InterviewSolution
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InterviewSolution
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There are two sets A and B each of which consists of three numbers in AP whose sum is 15 and where D and d are the common differences such that `D=1+d,dgt0.` If `p=7(q-p)`, where p and q are the product of the nymbers respectively in the two series. The value of `7D+8d` isA. 37B. 22C. 67D. 52 |
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Answer» Correct Answer - B Let `A={A-D,A,A+D},B={a-d,a,a+d}` according to the question, `A-D+A+A+D=15` `implies 3A=15` `implies A=15" " "…….(i)"` and `a-d+a+a+d=15` `implies a=5" "………(ii)"` and `D=1+d" " "……..(iii)"` `p=(A-D)A(A+D)` `p=A(A^(2)-D^(2))" " ......(iv)"` `p=5(25-D^(2))" " "........(v)"` Similarly, `q=5(25-d^(2))` Given that, `p=7(q-p)` `8p=7q` From Eqs. (iv) and (v), we get `8xx5(25-D^(2))=7xx5(25-d^(2))` `200-8D^(2)=175-7d^(2)` `25=8D^(2)-7d^(2)` `25=8(1+d)^(2)-7d^(2)" " [" from Eq. (iii) "]` `25=8+8d^(2)+16d-7d^(2)` `17-d^(2)-16d=0` `d^(2)+16d-17=0` `(d+17)(d-1)=0` `d=-17` or `d=1` `implies d=1 " " [:.dgt0]` `impliesD=2` `7D+8d=14+8=22` |
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