Q.

Consider an A.P. of positive integers, whose sum of the first three terms is 54 and the sum of the first twenty terms lies between 1600 and 1800. Then its 11th term is :          [2025]

1 84  
2 90  
3 108  
4 122  

Ans.

(2)

We have, a1+a2+a3=54

 3(a+d)=54  a+d=18

        S20=10[2a+19d]=10[36+17d]

As   1600<S20<1800

 1600<10(36+17d)<1800

 160<36+17d<180

 124<17d<144

 7517<d<8817

Since, A.P. is of positive integers.

Common difference will be a natural number

  d=8  a=10

  a11=10+10×8=90.