JEE MAIN - Mathematics (2021 - 26th February Morning Shift - No. 14)
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 4th, 6th and 8th terms is equal to :
30
32
26
35
Explanation
a, ar, ar2, .....
$${T_2} + {T_6} = {{25} \over 2} \Rightarrow ar(1 + {r^4}) = {{25} \over 2}$$
$${a^2}{r^2}{(1 + {r^4})^2} = {{625} \over 4}$$ .... (1)
$${T_3}.{T_5} = 25 \Rightarrow (a{r^2})(a{r^4}) = 25$$
$${a^2}{r^6} = 25$$ .....(2)
On dividing (1) by (2)
$${{{{(1 + {r^4})}^2}} \over {{r^4}}} = {{25} \over 4}$$
$$4{r^8} - 14{r^4} + 4 = 0$$
$$(4{r^4} - 1)({r^4} - 4) = 0$$
$${r^4} = {1 \over 4},4 \Rightarrow {r^4} = 4$$ (an increasing geometric series)
$${a^2}{r^6} = 25 \Rightarrow {(a{r^3})^2} = 25$$
$${T_4} + {T_6} + {T_8} = a{r^3} + a{r^5} + a{r^7}$$
$$ = a{r^3}(1 + {r^2} + {r^4})$$
$$ = 5(1 + 2 + 4) = 35$$
$${T_2} + {T_6} = {{25} \over 2} \Rightarrow ar(1 + {r^4}) = {{25} \over 2}$$
$${a^2}{r^2}{(1 + {r^4})^2} = {{625} \over 4}$$ .... (1)
$${T_3}.{T_5} = 25 \Rightarrow (a{r^2})(a{r^4}) = 25$$
$${a^2}{r^6} = 25$$ .....(2)
On dividing (1) by (2)
$${{{{(1 + {r^4})}^2}} \over {{r^4}}} = {{25} \over 4}$$
$$4{r^8} - 14{r^4} + 4 = 0$$
$$(4{r^4} - 1)({r^4} - 4) = 0$$
$${r^4} = {1 \over 4},4 \Rightarrow {r^4} = 4$$ (an increasing geometric series)
$${a^2}{r^6} = 25 \Rightarrow {(a{r^3})^2} = 25$$
$${T_4} + {T_6} + {T_8} = a{r^3} + a{r^5} + a{r^7}$$
$$ = a{r^3}(1 + {r^2} + {r^4})$$
$$ = 5(1 + 2 + 4) = 35$$
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