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Arithmetic Progressions - Model daily-life growth patterns using arithmetic progression concepts

Grade 10CBSE

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 the difference between any two consecutive terms is constant. This constant is called the common difference (dd).

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In daily life, growth patterns such as fixed annual salary increments, simple interest accumulations, and uniform physical stacking (like bricks or logs) can be modeled using AP.

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The first term is denoted by aa, the common difference by dd, the number of terms by nn, and the nthn^{th} term by ana_n.

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If a situation involves finding a value at a specific point in time or position, we use the nthn^{th} term formula.

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If a situation involves finding the total accumulation over a period (e.g., total savings, total number of objects), we use the Sum of nn terms formula.

📐Formulae

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) where l is the last term.S_n = \frac{n}{2} (a + l) \text{ where } l \text{ is the last term.}

d=an−an−1d = a_{n} - a_{n-1}

💡Examples

Problem 1:

A man starts his job with a monthly salary of ₹8,000₹ 8,000 and receives an annual increment of ₹500₹ 500. Find his monthly salary in the 15th15^{th} year.

Solution:

Here, the starting salary a=8000a = 8000. The annual increment is the common difference d=500d = 500. We need to find the salary in the 15th15^{th} year, so n=15n = 15. Using the formula an=a+(n−1)da_n = a + (n - 1)d: a15=8000+(15−1)500a_{15} = 8000 + (15 - 1)500 a15=8000+14×500a_{15} = 8000 + 14 \times 500 a15=8000+7000a_{15} = 8000 + 7000 a15=15000a_{15} = 15000

Explanation:

The salary follows an arithmetic progression because the increment is constant every year. The salary in the 15th15^{th} year is ₹15,000₹ 15,000.

Problem 2:

A student saves ₹5₹ 5 in the first week of a year and then increases her weekly savings by ₹1.75₹ 1.75 each week. In which week will her weekly savings be ₹20.75₹ 20.75?

Solution:

Given a=5a = 5, d=1.75d = 1.75, and an=20.75a_n = 20.75. We need to find nn. an=a+(n−1)da_n = a + (n - 1)d 20.75=5+(n−1)1.7520.75 = 5 + (n - 1)1.75 20.75−5=(n−1)1.7520.75 - 5 = (n - 1)1.75 15.75=(n−1)1.7515.75 = (n - 1)1.75 n−1=15.751.75n - 1 = \frac{15.75}{1.75} n−1=9n - 1 = 9 n=10n = 10

Explanation:

By identifying the initial savings and the constant increase, we set up an AP equation to solve for the specific time period (week) when the target amount is reached.

Problem 3:

In a flower bed, there are 2323 rose plants in the first row, 2121 in the second, 1919 in the third, and so on. There are 55 rose plants in the last row. How many rows are there in the flower bed?

Solution:

The sequence is 23,21,19,…,523, 21, 19, \dots, 5. Here a=23a = 23, d=21−23=−2d = 21 - 23 = -2, and an=5a_n = 5. an=a+(n−1)da_n = a + (n - 1)d 5=23+(n−1)(−2)5 = 23 + (n - 1)(-2) 5−23=−2(n−1)5 - 23 = -2(n - 1) −18=−2(n−1)-18 = -2(n - 1) 9=n−19 = n - 1 n=10n = 10

Explanation:

The number of plants decreases by a fixed amount, forming an AP with a negative common difference. There are 1010 rows in total.