MATH
posted by RAUNAK on .
AN AP has 21 term the sum of 10 ,11,12 term is 129 and sum of last 3 term is 237.find AP

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