Official solution
Note :
Sum of square of first n odd terms
12 + 32 + 52 + . . . . .+ n2 = \[{{n\left( {2n - 1} \right)\left( {2n + 1} \right)} \over 3}\]
Given,
12 + 2. 22 + 32 + 2.42 + 52 + 2.62 + . . . . . .
A = Sum of first 20 terms
\[\therefore\,\,\,\]A = 12 + 2.22 + 32 + 242 + 52 + 2.62 + . . . . . .20 terms
Arrange those terms this way,
A = [12 + 32 + 52 + . . . . . 10 terms] + [ 2.22 + 2.42 + 2.62 + . . . . 10 terms]
A = [ 12 + 32 + 52 + . . . . 10 terms ] + 2.2 [ 12 + 22 + 32 + . . . .10 terms ]
A = \[ {{10 \times \left( {2.10 - 1} \right)\left( {2.10 + 1} \right)} \over 3} + {2.2^2}\left[ {{{10 \times 11 \times 21} \over 6}} \right]\]
A = \[ {{10 \times 19 \times 21} \over 3} + 8 \times {{10 \times 11 \times 21} \over 6}\]
A =70 \[ \times \] 19 + 70 \[ \times \] 44
A = 70 \[ \times \] 63
B = Sum of first 40 terms
Arrange those terms this way.
B = [12+ 32 + 52 +. . . . 20 terms ] + [2.22 + 2.42 +. . . . . 20 terms ]
B = [12 + 32 + 52 + . . . . 20 terms] + 2.22 [12 + 22 + . . . 20 terms ]
B = \[{{20 \times 39 \times 41} \over 3} + \,\,8\,\, \times {{20 \times 21 \times 41} \over 6}\]
B = 260 \[ \times \] 41 + 560 \[ \times \] 41
B = 41 \[ \times \,\,\,820\]
\[\therefore\,\,\,\] B - 2A = 41 \[ \times \,\] 820 - 2 \[ \times \,\] 70 \[ \times \,\] 63 = 24800
Given that B - 2A = 100 \[\lambda \]
\[\therefore\,\,\,\] 100 \[\lambda \] = 24800
\[ \Rightarrow \,\,\,\lambda \] = 248