An A.P. has 21 terms ,if the sum of last three terms is 57 and the sum of middle three terms is 37. Find the A.P
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An A.P had 21 terms ;
•T19+T20+T21 = 57
(a+18d) + (a+19d) + (a+20d) = 57
3a + 57d = 57
(a+19d) = 19
•T10+T11+T12 = 37
(a+9d) + (a+10d) + (a+11d) = 37
3a+30d = 37
(a+10d) = 37/3
19-19d= (37/3)-10d
20/3 = 9d
d = 20/27
•T19+T20+T21 = 57
(a+18d) + (a+19d) + (a+20d) = 57
3a + 57d = 57
(a+19d) = 19
•T10+T11+T12 = 37
(a+9d) + (a+10d) + (a+11d) = 37
3a+30d = 37
(a+10d) = 37/3
19-19d= (37/3)-10d
20/3 = 9d
d = 20/27
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