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111. If a, b, c, d are in GP., show that (a² + b² + c²) (b² + c² + d²) = (ab + bc + cd)²

Here, a, b, c, d are in G.P. Let r be the common ratio of the G.P.

∴ b = ar, c = ar², d = ar³

Now, $left parenthesis straight a squared plus straight b squared plus straight c squared right parenthesis space left parenthesis straight b squared plus straight c squared plus straight d squared right parenthesis space equals space left parenthesis ab plus bc plus cd right parenthesis squared$

if $left parenthesis straight a squared plus straight a squared straight r squared plus straight a squared straight r to the power of 4 right parenthesis space left parenthesis straight a squared straight r squared plus straight a squared straight r to the power of 4 plus straight a squared straight r to the power of 6 right parenthesis space equals space left parenthesis straight a. space ar space plus space ar. space ar squared space plus space ar squared. space ar cubed right parenthesis squared$

or if $straight a squared left parenthesis 1 plus straight r squared plus straight r to the power of 4 right parenthesis straight a squared straight r squared space left parenthesis 1 plus straight r squared plus straight r to the power of 4 right parenthesis space equals space left parenthesis straight a squared straight r plus straight a squared straight r cubed plus straight a squared straight r to the power of 5 right parenthesis squared$

or if $space space space space space straight a to the power of 4 straight r squared left parenthesis 1 plus straight r squared plus straight r to the power of 4 right parenthesis squared space equals space left square bracket straight a squared straight r left parenthesis 1 plus straight r squared plus straight r to the power of 4 right parenthesis right square bracket squared$

or if $straight a to the power of 4 straight r squared left parenthesis 1 plus straight r squared plus straight r to the power of 4 right parenthesis squared space equals space straight a to the power of 4 straight r squared left parenthesis 1 plus straight r squared plus straight r to the power of 4 right parenthesis squared$

which is true.

Hence, if a, b, c d are in G.P. then $left parenthesis straight a squared plus straight b squared plus straight c squared right parenthesis space left parenthesis straight b squared plus straight c squared plus straight d squared right parenthesis space equals space left parenthesis ab plus bc plus cd right parenthesis squared$

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Long Answer Type

112.

If a, b, c are in G.P. and x, y are the arithmetic means of a, b and b, c respectively,

then prove that $straight a over straight x plus straight c over straight y equals 2$ and $space space 1 over straight x plus 1 over straight y equals 2 over straight b$

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Short Answer Type

113.

If a, b, c are in A.P. and x,y,z are in G.P., show that

$straight x to the power of straight b minus straight c end exponent. space straight y to the power of straight c minus straight a end exponent. space straight z to the power of straight a minus straight b end exponent space equals space 1$