Given, A={n∈N:n is a 3 -digit number } B={9k+2:k∈N}‌C={9k+l:k∈N} {101,109,...992} ⇒‌‌‌ Sum ‌=‌
100
2
[101+992]=‌
100×1093
2
‌ Similarly, ‌3‌-digit number of the form ‌9k+5‌ is ‌ ‌‌‌‌‌
100
2
[104+995]=‌
100×1099
2
‌ Their sum ‌=‌
100×1093
2
+‌
100×1099
2
=100×1096=400×274 Hence, we can say the value of I=5 as the second series of numbers obtained by set C is of the form 9k+5. ∴ Required value of I=5