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IF `a_(1),a_(2),a_(3),"...."a_(10)` be in AP and `h_(1),h_(2),h_(3),"...."h_(10)` be in HP. If `a_(1)=h_(1)=2` and `a_(10)=h_(10)=3`, then find value of `a_(4)h_(7)`. |
Answer» ` therefore a_(1),a_(2),a_(3),"...."a_(10)` are in AP. If d be the common difference, then `d=(a_(10)-a_(1))/(9)=(3-2)/(9)=(1)/(9)` `therefore a_(4)=a_(1)+3d=2+(3)/(9)=2+(1)/(3)=(7)/(3) "……..(i)"` and given ` h_(1),h_(2),h_(3),"...."h_(10)` are in HP. If D be common difference of corresponding AP. Then, ` D=((1)/h_(10)-(1)/h_(1))/(9)=((1)/(3)-(1)/(2))/(9)=-(1)/(54)` ` therefore (1)/h_(7)=(1)/h_(1)+6D=(1)/(2)-(6)/(54)=(1)/(2)-(1)/(9)=(7)/(18) implies h_(7)=(18)/(7)` Hence, ` a_(4)*h_(7)=(7)/(3)xx(18)/(7)=6` |
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