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If the first and the nth terms of a G.P., are `aa n db ,`respectively, and if `P`is hte product of the first `n`terms prove that `P^2=(a b)^ndot`A. abB. `(ab)^(n)`C. `(ab)^(n//2)`D. `(ab)^(2n)` |
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Answer» Correct Answer - B Let r be the common ratio of the G.P. Then, `b=n^(th)term=ar^(n-1)rArrr^(n-1)=(b)/(a)rArrr=((b)/(a))^((1)/(n-1))` and, P = Product of the first n terms `rArr" "P=a.ar.ar^(2) . . .ar^(n-1)=a^(n)r^(1+2+3+ . . .+(n-1)` `rArr" "P=a^(n)r^((n(n-1))/(2))=a^(n){((b)/(a))^((1)/(n-1))}^((n(n-1))/(2)` `rArr" "P=a^(n)((b)/(a))^(n//2)=a^(n//2)b^(n//2)=(ab)^(n//2)` `rArr" "P^(2)=[(ab)^(n//2)]^(2)=(ab)^(n)` |
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