1 votes 1 votes The natural numbers are divided into groups as $(1), (2,3,4), (5,6,7,8,9), \dots $ and so on. Then, the sum of the numbers in the $15 \text{th}$ group is equal to $6090$ $4941$ $6119$ $7471$ Quantitative Aptitude cat2021-set1 quantitative-aptitude number-systems + – soujanyareddy13 asked Jan 19, 2022 • retagged Mar 23, 2022 by Lakshman Bhaiya soujanyareddy13 2.7k points 727 views answer comment Share See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes We can write the number in the $ n^{\text{th}}$ group. The number in $1^{\text{st}}$ group $ = (1)$ The number in $2^{\text{nd}}$ group $ = (2, 3, 4)$ The number in $3^{\text{rd}}$ group $ = (5, 6, 7, 8, 9)$ $\vdots \quad \vdots \quad \vdots $ The number in $ n^{\text{th}}$ group $ = \left( n^{2} – (2n-2), n^{2} – (2n-1), \dots, n^{2} – 1, n^{2} \right)$ The number of element in $n^{\text{th}}$ group $ = 2n – 1$ So, the sum of numbers of $n^{\text{th}}$ group $S_{n} = n^{2} + (n^{2}-1) + (n^{2}-2) + \dots + \left( n^{2} – (2n-2) \right)$ $\Rightarrow S_{n} = (2n – 1)n^{2} – \left( 1 + 2 + 3 + \dots + (2n-2) \right)$ $\Rightarrow S_{n} = (2n-1)n^{2} + \frac{(2n-2)(2n-1)}{2}$ $\Rightarrow S_{n} = (2n-1)n^{2} + (n-1)(2n-1) $ $ \Rightarrow S_{n} = (2n-1)n^{2} + 2n^{2} – n – 2n + 1$ $\Rightarrow S_{n} = (2n-1)n^{2} + 2n^{2} – 3n + 1$ $\Rightarrow S_{n} = (2n-1) (n^{2} – n + 1) $ $\Rightarrow S_{n} = 2n^{3} – 3n^{2} + 3n – 1 $ $\Rightarrow S_{n} = n^{3} + n^{3} – 3n^{2} + 3n – 1$ $\Rightarrow S_{n} = n^{3} + (n-1)^{3} \quad [\because n^{3} – 3n^{2} + 3n – 1 = (n-1)^{3}]$ $\therefore$ The sum of the numbers in the $15^{\text{th}}$ group $ = 15^{3} + (15-1)^{3}$ $ \qquad \qquad \qquad = 15^{3} + 14^{3} = 3375 + 2744= 6119.$ Correct Answer $: \text{C}$ Anjana5051 answered Jan 30, 2022 • selected Mar 27, 2023 by Shoto Anjana5051 11.7k points comment Share See all 0 reply Please log in or register to add a comment.
0 votes 0 votes nth group starts with $(n-1)$$^{2}$ $+$ $1$ and ends with $n$$^{2}$ and every group consists $2n-1$ numbers $15th$ group starts with 196 and ends with $225$ $sum$ = $\frac{No of terms}{2}$ * $[first$ $term$+$last$ $term]$ = $\frac{29}{2} * [(196+1) + 225]$ = $6119$ Spiderman 🕷️ answered 6 hours ago Spiderman 🕷️ 18 points comment Share See all 0 reply Please log in or register to add a comment.