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Sequences and Series - Arithmetic Progression (A.P.)

Grade 11CBSE

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

🔑Concepts

Definition of Arithmetic Progression (A.P.): An A.P. is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant is called the common difference dd. Visually, if you plot the term number nn on the horizontal axis and the value of the term ana_n on the vertical axis, the points lie on a straight line, similar to a linear graph y=mx+cy = mx + c where the slope mm corresponds to the common difference dd.

Common Difference (dd): The common difference is calculated as d=anan1d = a_{n} - a_{n-1}. It can be positive, negative, or zero. If d>0d > 0, the sequence is increasing (like a rising staircase); if d<0d < 0, the sequence is decreasing (like a descending staircase); if d=0d = 0, the sequence is constant.

General Term (nn-th term): The nn-th term of an A.P. with first term aa and common difference dd is denoted by ana_n. It represents the value at the nn-th position. In a list of terms, this is the rule that allows you to find any term without listing all previous ones.

Sum of First nn Terms (SnS_n): This represents the total value obtained by adding the first nn terms of the progression. Conceptually, this sum is equivalent to the average of the first and last terms multiplied by the number of terms. If you visualize the terms as bars of increasing height, the sum is the total area of all bars.

Arithmetic Mean (A.M.): Given two numbers aa and bb, their Arithmetic Mean is a number AA such that a,A,ba, A, b are in A.P. Geometrically, AA is the midpoint between aa and bb on a number line. If nn numbers A1,A2,...,AnA_1, A_2, ..., A_n are inserted between aa and bb such that the resulting sequence is an A.P., these are called nn arithmetic means.

Properties of A.P.: If a constant is added to, subtracted from, multiplied by, or divided into (except zero) each term of an A.P., the resulting sequence is also an A.P. For example, shifting a 'staircase' up or down by a fixed height does not change the equal spacing between the steps.

Selection of Terms: For solving problems involving the sum or product of terms, it is often convenient to select terms symmetrically. For three terms in A.P., use (ad),a,(a+d)(a-d), a, (a+d). For four terms, use (a3d),(ad),(a+d),(a+3d)(a-3d), (a-d), (a+d), (a+3d).

📐Formulae

an=a+(n1)da_n = a + (n-1)d

d=anan1d = a_n - a_{n-1}

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

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

A=a+b2A = \frac{a+b}{2}

d=ban+1d = \frac{b-a}{n+1} (for inserting nn arithmetic means between aa and bb)

💡Examples

Problem 1:

Find the 15th term and the sum of the first 15 terms of the A.P.: 3,7,11,15,...3, 7, 11, 15, ...

Solution:

  1. Identify the first term and common difference: First term a=3a = 3 Common difference d=73=4d = 7 - 3 = 4

  2. Calculate the 15th term (a15a_{15}): a15=a+(151)da_{15} = a + (15-1)d a15=3+14(4)a_{15} = 3 + 14(4) a15=3+56=59a_{15} = 3 + 56 = 59

  3. Calculate the sum of the first 15 terms (S15S_{15}): Using the formula Sn=n2(a+l)S_n = \frac{n}{2}(a + l) where l=a15=59l = a_{15} = 59: S15=152(3+59)S_{15} = \frac{15}{2}(3 + 59) S15=152(62)S_{15} = \frac{15}{2}(62) S15=15×31=465S_{15} = 15 \times 31 = 465

Explanation:

This problem demonstrates the direct application of the general term formula and the sum formula. First, we extract the parameters aa and dd from the sequence, then find the specific term, and finally use that term as the 'last term' to simplify the sum calculation.

Problem 2:

Insert 3 arithmetic means between 8 and 24.

Solution:

  1. Let the three means be A1,A2,A3A_1, A_2, A_3. The sequence 8,A1,A2,A3,248, A_1, A_2, A_3, 24 is in A.P.
  2. Here, first term a=8a = 8, last term b=24b = 24, and number of means n=3n = 3.
  3. Calculate the common difference dd: d=ban+1=2483+1=164=4d = \frac{b-a}{n+1} = \frac{24-8}{3+1} = \frac{16}{4} = 4
  4. Find the means: A1=a+d=8+4=12A_1 = a + d = 8 + 4 = 12 A2=a+2d=8+8=16A_2 = a + 2d = 8 + 8 = 16 A3=a+3d=8+12=20A_3 = a + 3d = 8 + 12 = 20 The three arithmetic means are 12,16,12, 16, and 2020.

Explanation:

To insert nn means, we treat the entire set (including the boundaries) as an A.P. with n+2n+2 terms. We find the common difference that spans the gap between the two given numbers over the required number of steps.