In an A. P if S5+S7=167 and S10 =235 ,then find the first 3 term of A. P
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S5+S7=5/2(2a+4d) +7/2(2a+6d)=167
=5a+10d+7a+21d=167
=12a+31d=167 [let this be eqn 1]
similarly,
from S10=235, we get 2a+9d=47 {let this be equation 2}
by solving eqn 1 and 2, we get a=1 and d=5
therefore the ap is 1,6,11........
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