IIT JEE , AIEEE Forum - Mathematics , Algebra

tarangini

1] sum of 1st 2 terms of an

gp is 1 & every term is twice d sum of qll d successuve terms,then its 1st terms is ...

2] if s be d sum p d prod r d sum of reciprocals of n terms of gp, then p²=....

s\r

r\s

[r\s]ⁿ

[s\r]ⁿ

3]a,b,c, in gp & a^1\x=b^1\y=c^1\z, x,y,z are???

4 x+y\2,y,y+z\2 r in hp then x,y,z. r in....

ap

gp

hp

none????

vinu

1] given a(1+r) = 1 ;

sum (S) = a/(1-r) ;

now , u've a = 2[S-a] ;

so, r = 1/3 ;

hence,frm the first eq,......a = 3/4..ans.

2]

funky

2)
sum of first n terms of a gp is
a(r^n-1)/(r-1)
the terms in gp are
a,ar,ar^2,ar^3,...............ar^(n-1)
product of first n terms are a^n r^(1+2+3+4..........n-1)
or Pn=a^n*r^(n[n-1]/2)

Rn=1/a+1/ar+1/ar^2...........1/ar^(n-1) 1
dividing by r
Rn/r=1/ar+1/ar^2+1/ar^3................1/ar^n 2
subtracting 2-1
Rn([1/r]-1)=1/a([1/r^n]-1)
Rn=(1/ar^[n-1])*([r^n]-1/r-1)

hence on substituting the values we get p^2=[s/r]^n

3)let a^1/x=b^1/y=c^1/z=k
k^x=a,k^y=b,k^z=c
using prop of gp
b^2=ac
k^2y=k^x*k^z
k^2y=K^(x+z)
when bases are same powers can be equated
2y=x+z
x,y,z are in AP

4)
usin prop of HP [if a,b,c are in HP then 1/a + 1/c=2/b
2/(x+y) + 2/(y+z)=2/y
y(2y+x+z)=(x+y)*(y+z)
2y^2+xy+yz=xy+xz+y^2+yz
y^2=xz
x,y,z are in GP

the first sum has already been solved by vinu

funky

2)
sum of first n terms of a gp is given y
a(r^n-1)/(r-1)
first n terms of gp are
a,ar,ar^2,ar^3..................ar^n-1
product of n terms is
Pn=a^n r^(1+2+3.............(n-1)
Pn=a^n r^[n(n-1)/2]

Rn=1/a+1/ar+1/ar^2+1/ar^3................1/ar^n-1
divide both sides by r
Rn/r=1/ar+1/ar^2+1/ar^3......................1/ar^n-1
subtract second eq from first
Rn([1/r]-1)=(1/a) *([r^n]-1/r^n)
Rn=[1/ar^(n-1)]*([r^n]-1/r-1)

on substituting the above values p^2=[s/r]^n

3)let a^1/x=b^1/y=c^1/z=k
k^x=a,k^y=b,k^z=c
by gp property
b^2=ac
k^2y=k^(x+z)
when bases are same their powers can be equated
2y=x+z
x,y,z are in AP

4)by HP property 1/a +1/c =2/b
2/(x+y) + 2/(y+z) =2/y
y(2y+x+z) = (x+y)*(y+z)
2y^2+xy+yz=xy+xz+y^2+yz
y^2=xz
x,y,z are in GP

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