krit.club logo

Exploring some more Progressions - Summation to n terms-advanced

Grade 9CBSE

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

🔑Concepts

•

The sum of the first nn terms of an Arithmetic Progression (AP) is calculated using the first term aa, the common difference dd, and the number of terms nn.

•

If the last term ll (or ana_n) is known, the sum formula can be simplified using the first and last terms.

•

The nthn^{th} term of a sequence can be found if the formula for the sum of nn terms SnS_n is known: an=Sn−Sn−1a_n = S_n - S_{n-1} for n>1n > 1.

•

Special series sums for the first nn natural numbers, their squares, and their cubes are often used in advanced summation problems.

•

The sum of the first nn odd natural numbers is always a perfect square, given by n2n^2.

•

The sum of the first nn even natural numbers is given by n(n+1)n(n+1).

📐Formulae

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

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

∑k=1nk=n(n+1)2\sum_{k=1}^{n} k = \frac{n(n+1)}{2}

∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}

∑k=1nk3=[n(n+1)2]2\sum_{k=1}^{n} k^3 = \left[ \frac{n(n+1)}{2} \right]^2

an=Sn−Sn−1a_n = S_n - S_{n-1}

💡Examples

Problem 1:

Find the sum of the first 20 terms of the AP: 5,9,13,17,…5, 9, 13, 17, \dots

Solution:

Here, first term a=5a = 5, common difference d=9−5=4d = 9 - 5 = 4, and n=20n = 20. Using the formula Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1)d]: S20=202[2(5)+(20−1)4]S_{20} = \frac{20}{2} [2(5) + (20-1)4] S20=10[10+19×4]S_{20} = 10 [10 + 19 \times 4] S20=10[10+76]S_{20} = 10 [10 + 76] S20=10×86=860S_{20} = 10 \times 86 = 860

Explanation:

We identify the components of the AP and substitute them into the standard summation formula for nn terms.

Problem 2:

If the sum of the first nn terms of a progression is given by Sn=3n2+2nS_n = 3n^2 + 2n, find the 10th10^{th} term.

Solution:

The nthn^{th} term is given by an=Sn−Sn−1a_n = S_n - S_{n-1}. For n=10n = 10: S10=3(10)2+2(10)=300+20=320S_{10} = 3(10)^2 + 2(10) = 300 + 20 = 320 For n=9n = 9: S9=3(9)2+2(9)=3(81)+18=243+18=261S_9 = 3(9)^2 + 2(9) = 3(81) + 18 = 243 + 18 = 261 Calculating the 10th10^{th} term: 320−26159\begin{array}{r} 320 \\ -261 \\ \hline 59 \end{array} So, a10=59a_{10} = 59.

Explanation:

To find a specific term from a sum formula, we subtract the sum of (n−1)(n-1) terms from the sum of nn terms.

Problem 3:

Calculate the sum of the squares of the first 10 natural numbers.

Solution:

We use the special series formula for the sum of squares: ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} For n=10n = 10: S=10(10+1)(2×10+1)6S = \frac{10(10+1)(2 \times 10 + 1)}{6} S=10×11×216S = \frac{10 \times 11 \times 21}{6} S=23106=385S = \frac{2310}{6} = 385

Explanation:

This advanced formula allows us to find the sum of squares (12+22+⋯+1021^2 + 2^2 + \dots + 10^2) directly without adding each term manually.