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If the roots of the equation a(b-c) x^2+b(c-a)x+c(a-b)=0 are equal then prove that 1/a + 1/c = 2 /b Solution : Since roots of the equation are...

Squares and Cubes from Numbers 1 to 100: NUMBER SQUARE CUBE X X2 X3 1 1 1 2 4 8 3 9 27 4 16 64 5 25 125 6 36 216 7 49 343 8 64 512 9 81 729 10 100 1000 11 121 1331 12 144 1728 13 169 2197 14 196 2744 15 225 3375 16 256 4096 17 289 4913 18 324 5832 19 361 6859 20 400 8000 21 441 9261 22 484 10648 23 529 12167 24 576 13824 25 625 15625 26 676 17576 27 729 19683 28 784 21952 29 841 24389 30 900 27000 31 961 29791 32 1024 32768 33 1089 35937 34 1156 39304 35 1225 42875 36 1296 46656 37 1369 50653 38 1444 54872 39 1521 59319 40 1600 64000 41 1681 68921 42 1764 74088 43 1849 79507 44 1936 85184 45 2025 91125 46 2116 97336 47 2209 103823 48 2304 110592 49 2401 117649 50 2500 125000 51 2601 132651 52 2704 140608 53 2809 148877 54 2916 157464 55 3025 166375 56 3136 175616 57 3249 185193 58 3364 195112 59 3481 205379 60 3600 216000 61 3721 226981 62 3844 238328 63 3969 250047 64 4096 262144 65 4225 274625 66 4356 287496 67 4489 300763 68

Answer: Method 1: OA: 63/16 p, q, r, s are in HP Put x->1/x then roots in AP x^2 - 4x + A = 0 and x^2 - 6x + B = 0 Let roots be a-3d, a-d, a+d,...

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