An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first 11 terms is 33, then find the 4th term

Dear student
We know thatSn=n22a+n-1dSo,S11=1122a+11-1d33=1122a+10d6611=2a+10d2a+10d=6a+5d=3  ... 1Since every alternate term is an integer  and given sum is positivea+3d=2  ...2Subtracting  2 and 12d=1d=12a=2-32a=4-32a=12So, a4=a+3d=12+3×12=12+32=42=2
Regards

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