AP Police SI Mains 2019 Arithmetic and Reasoning

from given series we can say that,

$$\ T_n=\frac{n\ }{\left(2n-1\right)^2\left(2n+1\right)^2}=\frac{1\ }{8}\left(\ \frac{\ 1}{\left(2n-1\right)^2}-\frac{1\ }{\left(2n+1\right)^2}\right),\ where\ n=1,2,3,...,15.$$

So,$$S_n=$$

$$\ T_1+T_2+....+T_{15}$$

$$=\frac{1\ }{8}\left(\ \frac{\ 1}{1^2}-\frac{1\ }{3^2}\right)+\frac{1\ }{8}\left(\frac{1\ }{3^2}-\frac{1\ }{5^2}\right)+.....+\frac{1\ }{8}\left(\frac{1\ }{29^2}-\frac{1\ }{31^2}\right).$$

$$=\frac{1\ }{8}\left(1-\frac{1\ }{31^2}\right).$$

$$=\frac{1\ }{8}\left(\ \frac{\ 961-1}{961}\right).$$

$$=\frac{1\ }{8}\left(\ \frac{\ 960}{961}\right).$$

$$=\frac{\ 120}{961}.$$

B is correct choice.

$$5\sqrt{3\ }-2\sqrt{12\ }-\sqrt{32\ }+\sqrt{50\ }=5\sqrt{3\ }-4\sqrt{3\ }-4\sqrt{2\ }+5\sqrt{2\ }=\sqrt{3\ }+\sqrt{2\ }.$$

$$3+\sqrt{6\ }=\sqrt{3\ }\left(\sqrt{3\ }+\sqrt{2\ }\right).$$

So,$$x=\sqrt{3\ }.$$

So,$$\ \frac{\ x^4-1}{x^4+1}=\frac{9-1\ }{9+1}=\frac{8\ }{10}=\frac{4\ }{5}.$$

C is correct choice.

Person buys 2 mangoes and 4 oranges ,Total worth of$$=2\times5+4\times7=38.$$

A is correct choice.

$$x=\ \frac{\ 1}{\sqrt{13\ }-3}=\ \frac{\ \sqrt{13\ }+3}{13-9}=\ \frac{\ \sqrt{13\ }+3}{4}=1.65.$$

$$y=\ \frac{\ 1}{\sqrt{7\ }-\sqrt{\ 3}}=\ \frac{\ \sqrt{7\ }+\sqrt{3\ }}{7-3}=1.09.$$

$$z=\ \frac{\ 1}{\sqrt{6\ }-\sqrt{2\ }}=\ \frac{\ \sqrt{6\ }+\sqrt{2\ }}{4}=0.96.$$

So, $$x>y>z.$$

C is correct choice.

LCM(3,4,5,6,8)=120.

So, Smallest number that is divisible by 3,4,5,6,8 is 120.

Now, Smallest square numbers ,which are present on both side of 121, that can produce smallest difference are 100$$\left(10^2\right)$$ and 121$$\left(11^2\right)$$.

So, Smallest difference is (121-100)=21.

D is correct choice.

If x is rational number then 1/x is also a rational number.

A is correct choice.

mean proportion of b,c is$$\sqrt{bc\ }=8.$$

or,$$bc=64.$$

And, $$a:b=c:8.$$

then,$$a=\frac{\ bc}{8}=\frac{64\ }{8}=8.$$

So,$$abc=8\times64=2^9.$$

A is correct choice.

HCF of 513,1134 and 1215 is 27.

So, C is correct choice.

for an infinite series , SUM$$=\ \frac{\ a}{\left(1-r\right)}.$$

So, $$y=\ \frac{\ x}{\left(1-\ \frac{\ 1}{9}\right)}+\ \frac{\ \ \frac{\ 1}{2}}{\left(1-\ \frac{\ 1}{9}\right)}.$$

or,$$y=\ \frac{\ x}{\left(\frac{\ 8}{9}\right)}+\ \frac{\ \ \frac{\ 1}{2}}{\left(\frac{\ 8}{9}\right)}.$$

or,$$y=\ \frac{9\ x}{8}+\frac{\ \ 9}{16}.$$

Again,

$$x=\ \frac{\ \ \frac{\ 1}{2^2}}{\left(1-\ \frac{\ 1}{2}\right)}+1.$$

or,$$x=\frac{\ 3}{2}.$$

So,$$y=\frac{\ 27}{16}+\frac{\ 9}{16}=\frac{\ 36}{16}=\frac{\ 9}{4}=x^2.$$

C is correct choice.

possible values are (1,1),(1,2),(1,3) and (2,1).

So, A is correct choice.