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The next term of the G.P. `x ,x^2+2,a n dx^3+10`is`(729)/(16)`b. `6`c. `0`d. `54` |
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Answer» Correct Answer - C::D According to the question, `x,x^(2)+2" and "x^(3)+10` are in GP. So, `(x^(2)+2)^(2)=x(x^(3)+10)=0` `implies x^(4)+4+4x^(2)-x^(4)-10x=0` `implies 4x^(2)-10x+4=0` `implies 2x^(2)-5x+2=0` `implies 2x^(2)-4x-x+2=0` `implies 2x(x-2)-1(x-2)=0` `implies(x-2)(2x-1)=0` `implies x=2` or `x=(1)/(2)` For`x=2`,first 3 terms are2,6,18. So, 4th term of GP `=2*(3)^(3)=54` For `x=(1)/(2)`, first 3 terms are `(1)/(2), (9)/(4),(81)/(8)`. So, `T_(4)=(1)/(2)((9)(2))^(3)=(1)/(2)xx(729)/(16)` |
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