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A square is drawn by joining the mid-points of the sides of a given square. A third square is drawn inside the second square in the same way, and the process continues indefinitely. If a side of the first square is 16 cm., determine the sum of the areas of all the squares.
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#6 {main}</pre>




Let ABCD be the given square of side 16cm as shown in the figure. Let EFGH be the second square drawn by joining the mid-points of the sides of given square ABCD.

         BF = EB = 8


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#6 {main}</pre>

rightwards double arrow               space space EF space equals space square root of 128 space equals space 8 square root of 2

Area of first square = <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>

Area of 2nd square = space space left parenthesis 8 square root of 2 right parenthesis squared space equals space 128 space sq. space cm.

Now, side of the third square = square root of left parenthesis 4 square root of 2 right parenthesis squared plus left parenthesis 4 square root of 2 right parenthesis squared space end root space equals space square root of 32 plus 32 end root space equals space 8

∴  Area of third square = left parenthesis 8 right parenthesis squared space equals space 64 sq. cm. Since the number of squares is infinite

∴  Sum of areas of all such squares = 256 + 128 + 64 + .......

                        equals fraction numerator 256 over denominator 1 minus begin display style 1 half end style end fraction                             open curly brackets because space straight a space equals space 256 comma space space straight r space equals space 128 over 256 space equals space 1 half close curly brackets

                   space space space space space space equals fraction numerator 256 over denominator begin display style 1 half end style end fraction space equals space 512



      
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An odd number of stones lie along a straight path, the distance between any two consecutive stones being 10 m. The stones are to the collected at the place where the middle stone lies. A man can carry only one stone at a time. He starts carrying the stones beginning from the extreme stone. If he covers a path of 3 km, how many stones are there ?

Let the total number of stones be 2n + 1

∴  There are n stones on each side of the middle stone. Let the man starts collecting stones from the extreme left stone.  The distance covered by man to bring the extreme left stone to the middle stone = 10n metres. The distance covered by man to bring the (n - 1)th stone on left to the middle stone = 2cross times 10 (n - 1) metres.

∴   The total distance covered by the man to bring all the stones from left to the middle stone 

                = 10n + 20(n - 1) + 20(n - 2) + ......... + 20

Similarly, the total distance covered by the man to bring all the stones from the right to the middle stone 

                                               = 20 + 20 (n - 1) + ......... + 20

∴     Total distance covered by the man to bring all the stones to the middle stone

                                              = 30n + 40 (n - 1) + 40 (n - 2) +........ + 40 (1)

                                            = 40 [n + (n - 1) + (n - 2) + ....... + 1] - 10n = 40 sum from straight k equals 1 to straight n of straight k minus 10 straight n

                                           equals fraction numerator 40 straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction minus 10 straight n

But the distance covered is 3 km or 3000 metres.

∴    fraction numerator 40 straight n left parenthesis straight n plus 1 right parenthesis over denominator 2 end fraction minus 10 straight n space equals space 3000 space space space space rightwards double arrow space space space 20 straight n squared plus 20 straight n minus 10 straight n space equals space 3000

rightwards double arrow     20 straight n squared plus 10 straight n minus 3000 space equals space 0     rightwards double arrow     2 straight n squared plus straight n minus 300 space equals space 0

rightwards double arrow                         space space space space space straight n equals space fraction numerator negative 1 plus-or-minus square root of 1 plus 4 left parenthesis 2 right parenthesis left parenthesis 300 right parenthesis end root over denominator 2 left parenthesis 2 right parenthesis end fraction space equals space fraction numerator negative 1 plus-or-minus square root of 2401 over denominator 4 end fraction space equals space fraction numerator negative 1 plus-or-minus 49 over denominator 4 end fraction space equals space 48 over 4 comma space fraction numerator negative 50 over denominator 4 end fraction

rightwards double arrow                             n = 12

∴      Number of stones = 2(12) + 1 = 25

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The income of a person is Rs. 3,00,000 in first year and he receives an increase of Rs. 10,000 to his income per year for next 19 years. Find the total amount, he received in 20 years. 

Here,     a = 3,00,000,  d = 10,000, n = 20


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#6 {main}</pre> open curly brackets By space using space straight S subscript straight n space equals space straight n over 2 left parenthesis 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d close curly brackets

                                     = 10 (7,90,000) = 79,00,000.

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