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rautela

if p,q,r,s are 4 consecutive terms of an ap then pth , qth,rth ,sth terms of a gp are in

ap,gp,hp,none of these

3^(x-1) +3^(x-2)+3^(x-3)+.=2(5^2+5+5^( -1)+ ...)

find x

if x^2(y+z) ,y^2(z+x),z^2(x+y) are in ap then x,y,z are in

puneet

hii

i ll answer this one ..

3^(x-1) +3^(x-2)+3^(x-3)+.=2(5^2+5+5^( -1)+ ...)

find x

3^x ( 3^-1 + 3^-2 + ... ) = 2. (5^2 + 5 + ..)

3^x( 1/3 / ( 1- 1/3)) = 2. 5^2( 1 - 1/5)

3^x = 125

so solve for x now

cheers

Decoder

hi bhaiya...
how to do the first one..

Avinash_Bhat

Solution to 1st question.

Let :
            p = k - 3d
            q = k - d
            r = k + d
            s = k + 3d

For G.P. pth term = ac^p-1 = ac^{k - 3d - 1} = ac^{k - 1} / c^3d
               qth term = ac^q-1 = ac^{k - d - 1} = ac^{k - 1} / c^d
               rth term = ac^r-1 = ac^{k + d - 1} = ac^{k - 1}c^d
               sth term = ac^s-1 = ac^{k + 3d - 1} = ac^{k - 1} c^3d

These are in G.P. with common ratio c^2d.

Avinash_Bhat

Solution to 3rd question.

x²(y + z) , y²(x + z) , z²(x + y)       are in A.P.

x²(y + z) / xyz , y²(x + z) / xyz , z²(x + y) / xyz     are in A.P.

x(1/y + 1/z) , y(1/x + 1/z) , z(1/x + 1/y)      are in A.P.

x(1/y + 1/z) + 1 , y(1/x + 1/z) + 1 , z(1/x + 1/y) + 1    are in A.P.

x(1/y + 1/z) + x/x , y(1/x + 1/z) + y/y , z(1/x + 1/y) + z/z    are in A.P.

x(1/x + 1/y + 1/z) , y(1/x + 1/y + 1/z) , z(1/x + 1/y + 1/z)    are in A.P.

x , y , z    are in A.P.

ayshwarya

hi puneet check ur sol. 1st u 've commited a mistake its actually coming as
3^x=117
solve thisplzzzzzzzzzzzzz

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