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Sequences and Series - Sequences and Series

Grade 11CBSE

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

🔑Concepts

A sequence is a collection of numbers arranged in a specific order, such as a1,a2,a3,...,ana_1, a_2, a_3, ..., a_n. A series is the expression obtained by adding the terms of a sequence, often represented using the sigma notation ai\sum a_i. Visually, a sequence can be represented as discrete dots on a coordinate plane where the x-axis represents the term number and the y-axis represents the term value.

An Arithmetic Progression (AP) is a sequence in which the difference between any two consecutive terms is a constant called the common difference dd. Visually, if you plot the terms of an AP against their position nn, the points will lie on a straight line with a slope equal to dd. If d>0d > 0, the line slopes upward; if d<0d < 0, it slopes downward.

The Arithmetic Mean (AM) of two numbers aa and bb is the value A=a+b2A = \frac{a+b}{2}. Geometrically, the AM represents the midpoint between two points on a number line. If nn numbers are inserted between aa and bb such that the resulting sequence is an AP, these are called nn arithmetic means.

A Geometric Progression (GP) is a sequence where each term is obtained by multiplying the preceding term by a constant non-zero number rr, known as the common ratio. Visually, the terms of a GP follow an exponential curve. If r>1r > 1, the sequence grows rapidly; if 0<r<10 < r < 1, the sequence decays and the points approach the x-axis asymptotically.

The Geometric Mean (GM) for two positive numbers aa and bb is given by G=abG = \sqrt{ab}. In a geometric context, if a semicircle is drawn with a diameter equal to a+ba + b, the length of the perpendicular segment from the meeting point of aa and bb to the circle's boundary is the GM.

The relationship between AM and GM states that for any two positive real numbers, the Arithmetic Mean is always greater than or equal to the Geometric Mean (AGA \ge G). This inequality is a fundamental property used to find minimum or maximum values in various mathematical problems.

The sum of the first nn terms of a sequence is denoted by SnS_n. For an AP, this sum represents the total accumulated value which grows quadratically with nn. For a GP, the sum depends on whether the ratio rr is greater or less than 1, representing how quickly the total magnitude of the series is scaling.

📐Formulae

nthn^{th} term of an AP: an=a+(n1)da_n = a + (n-1)d

Sum of first nn terms of an AP: Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d] or Sn=n2[a+l]S_n = \frac{n}{2}[a + l], where ll is the last term.

Arithmetic Mean (AM) of aa and bb: A=a+b2A = \frac{a+b}{2}

nthn^{th} term of a GP: an=arn1a_n = ar^{n-1}

Sum of first nn terms of a GP: Sn=a(rn1)r1S_n = \frac{a(r^n - 1)}{r-1} if r1r \neq 1

Geometric Mean (GM) of aa and bb: G=abG = \sqrt{ab}

Relationship between AM and GM: AGA \ge G

💡Examples

Problem 1:

Find the 10th10^{th} term and the sum of the first 10 terms of the arithmetic progression: 2,5,8,11,...2, 5, 8, 11, ...

Solution:

  1. Identify the first term a=2a = 2 and calculate the common difference d=52=3d = 5 - 2 = 3.
  2. To find the 10th10^{th} term (a10a_{10}), use the formula an=a+(n1)da_n = a + (n-1)d: a10=2+(101)3=2+27=29a_{10} = 2 + (10-1)3 = 2 + 27 = 29.
  3. To find the sum of the first 10 terms (S10S_{10}), use the formula Sn=n2[a+l]S_n = \frac{n}{2}[a + l] where ll is the 10th10^{th} term: S10=102[2+29]=5×31=155S_{10} = \frac{10}{2}[2 + 29] = 5 \times 31 = 155.

Explanation:

We first identify the parameters of the AP and then apply the standard formulas for the general term and the sum of an arithmetic series.

Problem 2:

In a Geometric Progression, the 3rd3^{rd} term is 24 and the 6th6^{th} term is 192. Find the 10th10^{th} term.

Solution:

  1. Let the first term be aa and common ratio be rr. According to the GP formula an=arn1a_n = ar^{n-1}: a3=ar2=24a_3 = ar^2 = 24 ... (i) a6=ar5=192a_6 = ar^5 = 192 ... (ii)
  2. Divide equation (ii) by (i): ar5ar2=19224    r3=8    r=2\frac{ar^5}{ar^2} = \frac{192}{24} \implies r^3 = 8 \implies r = 2.
  3. Substitute r=2r = 2 into (i): a(2)2=24    4a=24    a=6a(2)^2 = 24 \implies 4a = 24 \implies a = 6.
  4. Find the 10th10^{th} term: a10=ar9=6×(2)9=6×512=3072a_{10} = ar^9 = 6 \times (2)^9 = 6 \times 512 = 3072.

Explanation:

By setting up a system of equations based on the GP term formula, we solve for the ratio rr first, then the first term aa, and finally calculate the required term.