IIT JEE , AIEEE Forum - Mathematics , Algebra - Two salutes are waiting .... reply quickly....

amay

If the first and 2n+1 th terms of an AP,GP,HP of positive terms are equal,and their n+1 th terms are a,b,c, respectively, then:

a)a=b=c b)a>=b>=c c)a+c=2b d)ac=b^2

Q2)if b-c, 2b-

, b-a are in AP, then a-

/2, b-

/2, c-

/2 are in:

a)AP b)GP c)HP d) none

give a quick reply

kghedriu

hii, i think i got the answer..

A1)

let tn represent n th term....

say t1=x and t(2n+1)=y

so, x,a,y---------AP

x,b,y---------GP

x,c,y----------HP

hence, a>=b>=c and b^2=ac i.e (b,d)

(.: x,y,z are positive and apply am>=gm>=hm)

ionic

1.clearly every odd term of progressions are equal.
also the AM GM HM are infact n+1 th term (for 2n+1 terms)
which follow a simple relation
GM^2=AM.GM
hence correct answer is
b^2=a.c

kghedriu

say a-

/2 = x , b-

/2=y and c-

/2=z

so, we r given that....

y-z, 2y, y-x are in HP

so, 2y= 2(y-z)(y-x)/2y-z-x

i.e y^2=xz...hence tha ans is (b)

kghedriu

hey ppl and amay, if my ans is right, pls gimme ur rate..

Arjun

for the second qustion...

let a-

/2=p

b-alpha/2=q

c-alpha/2=r

rearranging the given series we get q-r,2q,q-p in A.P

then 4q=q-r+q-p

4q=2q-p-r

2q=p+r

therfore the answer is they r in AP

amay

Thanx a lot.

all the three answers were correct.

also try the two qns posted by me in other topic.

as promised, i gave you votes.

amay

arjun, sorry

in your sol we get 2q= -p-r so, kghedrui's ans is right.

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