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:
Solutions · 1
Official solution
S3=3a+3d=54⇒a+d=18.
\[S_{20} = 10(2a+19d) = 10(36+17d), \qquad 1600 < 10(36+17d) < 1800 \implies 124 < 17d < 144.\]
So 7.3<d<8.5, and since the terms are positive integers, d=8⇒a=10.
a11=a+10d=10+80=90.