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The sum of first three terms of A.P. is 21 and their products is 231. Find the numbers |
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Answer» Let the No. a-d,a and a+dSum: a-d+a+a+d=213a=21, a=7product: a(a-d)(a+d)=231a(a2-d2)=2317(49-d2)=23149-d2=33d2=16, d={tex}\\pm{/tex}4So numbers are:3,7,11 or 11,7,3\xa0 Let the AP is (a-d), a, (a+d)(a-d)+a+(a+d) =213a=21a=7(a-d)×a×(a+d)=231(a×a-d×d) ×a=231(7×7-d×d)×7=231(7×7-d×d)=231÷7(7×7-d×d)=33d×d=49-33d×d=16d=4Hence, the A. P. is 3, 7, 11. the A. P. is 3, 7, 11. A=3 and d=4...therfore AP is 3,7,11.......And three numbers are 3,7 and 11 |
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