JEE Main 2025 Apr 3 Shift 1, Mathematics Q7: Summation of Series

Successive differences: $2,8,14,20,26,\dots$ form an AP, so the second differences are constant. Hence the $n$-th term is a quadratic: $T_n=a{n}^2+bn+c.$

Using $T_1=1,T_2=3,T_3=11$:

$a+b+c=1,4a+2b+c=3,9a+3b+c=11.$

Solving gives $a=3,b=-7,c=5$, so $T_n=3n^2-7n+5.$

Sum to 20 terms:

$S_{20}=3\cdot \frac{20\cdot 21\cdot 41}{6}-7\cdot \frac{20\cdot 21}{2}+5\cdot 20=3\cdot 2870-7\cdot 210+100=8610-1470+100=7240.$