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If first term of a GP is a, third term is b and `(n+1)th` term is c. The `(2n+1)th` term of a GP isA. `csqrt((b)/(a))`B. `(bc)/(a)`C. `abc`D. `(c^(2))/(a)` |
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Answer» Let common ratio =r ` therefore b=ar^(2) implies r=sqrt((b)/(a))` Also, `c=ar^(n) implies r^(n)=(c)/(a)` `therefore t_(2n+1)= ar^(2n)=a(r^(n))^(2)=a((c)/(a))^(2)=(c^(2))/(a)` Hence, (d) is the correct answer. |
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