as geometric sequence help please

[tanmaydube]

q1) If x, y and z are the first three terms of a geometric sequence, show that x^2, y^2 and z^2 form another geometric sequence.

q2)Different numbers x,y and z are the first three terms of a geometric progression with the common ratio r, and also the first, second and forth terms of an arithmetic progression.
(a)Find the value of r.(ans=2)
(b)Find which term of the arithmetic progression will next be equal to a term of the geometric progression.(ans=8th)

q3) Consider the geometric progression
q^n-1 + q^n-2p + q^n-3p^2+.....+qp^n-2+p^n-1
(a) Find the common ratio and the numbers of terms.(ans=p/q , n)
(b) Show that the sum of the series is equal to ((q^n-p^n)/q-p) (ans= np^n-1)
(c) By considering the limit as q→p deduce expressions for f'(p) in the case (f inverse of p) in this cases
(i) f(x)=x^n (ans= np^n-1), (ii) f(x)=x^-n (ans= -np^-(n+1) ), for all positive integers n.

THANK YOU FOR HELPING!

 

[Nikesh]

Answer to question 1

When x, y , z are in geometric sequence we get,
y/x = z/y

To show x^2 , y^2 , z^2 in G.P ,
2ndterm(t2)/t1 = t3/t2
i.e. y^2/x^2 = z^2 / Y^2

your answer will be y/x= z/y
thats it you got the equation equal to the given one.........Thus they are too in geometric sequence

 

[Nikesh]

Question 2 and 3 are not so cleared.........Is the given questions correct, please check the question and post a reply

 

[mrpaudel]

QN2) a) x, y and z are 1st 2nd and 4th term in A.P so, a+d/a = a+3d/a+d ....and..get d=0 and d=a.....d=0 is not possible....whereas d=a is valid..so put Since r=a+d/a= a+a/a = 2...

b) and..i didnot get this question..., i thot it was a+(n-1)d=ar^(n-1) ...and i got 2n=2^n...and...for n=8..it doesnot satisfy this one....so i thought i didnot get the question properly...!!

QN3) can u scan the question and put it here...!! difficult to read..!!

 

[Nikesh]

QN2) b. now you have value of d and r so find next value in GP after z
i.e. In G.P. t4=ar^3 which gives 4th term = 8x (Since a=x and r=2)
now
In A.P
tn = a + (n-1)d
for a=d=x
tn = x + (n-1)x
that gives tn=nx
By question
tn=8x (next term in A.P which is equal to a term in G.P, all three given terms of GP are already matched in A.P as 1st 2nd and 4th term in AP)
or, nx=8x
it gives n=8.......