Sum of Numbers in a Grouped Sequence

Number of terms in $k$-th group is $2k - 1$.

• Group 1 has 1 term
• Group 2 has 3 terms
• Group 3 has 5 terms

Total number of terms in the first 14 groups: $1 + 3 + 5 + \dots + (2 \times 14 - 1) = 14^2 = 196$

So the 15th group starts with the $197$-th natural number, which is $197$.

The 15th group has $2(15) - 1 = 29$ terms. These terms form an AP with first term $a = 197$, common difference $d = 1$, and $n = 29$ terms.

Sum of terms in 15th group: $S_{29} = \frac{29}{2} [2(197) + (29 - 1)(1)] = \frac{29}{2} [394 + 28] = \frac{29}{2} \times 422 = 29 \times 211 = 6119$.