Question

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 the fourth term is ?

Answer 1

Answer:

Dear student

We know thatSn=n2[2a+(n−1)d]So,S11=112[2a+(11−1)d]33=112[2a+10d]6611=2a+10d2a+10d=6a+5d=3 ... (1)Since every alternate term is an integer and given sum is positivea+3d=2 ...(2)Subtracting (2) and (1)2d=1d=12a=2−32a=4−32a=12So, a4=a+3d=12+3×(12)=12+32=42=2

hope this will help you

Answer 2

Refer to attachment.

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