Question

If the sum of the first 10 positions in class $(1 \frac{3}{5} ) {}^{2} + (2 \frac{2}{5} ) {}^{2} + (3 \frac{1}{5} ) {}^{2} + {4}^{2} + (4 \frac{4}{5} ) {}^{2} + .........$is $\frac{16}{5} m$ then the value of m is.



(a) 102. (b) 101

(c) 100. (d) 99

Answer 1

$\bold{\huge{\underline{Question}}}$


If the sum of the first 10 positions in class $(1 \frac{3}{5} ) {}^{2} + (2 \frac{2}{5} ) {}^{2} + (3 \frac{1}{5} ) {}^{2} + {4}^{2} + (4 \frac{4}{5} ) {}^{2} + .........$is $\frac{16}{5} m$ then the value of m is.

(a) 102. (b) 101

(c) 100. (d) 99





$\bold{\huge{\underline{Solution-}}}$



Let the sum of the first 10 positions is $S_{10}$



then,

$S_{10} = \frac{16}{5} m \\ \\ \\ = (1 \frac{1}{3} ) {}^{2} + (2 \frac{2}{5} ) {}^{2} + (3 \frac{1}{5} ) { }^{2} + {4}^{2} + (4 \frac{4}{5}) {}^{2} + ..........10 \: positions \: \\ \\ \\ \\ = ( \frac{8}{5} ) {}^{2} + ( \frac{12}{5} ) {}^{2} + ( \frac{16}{5} ) {}^{2} + {4}^{2} + ( \frac{24}{5} ) {}^{2} + ........10positions$




$= \frac{1}{ {5}^{2} } [(8) {}^{2} + (12) {}^{2} + (16) {}^{2} + (20) {}^{2} + (24) {}^{2} + 10....positions ] \\ \\ \\ \\ = \frac{ {4}^{2} }{ {5}^{2} } [ {2}^{2} + {3}^{2} + {4}^{2} + {5}^{2} + {6}^{2} + ..........10positions ] \\ \\ \\ \\ \\ = \frac{16}{25} [( {1}^{2} + {2}^{2} + {3}^{2} + {4}^{2} + {5}^{2} + {6}^{2} + ........... {11}^{2} ) { - 1}^{2} ] \\ \\ \\ \\ = \frac{16}{25} [ \frac{11 \times (11 + 1)(2 \times 11 + 1)}{6} - 1] \\ \\ \\ \\ = \frac{16}{25} [ \frac{11 \times 12 \times 23}{6} - 1 ] \\ \\ \\ \\ = \frac{16}{25} [506 - 1] \\ \\ \\ \\ = \frac{16}{25} \times 505$



it Means,


$\frac{16}{5} m = \frac{16}{25 } \times 505 \\ \\ \\ \\ m = \frac{16}{25} \times 505 \times \frac{5}{16} \\ \\ \\ \\ = 101$






$\bold{so, \:the\: option\: b\: is\: correct\: answer }$

Answer 2

Answer:

Step-by-step explanation:

Let the sum of the first 10 positions is S_{10}

then,  

S_{10} =  \frac{16}{5} m \\  \\  \\  = (1 \frac{1}{3} ) {}^{2}  + (2 \frac{2}{5} ) {}^{2}  + (3 \frac{1}{5} ) { }^{2}  +  {4}^{2}  + (4 \frac{4}{5})  {}^{2}  + ..........10 \: positions \:  \\  \\  \\  \\  = ( \frac{8}{5} ) {}^{2}  + ( \frac{12}{5} ) {}^{2}  + ( \frac{16}{5} ) {}^{2}  +  {4}^{2}  + ( \frac{24}{5} ) {}^{2}  + ........10positions

=  \frac{1}{ {5}^{2} } [(8) {}^{2} + (12) {}^{2} + (16) {}^{2}  + (20) {}^{2}  + (24) {}^{2}  + 10....positions  ] \\  \\  \\  \\  =  \frac{ {4}^{2} }{ {5}^{2} } [ {2}^{2}  +  {3}^{2}  +  {4}^{2} +  {5}^{2} +  {6}^{2}  + ..........10positions  ] \\  \\  \\  \\  \\  =  \frac{16}{25} [( {1}^{2}  +  {2}^{2} +  {3}^{2}  +  {4}^{2}  +  {5}^{2}  +  {6}^{2}   + ........... {11}^{2} ) { - 1}^{2} ] \\  \\  \\  \\  =  \frac{16}{25} [ \frac{11 \times (11 + 1)(2 \times 11 + 1)}{6}  - 1] \\  \\  \\  \\  =  \frac{16}{25} [ \frac{11 \times 12 \times 23}{6} - 1 ] \\  \\  \\  \\  =  \frac{16}{25} [506 - 1] \\  \\  \\  \\  =  \frac{16}{25}  \times 505

it Means,  

\frac{16}{5} m =  \frac{16}{25 }  \times 505 \\  \\  \\  \\ m =  \frac{16}{25}  \times 505 \times  \frac{5}{16}  \\  \\  \\  \\  = 101