krit.club logo

Arithmetic Progressions - Sum of First n Terms of an AP

Grade 10CBSE

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 denoted by SnS_n. It represents the total value obtained by adding the first nn terms of the sequence: a1+a2+a3+...+ana_1 + a_2 + a_3 + ... + a_n. Visually, if you imagine each term as a vertical bar with a height equal to its value, SnS_n represents the total area covered by the first nn bars, forming a staircase-like shape.

The sum of an AP depends on three primary variables: the first term (aa), the common difference (dd), and the number of terms (nn). If the common difference dd is positive, the 'staircase' of terms rises upward, and the sum increases rapidly. If dd is negative, the terms decrease, and the 'staircase' slopes downward.

A useful relationship exists between the sum of terms and the general term (ana_n): the nn-th term of an AP can be found by subtracting the sum of the first (n1)(n-1) terms from the sum of the first nn terms, expressed as an=SnSn1a_n = S_n - S_{n-1}. This is helpful when you are given a formula for SnS_n in terms of nn.

The sum formula Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d] is a quadratic expression in terms of nn. When you plot the values of SnS_n against nn on a coordinate plane, the resulting points lie on a parabolic curve that passes through the origin (0,0)(0,0), because the sum of zero terms is always zero.

The sum of the first nn terms can also be viewed as nn times the average of the first and last terms. Visually, this is equivalent to taking the 'staircase' of terms and rearranging it into a rectangle with a width of nn and a height equal to the average of the first term aa and the last term ll, which is a+l2\frac{a + l}{2}.

Special case: The sum of the first nn positive integers (1,2,3,...,n1, 2, 3, ..., n) is a common calculation where a=1a=1 and d=1d=1. This simplifies to the formula Sn=n(n+1)2S_n = \frac{n(n + 1)}{2}, which is often used in probability and series problems.

📐Formulae

Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n - 1)d]

Sn=n2(a+l)S_n = \frac{n}{2}(a + l) where ll is the last term (ana_n)

an=SnSn1a_n = S_n - S_{n-1}

l=an=a+(n1)dl = a_n = a + (n - 1)d

Sn=n(n+1)2S_n = \frac{n(n + 1)}{2} (Sum of first nn natural numbers)

💡Examples

Problem 1:

Find the sum of the first 22 terms of the AP: 8,3,2,...8, 3, -2, ...

Solution:

Step 1: Identify the given values. Here, the first term a=8a = 8, the common difference d=38=5d = 3 - 8 = -5, and the number of terms n=22n = 22. Step 2: Use the formula Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n - 1)d]. Step 3: Substitute the values: S22=222[2(8)+(221)(5)]S_{22} = \frac{22}{2}[2(8) + (22 - 1)(-5)]. Step 4: Simplify inside the brackets: S22=11[16+(21)(5)]=11[16105]S_{22} = 11[16 + (21)(-5)] = 11[16 - 105]. Step 5: Final calculation: S22=11(89)=979S_{22} = 11(-89) = -979.

Explanation:

Since we know the first term, the common difference, and the total number of terms, we use the standard SnS_n formula involving aa and dd.

Problem 2:

In an AP, the first term a=5a = 5, the last term l=45l = 45, and the sum Sn=400S_n = 400. Find the number of terms nn.

Solution:

Step 1: Identify the given values: a=5a = 5, l=45l = 45, and Sn=400S_n = 400. Step 2: Use the simplified sum formula Sn=n2(a+l)S_n = \frac{n}{2}(a + l). Step 3: Substitute the known values: 400=n2(5+45)400 = \frac{n}{2}(5 + 45). Step 4: Solve for nn: 400=n2(50)400=25n400 = \frac{n}{2}(50) \Rightarrow 400 = 25n. Step 5: n=40025=16n = \frac{400}{25} = 16.

Explanation:

When the first and last terms are provided, it is much faster to use the formula Sn=n2(a+l)S_n = \frac{n}{2}(a + l) to find the missing variable nn.