The natural numbers are divided into groups as (1), $(2,3,4),(5,6,7,8,9), \ldots .$. and so on. Then, the sum of the numbers in the 15th group is equal to

Let's start by understanding the pattern in how the natural numbers are grouped:

Group 1: (1) → 1 number

Group 2: (2, 3, 4) → 3 numbers

Group 3: (5, 6, 7, 8, 9) → 5 numbers

Each group contains consecutive natural numbers, and the number of elements in each group follows the pattern: 1, 3, 5, 7, ... (odd numbers).

So the nth group contains $(2n - 1)$ numbers.

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Using our pattern: 14th group has $(2 × 14 - 1) = 27$ numbers

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We need to count all numbers from groups 1 through 14.

Total numbers used = $1 + 3 + 5 + 7 + ... + 27$

This is the sum of the first 14 odd numbers. There's a beautiful formula here:

Sum of first n odd numbers = $n^2$

Why this works: The sequence 1, 3, 5, 7, ... represents odd numbers. When you add the first n odd numbers, you always get $n^2$. For example: $1 = 1^2$, $1+3 = 4 = 2^2$, $1+3+5 = 9 = 3^2$.

Therefore: Numbers used in first 14 groups = $14^2 = 196$

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Since 196 natural numbers have been used in the first 14 groups, the 15th group starts with the 197th natural number.

First number of 15th group = 197

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Using our pattern: 15th group has $(2 × 15 - 1) = 29$ numbers

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The 15th group contains 29 consecutive numbers starting from 197:

$(197, 198, 199, ..., 225)$

This is an arithmetic progression (AP) with:

First term $(a) = 197$

Number of terms $(n) = 29$

Common difference $(d) = 1$

AP Sum Formula: $S = \tfrac{n}{2} × [2a + (n-1)d]$

$S = \tfrac{29}{2} × [2(197) + (29-1)(1)]$

$S = \tfrac{29}{2} × [394 + 28]$

$S = \tfrac{29}{2} × 422$

$S = 29 × 211$

$S = 6119$

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Therefore, the sum of numbers in the 15th group is 6119.