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Sequences and Progressions - Derive and apply sum of first n natural numbers in problem solving

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A sequence of natural numbers 1,2,3,…,n1, 2, 3, \dots, n forms an Arithmetic Progression (AP) where the first term a=1a = 1 and the common difference d=1d = 1.

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To derive the sum, let Sn=1+2+3+⋯+nS_n = 1 + 2 + 3 + \dots + n. Writing the sequence in reverse, we get Sn=n+(n−1)+⋯+2+1S_n = n + (n-1) + \dots + 2 + 1. Adding these two equations term by term gives 2Sn=(n+1)+(n+1)+⋯+(n+1)2S_n = (n+1) + (n+1) + \dots + (n+1) (nn times). Therefore, 2Sn=n(n+1)2S_n = n(n+1), which leads to Sn=n(n+1)2S_n = \frac{n(n+1)}{2}.

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The sum of the first nn natural numbers is the same as finding the sum of an AP where the first term aa is 11 and the last term ll is nn.

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This formula can be used to find the sum of any range of consecutive natural numbers by calculating the difference between two sums starting from 11.

📐Formulae

Sn=1+2+3+⋯+n=n(n+1)2S_n = 1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}

Sn=n2[a+l]S_n = \frac{n}{2}[a + l]

Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d]

💡Examples

Problem 1:

Find the sum of the first 5050 natural numbers.

Solution:

Given n=50n = 50. Using the formula: S50=n(n+1)2S_{50} = \frac{n(n+1)}{2} S50=50(50+1)2S_{50} = \frac{50(50+1)}{2} S50=25×51S_{50} = 25 \times 51 S50=1275S_{50} = 1275

Explanation:

We identify the total number of terms nn as 5050 and substitute it into the sum formula for natural numbers.

Problem 2:

Calculate the sum of natural numbers from 1111 to 3030.

Solution:

The sum of numbers from 1111 to 3030 is calculated as: Sum=S30−S10\text{Sum} = S_{30} - S_{10} Calculating S30S_{30}: S30=30(31)2=15×31=465S_{30} = \frac{30(31)}{2} = 15 \times 31 = 465 Calculating S10S_{10}: S10=10(11)2=5×11=55S_{10} = \frac{10(11)}{2} = 5 \times 11 = 55 Final calculation: 465−55410\begin{array}{r} 465 \\ - 55 \\ \hline 410 \end{array}

Explanation:

To find the sum of a specific range, we find the sum of the first 3030 natural numbers and subtract the sum of the first 1010 natural numbers (the numbers we don't want).

Problem 3:

If the sum of the first nn natural numbers is 210210, find the value of nn.

Solution:

Given Sn=210S_n = 210. Using the formula: n(n+1)2=210\frac{n(n+1)}{2} = 210 n(n+1)=420n(n+1) = 420 n2+n−420=0n^2 + n - 420 = 0 Factoring the quadratic equation: (n+21)(n−20)=0(n + 21)(n - 20) = 0 Since nn must be a positive natural number, n=20n = 20.

Explanation:

We set up a quadratic equation by equating the sum formula to the given value and solve for the positive integer nn.