In an increasing geometric series, the sum of the second and the sixth term is ${{25} \over 2}$ and the product of the third and fifth term is 25. Then, the sum of 4^th, 6^th and 8^th terms is equal to :

a, ar, ar², .....<br><br>${T_2} + {T_6} = {{25} \over 2} \Rightarrow ar(1 + {r^4}) = {{25} \over 2}$<br><br>${a^2}{r^2}{(1 + {r^4})^2} = {{625} \over 4}$ .... (1)<br><br>${T_3}.{T_5} = 25 \Rightarrow (a{r^2})(a{r^4}) = 25$<br><br>${a^2}{r^6} = 25$ .....(2)<br><br>On dividing (1) by (2)<br><br>${{{{(1 + {r^4})}^2}} \over {{r^4}}} = {{25} \over 4}$<br><br>$4{r^8} - 14{r^4} + 4 = 0$<br><br>$(4{r^4} - 1)({r^4} - 4) = 0$<br><br>${r^4} = {1 \over 4},4 \Rightarrow {r^4} = 4$ (an increasing geometric series)<br><br>${a^2}{r^6} = 25 \Rightarrow {(a{r^3})^2} = 25$<br><br>${T_4} + {T_6} + {T_8} = a{r^3} + a{r^5} + a{r^7}$<br><br>$= a{r^3}(1 + {r^2} + {r^4})$<br><br>$= 5(1 + 2 + 4) = 35$