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Find AP whose sum and product of 3 adjacent terms are 39 and 1560 respectively nth term of a sequence is given by the formula an = 10-3n ,find the sum of its 20 terms . |
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Answer» Step-by-step explanation: Given: Sum of 3 adjacent terms=39 Product of 3 adjacent terms=1560 an=10-3n To find: A.P FORMED Sum of its 20 terms solution: Let 3 adjacent terms be a-d, a, a+d sum=a-d+a+a+d=3a but it is given that sum = 39 therefore, 3a=39 a=13 - - - - 1 Product= (a-d)(a)(a+d) =(a^2-ad)(a+d) =a^3+a^2d-a^2d-ad^2 =a^3-ad^2 but it is given that product = 1560 therefore, a^3-ad^2=1560 a(a^2-d^2)=1560- - - - 2 from 1 a= 13 putting in 2, we get, 13(13^2-d^2)=1560 169-d^2=1560/13 -d^2=1560/13-169 -d^2=1560/13-2197/13 -d^2=-637/13 d^2=49 d=7 Now A.P formed: =a,a+d,a+2d- - - =13,20,27- - - sum of its 20 terms=n/2(2a+(n-1)d) here n=20 (given) a=13(find above) d=7(find above) putting values we get =20/2(2*13+(20-1)7) =10(26+133) =10*159 =1590 |
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