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Permutations And Combinations

Question
CBSEENMA11013018

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of the 7th and (m – l)th numbers is 5 : 9. Find the value of m.

Solution

Let straight A subscript 1 comma space straight A subscript 2 comma space straight A subscript 3 comma........... straight A subscript straight m be m A.M.'s between 1 and 31.

∴ 1, straight A subscript 1 comma space straight A subscript 2 comma space straight A subscript 3 comma space space....... comma space straight A subscript straight m comma space 31 are in A.P.
Here,                    a = 1,
Let d be the common difference
straight t subscript straight m plus 2 end subscript space equals space 31  or    a + (m+ 2 - 1)d = 31    or     1 + (m + 1)d = 31    or     <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>
straight A subscript 7 space equals space straight t subscript 8 space equals space straight a plus 7 straight d space equals space 1 plus 7 open parentheses fraction numerator 30 over denominator straight m plus 1 end fraction close parentheses space equals space fraction numerator straight m plus 1 plus 210 over denominator straight m plus 1 end fraction space equals space fraction numerator straight m plus 211 over denominator straight m plus 1 end fraction
straight A subscript straight m minus 1 end subscript space equals space straight t subscript straight m space equals space straight a space plus space left parenthesis straight m minus 1 right parenthesis straight d space equals space 1 plus left parenthesis straight m minus 1 right parenthesis open parentheses fraction numerator 30 over denominator straight m plus 1 end fraction close parentheses space equals space fraction numerator straight m plus 1 plus 30 straight m minus 30 over denominator straight m plus 1 end fraction equals fraction numerator 31 straight m minus 29 over denominator straight m plus 1 end fraction
Now,    straight A subscript 7 over straight A subscript straight m minus 1 end subscript space equals space 5 over 9               or                       space space fraction numerator begin display style fraction numerator straight m plus 211 over denominator straight m plus 1 end fraction end style over denominator begin display style fraction numerator 31 straight m minus 29 over denominator straight m plus 1 end fraction end style end fraction equals space 5 over 9
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#6 {main}</pre>            or                        9m + 1899 = 155m - 145
or             146m = 2044             or                                      m = 14
Hence,    m = 14

Some More Questions From Permutations and Combinations Chapter

Determine K, so that K + 2, 4K – 6 and 3K – 2 are three consecutive terms of an A.P.