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Sequences and Progressions - Find nth term of arithmetic progressions and interpret AP in practical contexts

Grade 9CBSE

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

🔑Concepts

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An Arithmetic Progression (AP) is a sequence of numbers in which each term is obtained by adding a fixed number dd to the preceding term, except the first term aa.

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The fixed number dd is called the common difference. It can be positive, negative, or zero. It is calculated as d=a2−a1=a3−a2=…d = a_2 - a_1 = a_3 - a_2 = \dots

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The first term is usually denoted by aa or a1a_1, and the nthn^{th} term is denoted by ana_n.

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A sequence is an AP if the difference between any two consecutive terms (ak+1−aka_{k+1} - a_k) is constant for all values of kk.

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In practical contexts, the first term aa represents the starting value (e.g., initial salary, initial distance), and the common difference dd represents the constant rate of change (e.g., annual increment, constant speed increase).

📐Formulae

an=a+(n−1)da_n = a + (n - 1)d

d=an−an−1d = a_n - a_{n-1}

l=a+(n−1)d (where l is the last term)l = a + (n - 1)d \text{ (where } l \text{ is the last term)}

💡Examples

Problem 1:

Find the 15th15^{th} term of the arithmetic progression: 3,8,13,18,…3, 8, 13, 18, \dots

Solution:

Given AP: 3,8,13,18,…3, 8, 13, 18, \dots First term a=3a = 3 Common difference d=8−3=5d = 8 - 3 = 5 We need to find a15a_{15}, so n=15n = 15. Using the formula an=a+(n−1)da_n = a + (n - 1)d: a15=3+(15−1)5a_{15} = 3 + (15 - 1)5 a15=3+(14×5)a_{15} = 3 + (14 \times 5) a15=3+70a_{15} = 3 + 70 a15=73a_{15} = 73

Explanation:

To find a specific term in an AP, identify the first term and common difference, then substitute them into the general term formula an=a+(n−1)da_n = a + (n-1)d.

Problem 2:

A manufacturer of TV sets produced 600600 sets in the third year and 700700 sets in the seventh year. Assuming that the production increases uniformly by a fixed number every year, find the production in the 10th10^{th} year.

Solution:

Let the production in the first year be aa and the fixed annual increase be dd. Production in the 3rd3^{rd} year (a3a_3) = 600600 Production in the 7th7^{th} year (a7a_7) = 700700 Using an=a+(n−1)da_n = a + (n - 1)d:

  1. a+2d=600a + 2d = 600
  2. a+6d=700a + 6d = 700 Subtracting (1) from (2): a+6d=700−(a+2d=600)4d=100\begin{array}{r} a + 6d = 700 \\ -(a + 2d = 600) \\ \hline 4d = 100 \end{array} d=1004=25d = \frac{100}{4} = 25 Substitute d=25d = 25 in (1): a+2(25)=600⇒a+50=600⇒a=550a + 2(25) = 600 \Rightarrow a + 50 = 600 \Rightarrow a = 550 Now, find production in the 10th10^{th} year (a10a_{10}): a10=a+9da_{10} = a + 9d a10=550+9(25)a_{10} = 550 + 9(25) a10=550+225=775a_{10} = 550 + 225 = 775

Explanation:

This is a practical application where production follows an AP. We use the given information to create two linear equations, solve for aa and dd, and then find the required term.

Problem 3:

Reena saves ₹32₹ 32 during the first month, ₹36₹ 36 in the second month, and ₹40₹ 40 in the third month. If she continues to save in this manner, in which month will she save ₹200₹ 200?

Solution:

The savings follow an AP: 32,36,40,…32, 36, 40, \dots Here a=32a = 32 and d=36−32=4d = 36 - 32 = 4. We are given the nthn^{th} term an=200a_n = 200 and we need to find nn. an=a+(n−1)da_n = a + (n - 1)d 200=32+(n−1)4200 = 32 + (n - 1)4 Subtracting 3232 from both sides: 200−32168\begin{array}{r} 200 \\ - 32 \\ \hline 168 \end{array} 168=(n−1)4168 = (n - 1)4 Divide by 44: 1684=n−1\frac{168}{4} = n - 1 42=n−142 = n - 1 n=43n = 43

Explanation:

In this context, the month index nn is the unknown. We solve the linear equation for nn to determine when the savings goal is reached.