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Explanation:

Here, 
1^st group has 1 integer
2^nd group has 3 integers
3^rd group has 5 integers
∴ nth groups will have (2n – 1) integers.

Total inters used till
1^st group = 1 = 1²
2^nd group = 1 + 3 = 2²
3^rd group = 1 + 3 + 5 = 9 = 3²

∴ Total integers used till n^th group = n²

∴ Sum of integers in 15^th group = Sum of all integers used till 15^th group - Sum of all integers used till14^th group

= (1 + 2 + 3 + … + 15²) - (1 + 2 + 3 + … + 14²)

= $\frac{225\times 226}{2}$ - $\frac{196\times 197}{2}$

= 25425 – 19306 = 6119

Hence, option (b).