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Let `a_1, a_2, ,a_(10)`be in A.P. and `h_1, h_2, h_(10)`be in H.P. If `a_1=h_1=2a n da_(10)=h_(10)=3,t h e na_4h_7`is`2`b. `3`c. `5`d. `6`A. 2B. 3C. 5D. 6 |
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Answer» Correct Answer - D Since, `a_(1), a_(2), a_(3), ..., a_(10)` are in AP Now, `a_(10) = a_(1) + 9d` `rArr 3 = 2 + 9d` `rArr d = 1//9 and a_(4) = a_(1) + 3d` `rArr a_(4) = 2 +3 (1//9) = 2 + 1//3 = 7//3` Also, `h_(1), h_(2), h_(3),...,h_(10)` are in HP `rArr (1)/(h_(1)), (1)/(h_(2)), (1)/(h_(3)) ,..., (1)/(h_(10))`are in AP Given, `h_(1) - 2, h_(10) -3` `:. (1)/(h_(10)) = (1)/(h_(1)) + 9d_(1) rArr (1)/(3) = (1)/(2) + 9d_(1)` `rArr - (1)/(6) = 9d_(1)` `rArr d_(1) = -(1)/(54) and (1)/(h_(7)) = (1)/(h_(1)) + 6d_(1)` `rArr (1)/(h_(7)) = (1)/(2) + (6xx1)/(-54)` `rArr (1)/(h_(7)) = (1)/(2) - (1)/(9) rArr h_(7) = (18)/(7)` `:. a_(4) h_(7) = (7)/(3) xx (18)/(7) = 6` |
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