The ratio of the sum of first three terms is to that of first 6

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 Multiple Choice QuestionsShort Answer Type

81.

How many terms of the sequence

space space square root of 3 comma space 3 comma space 3 square root of 3 comma space........ must be taken to make the sum <pre>uncaught exception: <b>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</b><br /><br />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<br />#0 [internal function]: _hx_error_handler(2, 'mkdir(): Permis...', '/home/config_ad...', 56, Array)
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#6 {main}</pre>?

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82.

Sum the following series:

7 + 77 + 777 + ......... to n terms

84 Views

83.

Sum the following series:

0.5 + 0.55 + 0.555.......... to n terms



1078 Views

84.

Given a G.P. with a = 729 and 7th term 64, determine S7.

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85.

Determine the number of terms in a GP., if t1 = 3, tn = 96 and Sn = 189.

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86.

Find the least value of n for which the sum 1 + 3 + 32+ ...... to n terms is greater than 7000.

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87.

The ratio of the sum of first three terms is to that of first 6 terms of a G.P. is 125 : 152. Find the common ratio.


Let a be the first term and r be the common ratio

According to the given equation,

straight S subscript 3 over straight S subscript 6 space equals space 125 over 152 space space space space space space space rightwards double arrow space space space space space fraction numerator begin display style fraction numerator straight a left parenthesis straight r cubed minus 1 right parenthesis over denominator straight r minus 1 end fraction end style over denominator begin display style fraction numerator straight a left parenthesis straight r to the power of 6 minus 1 right parenthesis over denominator straight r minus 1 end fraction end style end fraction space equals space 125 over 152 space space space space space space space space space space space rightwards double arrow space space space space fraction numerator straight r cubed minus 1 over denominator straight r to the power of 6 minus 1 end fraction space equals space 125 over 152

rightwards double arrow space fraction numerator straight r cubed minus 1 over denominator left parenthesis straight r cubed minus 1 right parenthesis left parenthesis straight r cubed plus 1 right parenthesis end fraction space equals space 125 over 152 space space space space rightwards double arrow space space space fraction numerator 1 over denominator straight r cubed plus 1 end fraction space equals space 125 over 152 space space space space rightwards double arrow space space space space 125 straight r cubed space plus space 125 space equals space 152


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#6 {main}</pre>

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88.

If S1, S2 and S3 be respectively the sums of n, 2n and 3n terms of a GP., prove that S1 (S3 – S2) = (S2 – S1)2

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 Multiple Choice QuestionsLong Answer Type

89.

Let S be the sum, P the product and R the sum of the reciprocals of n terms in a G.P. Prove that straight P squared straight R to the power of straight n space equals space straight S to the power of straight n.

131 Views

 Multiple Choice QuestionsShort Answer Type

90.

Evaluate:

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#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...')
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#6 {main}</pre>

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