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Predicting What Comes Next: Exploring Sequences - Arithmetic Progressions

Grade 9CBSE

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

🔑Concepts

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A Sequence is an ordered list of numbers following a specific pattern or rule.

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An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is always constant. This constant value is known as the common difference (dd).

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The first term of an AP is denoted by aa, and the terms are generally represented as a,a+d,a+2d,a+3d,…a, a+d, a+2d, a+3d, \dots.

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If the common difference dd is positive (d>0d > 0), the sequence is increasing; if dd is negative (d<0d < 0), the sequence is decreasing; if d=0d = 0, all terms are the same.

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The nn-th term of an AP is also called the general term.

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A finite AP has a last term, often denoted by ll.

📐Formulae

d=an−an−1d = a_{n} - a_{n-1}

an=a+(n−1)da_n = a + (n - 1)d

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)

💡Examples

Problem 1:

Find the 20th20^{th} term of the AP: 3,7,11,15,…3, 7, 11, 15, \dots

Solution:

a=3,d=7−3=4,n=20a = 3, d = 7 - 3 = 4, n = 20 a20=a+(20−1)da_{20} = a + (20 - 1)d a20=3+(19×4)a_{20} = 3 + (19 \times 4) a20=3+76=79a_{20} = 3 + 76 = 79

Explanation:

Identify the first term aa and the common difference dd. Substitute these values into the general term formula an=a+(n−1)da_n = a + (n-1)d for n=20n=20.

Problem 2:

Calculate the sum of the first 1010 terms of the sequence: 2,5,8,…2, 5, 8, \dots

Solution:

a=2,d=5−2=3,n=10a = 2, d = 5 - 2 = 3, n = 10 S10=102[2(2)+(10−1)3]S_{10} = \frac{10}{2} [2(2) + (10 - 1)3] S10=5[4+(9×3)]S_{10} = 5 [4 + (9 \times 3)] S10=5[4+27]=5×31=155S_{10} = 5 [4 + 27] = 5 \times 31 = 155

Explanation:

Use the sum formula Sn=n2[2a+(n−1)d]S_n = \frac{n}{2} [2a + (n-1)d] where aa is the first term, dd is the common difference, and nn is the number of terms.

Problem 3:

Which term of the AP: 21,18,15,…21, 18, 15, \dots is −81-81?

Solution:

a=21,d=18−21=−3,an=−81a = 21, d = 18 - 21 = -3, a_n = -81 −81=21+(n−1)(−3)-81 = 21 + (n - 1)(-3) −81−21=(n−1)(−3)-81 - 21 = (n - 1)(-3) −102=(n−1)(−3)-102 = (n - 1)(-3) n−1=−102−3=34n - 1 = \frac{-102}{-3} = 34 n=34+1=35n = 34 + 1 = 35

Explanation:

Set the nn-th term formula equal to −81-81 and solve for nn. Note that the common difference dd is negative because the sequence is decreasing.