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Algebra - Arithmetic Progression (AP)

Grade 10ICSE

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

πŸ”‘Concepts

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Definition of Arithmetic Progression (AP): An Arithmetic Progression 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, you can imagine an AP as a ladder where each rung is placed at an equal vertical distance from the previous one.

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First Term and Common Difference: The first term of an AP is denoted by aa. The common difference dd can be positive (resulting in an increasing sequence like a rising staircase), negative (a decreasing sequence like a descending staircase), or zero (a constant sequence where all terms are identical).

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General Term (nthn^{th} term): The nthn^{th} term, denoted as ana_n or TnT_n, allows you to calculate the value of any position in the sequence. Conceptually, this is like starting at the first point on a number line and taking nβˆ’1n-1 equal-sized jumps of length dd.

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Sum of nn Terms (SnS_n): This represents the total of the first nn terms of the AP. Visually, if you represent each term as a vertical column of blocks, the sum is the total number of blocks used, which forms a shape similar to a trapezoid.

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Arithmetic Mean: When three numbers a,b,ca, b, c are in AP, the middle term bb is called the arithmetic mean of aa and cc. Geometrically, on a 1D number line, bb is the exact midpoint between the points aa and cc, calculated as b=a+c2b = \frac{a+c}{2}.

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Properties of AP: If a constant is added to, subtracted from, multiplied by, or divided into (except by zero) each term of an AP, the resulting sequence remains an AP. This reflects a uniform transformation across the entire linear pattern of the sequence.

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Selection of Terms: For solving problems involving a specific number of terms in AP, it is strategically useful to pick them symmetrically. For 3 terms, use (aβˆ’d),a,(a+d)(a-d), a, (a+d); for 4 terms, use (aβˆ’3d),(aβˆ’d),(a+d),(a+3d)(a-3d), (a-d), (a+d), (a+3d). This visual symmetry ensures that the sum of terms eliminates the variable dd.

πŸ“Formulae

d=anβˆ’anβˆ’1d = a_n - a_{n-1}

an=a+(nβˆ’1)da_n = a + (n-1)d

Sn=n2[2a+(nβˆ’1)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

ArithmeticΒ Mean=a+b2\text{Arithmetic Mean} = \frac{a+b}{2}

nthΒ termΒ fromΒ theΒ end=lβˆ’(nβˆ’1)dn^{th} \text{ term from the end} = l - (n-1)d

an=Snβˆ’Snβˆ’1a_n = S_n - S_{n-1}

πŸ’‘Examples

Problem 1:

Find the 20th20^{th} term and the sum of the first 2020 terms of the AP: 5,9,13,17,…5, 9, 13, 17, \dots

Solution:

  1. Identify the given values: First term a=5a = 5, Common difference d=9βˆ’5=4d = 9 - 5 = 4, and n=20n = 20.
  2. To find the 20th20^{th} term (a20a_{20}): a20=a+(nβˆ’1)d=5+(20βˆ’1)4a_{20} = a + (n-1)d = 5 + (20-1)4 a20=5+(19Γ—4)=5+76=81a_{20} = 5 + (19 \times 4) = 5 + 76 = 81
  3. To find the sum of the first 2020 terms (S20S_{20}): S20=n2[2a+(nβˆ’1)d]=202[2(5)+(19Γ—4)]S_{20} = \frac{n}{2} [2a + (n-1)d] = \frac{20}{2} [2(5) + (19 \times 4)] S20=10[10+76]=10Γ—86=860S_{20} = 10 [10 + 76] = 10 \times 86 = 860

Explanation:

We first identify the starting point (aa) and the constant increase (dd). We then use the general term formula to find the value at a specific position and the sum formula to aggregate all values up to that position.

Problem 2:

How many terms of the AP 24,21,18,…24, 21, 18, \dots must be taken so that their sum is 7878?

Solution:

  1. Given: a=24a = 24, d=21βˆ’24=βˆ’3d = 21 - 24 = -3, and Sn=78S_n = 78.
  2. Use the sum formula: Sn=n2[2a+(nβˆ’1)d]S_n = \frac{n}{2} [2a + (n-1)d] 78=n2[2(24)+(nβˆ’1)(βˆ’3)]78 = \frac{n}{2} [2(24) + (n-1)(-3)] 156=n[48βˆ’3n+3]156 = n [48 - 3n + 3] 156=n[51βˆ’3n]=51nβˆ’3n2156 = n [51 - 3n] = 51n - 3n^2
  3. Form a quadratic equation: 3n2βˆ’51n+156=03n^2 - 51n + 156 = 0 Divide by 33: n2βˆ’17n+52=0n^2 - 17n + 52 = 0
  4. Factorize: (nβˆ’4)(nβˆ’13)=0(n-4)(n-13) = 0 n=4n = 4 or n=13n = 13.

Explanation:

By substituting the known values into the sum formula, we generate a quadratic equation. The two positive integer solutions n=4n=4 and n=13n=13 mean that both the first 4 terms and the first 13 terms sum to 78. This happens because the terms become negative after a certain point, and the sum of terms from the 5th5^{th} to the 13th13^{th} is zero.