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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