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

Step-by-step explanation:

Let  aa be the first term  and dd be their common difference of the AP. 

Then, Sum to nn terms of  an AP =n2(2a+(n−1)d)=n2(2a+(n−1)d)

Given, S11=33S11=33

=>n2(2a+(n−1)d)=33=>n2(2a+(n−1)d)=33

=>112(2a+(11−1)d)=33=>112(2a+(11−1)d)=33

Solving we get 

a+5d=3a+5d=3

As aa is a fraction, dd should be the same fraction, so that adding a+da+d gives an integer second term. 

So, a+5a=3a+5a=3

=>a=12=d=>a=12=d 

nthnth term of the AP Tn=a+(n−1)dTn=a+(n−1)d

So, T4=12+(4−1)12=2