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Engineering
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 If the sum of the first ten terms of the series
 \left ( 1\frac{3}{5} \right )^{2}+\left ( 2\frac{2}{5} \right )^{2}+\left ( 3\frac{1}{5} \right )^{2}+4^{2}+\left ( 4\frac{4}{5} \right )^{2}+.........is\frac{16}{5}m

then m is equal to:

  • Option 1)

    102

  • Option 2)

    101

  • Option 3)

    100

  • Option 4)

    99

 

As we learnt in 

Summation of series of natural numbers -

\sum_{k=1}^{n}K^{2}= \frac{1}{6}n\left ( n+1 \right )\left ( 2n+1 \right )
 

- wherein

Sum of  squares of first n natural numbers

1^{2}+2^{2}+3^{2}+4^{2}+------+n^{2}= \frac{n(n+1)\left ( 2n+1 \right )}{6}

 

 \left ( 1\frac{3}{5} \right )^{2}+\left ( 2\frac{2}{5} \right )^{2}+\left ( 3\frac{3}{5} \right )^{2}+4^{2}+\left (4\frac{4}{5} \right )^{2}----is \frac{16}{5}n

\Rightarrow \left ( \frac{8}{5} \right )^{2}+\left ( \frac{12}{5} \right )^{2}+\left ( \frac{16}{5} \right )^{2}+\left ( \frac{20}{5} \right )^{2}+ -----

\therefore\frac{1}{25} \left [ 8^{2}+12^{2}+16^{2}+20^{2}+ --- \right ]

\therefore\frac{1}{25}\times4^{}2 \left [ 2^{2}+3^{2}+4^{2}+5^{2}+--- \right ]

=\frac{16}{25} \left [ 1^{2}+2^{2}+3^{}2+-----11^{2}-1\right ]

\Rightarrow \frac{16}{25}\left [ \frac{11\times12\times23}{6}-1 \right ]=\frac{16}{5}n

\therefore 11\times46-1=5n

      506-1=5n

      505=5n

\therefore n=101

 

 


Option 1)

102

Incorrect option

Option 2)

101

Correct option

Option 3)

100

Incorrect option

Option 4)

99

Incorrect option

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Engineering
199 Views   |  

If the 2nd, 5th and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is :

 

  • Option 1)

    \frac{8}{5}

  • Option 2)

    \frac{4}{3}

  • Option 3)

    1

  • Option 4)

    \frac{7}{4}

 

As we learnt in 

Common ratio of a GP (r) -

The ratio of two consecutive terms of a GP

- wherein

eg: in 2, 4, 8, 16, - - - - - - -

r = 2

and in 100, 10, 1, 1/10 - - - - - - -

r = 1/10

 


 {A}, {B},{C} are in  G.P where {A}, {B},{C}  are Terms of an A.P

 

Let first term is a and common difference is d, and common ratio be r then

A = a+d

B = a+4d

C= a+8d

\therefore \: \frac{B}{A}=\frac{C}{B}=\frac{r}{1}

\frac{a+4d}{a+d} =\frac{a+8d}{a+4d}=\frac{r}{1}

\therefore \frac{a+4d+a+d}{a+4d-a-d}=\frac{r+1}{r-1}

\Rightarrow \frac{2a+5d}{3d}=\frac{r+1}{r-1}-----(i)

    \frac{a+8d+a+4d}{a+8d-a-4d} =\frac{r+1}{r-1}

\Rightarrow \frac{2a+12d}{4d}=\frac{r+1}{r-1}------(ii)

from (i) and (ii)

\frac{2a+5d}{2a+12d}=\frac{3}{4}

8a+20d=6a+36d

2a=16d

a=8d

\Rightarrow \frac{r+1}{r-1}=\frac{2\times8d+5d}{3d}=\frac{16d+5d}{3d}=\frac{21}{3}=7

\therefore r+1=7r-7

8=6r

r=\frac{4}{3}


Option 1)

\frac{8}{5}

Incorrect option

Option 2)

\frac{4}{3}

Correct option

Option 3)

1

Incorrect option

Option 4)

\frac{7}{4}

Incorrect option

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