Posted by RAUNAK on Saturday, February 15, 2014 at 1:26am.
AN AP has 21 term the sum of 10 ,11,12 term is 129 and sum of last 3 term is 237.find AP

MATH  Steve, Saturday, February 15, 2014 at 6:32am
T10+T11+T12 = (a+9d)+(a+10d)+(a+11d), so
3a+30d = 129
Similarly, for the last three terms,
3a+57d = 237
a=3
d=4
and the sequence is
3,7,11,...,83
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