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If two G.M.'s $<pre>uncaught exception: mkdir(): Permission denied (errno: 2) in /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php at line #56mkdir(): Permission denied in file: /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php line 56 #0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array) #1 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/util/sys/Store.class.php(56): mkdir('/home/config_ad...', 493) #2 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/FolderTreeStorageAndCache.class.php(110): com_wiris_util_sys_Store->mkdirs() #3 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/RenderImpl.class.php(231): com_wiris_plugin_impl_FolderTreeStorageAndCache->codeDigest('mml=<math xmlns...') #4 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/lib/com/wiris/plugin/impl/TextServiceImpl.class.php(59): com_wiris_plugin_impl_RenderImpl->computeDigest(NULL, Array) #5 /home/config_admin/public/felixventures.in/public/application/css/plugins/tiny_mce_wiris/integration/service.php(19): com_wiris_plugin_impl_TextServiceImpl->service('mathml2accessib...', Array) #6 {main}</pre>$ and one A.M. 'A' be inserted between two numbers, show that

$2 straight A space equals space fraction numerator straight g subscript 1 superscript 2 over denominator straight g subscript 2 end fraction space plus space fraction numerator straight g subscript 2 superscript 2 over denominator straight g subscript 1 end fraction$

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102.
If a be the A.M. and x, y be the two G.M.'s between b and c, show that

$straight x cubed plus straight y cubed space equals space 2 abc$

Since a is the A.M. between b and c

∴ $space space straight a space equals space fraction numerator straight b plus straight c over denominator 2 end fraction$ ...(i)

Since x, y are two G.M.'s between b and c

∴          b, x, y, c are in G.P.

Let r be the common ratio

                       $straight x equals br comma space space straight y space equals space br squared comma space space straight c space equals space br cubed$

                 L.H.S. = $straight x cubed plus straight y cubed space equals space left parenthesis br right parenthesis cubed space plus space left parenthesis br squared right parenthesis cubed space equals space straight b cubed straight r cubed space plus space straight b cubed straight r to the power of 6$

                           $space space equals straight b squared left parenthesis br cubed right parenthesis space plus space straight b left parenthesis br cubed right parenthesis squared space equals space straight b squared straight c plus bc squared$                      $open curly brackets because space straight c space equals space br cubed close curly brackets$

                             = bc (b + c) = bc(2a)                                 {By using (i)}

                             = 2abc = R.H.S.

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103.

The sum of first three terms of a G.P. is $13 over 12$ and their product is -1. Find the terms.

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104.

The product of three numbers in G.P. is 125 and the sum of their products taken in pairs is $87 1 half.$ Find them.

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Long Answer Type

105.

The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

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Short Answer Type

106. Find four numbers forming a GP. in which the third term is greater than the first by 9 and the second term is greater than the fourth by 18.

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Long Answer Type

107. The sum of three numbers in GP. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression. Find the numbers.

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Short Answer Type

108. The sum of three numbers which are consecutive terms of an A.P. is 21. If the second number is reduced by 1 and the third is increased by 1, we obtain three consecutive terms of a GP. Find the numbers.

725 Views