(1+b)/(1-b)

-1-a+b+ab=1+b+a+ab 2a=-2 a=-1 Putting -1 insteadb a we get 1+b== b=-1 so a=b

andy/x=z/y what we has to prove