$$y = \left( {1 - {1 \over 2}} \right){x^2} + \left( {1 - {1 \over 3}} \right){x^3} + ....$$

$$ = ({x^2} + {x^3} + {x^4} + ......) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + ....} \right)$$

$$ = {{{x^2}} \over {1 - x}} + x - \left( {x + {{{x^2}} \over 2} + {{{x^3}} \over 3} + ....} \right)$$

$$ = {x \over {1 - x}} + \ln (1 - x)$$

$$x = {1 \over 2} \Rightarrow y = 1 - \ln 2$$

$${e^{1 + y}} = {e^{1 + 1 - \ln 2}}$$

$$ = {e^{2 - \ln 2}} = {{{e^2}} \over 2}$$

$${{2(1 + 2 + 3 + .... + y)} \over {3(1 + 2 + 3 + .... + y)}} = {4 \over {{{\log }_{10}}x}}$$

$$ \Rightarrow {\log _{10}}x = 6 \Rightarrow x = {10^6}$$

Now,

$$y = ({\log _{10}}x) + \left( {{{\log }_{10}}{x^{{1 \over 3}}}} \right) + \left( {{{\log }_{10}}{x^{{1 \over 9}}}} \right) + ....\infty $$

$$ = \left( {1 + {1 \over 3} + {1 \over 9} + ....\infty } \right){\log _{10}}x$$

$$ = \left( {{1 \over {1 - {1 \over 3}}}} \right){\log _{10}}x = 9$$

So, (x, y) = (10⁶, 9)

Let $$t = {3 \over 2}{x^2} + {5 \over 3}{x^3} + {7 \over 4}{x^4} + ......\infty $$

$$ = \left( {2 - {1 \over 2}} \right){x^2} + \left( {2 - {1 \over 3}} \right){x^3} + \left( {2 - {1 \over 4}} \right){x^4} + ......\infty $$

$$ = 2({x^2} + {x^3} + {x^4} + .....\infty ) - \left( {{{{x^2}} \over 2} + {{{x^3}} \over 3} + {{{x^4}} \over 4} + .....\infty } \right)$$

$$ = {{2{x^2}} \over {1 - x}} - (\ln (1 - x) - x)$$

$$ \Rightarrow t = {{2{x^2}} \over {1 - x}} + x - \ln (1 - x)$$

$$ \Rightarrow t = {{x(1 + x)} \over {1 - x}} - \ln (1 - x)$$

Sum of infinite terms :

$${a \over {1 - r}} = 15$$ ..... (i)

Series formed by square of terms :

a², a²r², a²r⁴, a²r⁶ .......

Sum = $${{{a^2}} \over {1 - {r^2}}} = 150$$

$$ \Rightarrow {a \over {1 - r}}.{a \over {1 + r}} = 150 \Rightarrow 15.{a \over {1 + r}} = 150$$

$$ \Rightarrow {a \over {1 + r}} = 10$$ ...... (ii)

by (i) and (ii), a = 12; r = $${1 \over 5}$$

Now, series : ar², ar⁴, ar⁶

Sum = $${{a{r^2}} \over {1 - {r^2}}} = {{12.\left( {{1 \over {25}}} \right)} \over {1 - {1 \over {25}}}} = {1 \over 2}$$