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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 :

JEE · Math · previous-year question

  1. A.30
  2. B.32
  3. C.26
  4. D.35correct

Answer

D. 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) $${a^2}{r^6} = 25 \Rightarrow {(a{r^3})^2} = 25$$ $${T_4} + {T_6} + {T_8} = a{r^3} + a{r^5} + a{r^7}$$ $$ = a{r^3}(1 + {r^2} + {r^4})$$ $$ = 5(1 + 2 + 4) = 35$$

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