CAT Exam 2021 Slot 1 Solved Paper

The first number in each group: 1, 2, 5, 10, 17.....
Their common difference is in Arithmetic Progression. Hence, the general term of the series can be expressed as a quadratic equation.
Let the quadratic equation of the general term be $\mathit{a}{\mathit{x}}^2+\mathit{b}\mathit{x}+\mathit{c}$
1st term $=\mathit{a}+\mathit{b}+\mathit{c}=1$
2nd term $=4\mathit{a}+2\mathit{b}+\mathit{c}=2$
3rd term $=9\mathit{a}+3\mathit{b}+\mathit{c}=5$
Solving the equations, we get $\mathit{a}=1,\mathit{b}=-2,\mathit{c}=2.$
Hence, the first term in the 15th group will be ${\left(15\right)}^2-{\left(15\right)}^2+2=197$
We can see that the number of terms in each group is 1, 3, 5, 7.... and so on. These are of the form 2n-1. Hence, the number of terms in15th group will be 29. Hence, the last term in the 15th group will be $197+29-1=225.$
Sum of terms in group $15=\frac{29}{2}\left(197+225\right)=6119$
Alternatively,
The final term in each group is the square of the group number.
In the first group 1, second group 4, ............
The final element of the 14th group is ${\left(14\right)}^2$ , similarly for the 15th group this is : ${\left(15\right)}^2=225$
Each group contains all the consecutive elements in this range.
Hence the 15th group the elements are:
(197, 198, ................................225).
This is an Arithmetic Progression with a common difference of 1 and the number of element 29.
Hence the sum is given by : $\frac{\mathrm{n}}{2}\left(\mathrm{first}\mathrm{term}+\mathrm{last}\mathrm{term}\right)$
$=\frac{29}{2}\times \left(197+225\right)$

$=6119$