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Which term of the `GP sqrt(3), 1/sqrt(3), 1/(3sqrt(3)), 1/(9sqrt(3))`, ... Is `1/(729sqrt(3))` ? |
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Answer» In the given `GP`, we have `a=sqrt(3)` and `r=(1/sqrt(3)xx1/sqrt(3))=1/3`. Let its nth term be `1/(729sqrt(3))`. Then, `T_(n)=1/(729sqrt(3))rArr ar^((n-1))=1/(729 sqrt(3))` `rArr sqrt(3)xx(1/3)^((n-1))=1/(729sqrt(3))` `rArr 1/3^((n-1))=1/((729xx3))=1/2187=1/3^(7)` `rArr n-1=7 rArr n=8`. Hence, the 8th term of the given GP is `1/(729sqrt(3))` |
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