10 Math Arithmetic Progression

Finish Line & Beyond

ARITHMETIC PROGRESSIONS

General Term of AP nth Term of an AP Sum of n Terms of an AP

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Finish Line & Beyond Arithmetic Progression ( AP ): An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is called the common difference of the AP. This can be positive or negative.

Consider the following lists of numbers : (i) 1, 2, 3, 4, . . . (ii) 100, 70, 40, 10, . . . (iii) – 3, –2, –1, 0, . . . (iv) 3, 3, 3, 3, . . . (v) –1.0, –1.5, –2.0, –2.5, . . . Each of the numbers in the list is called a term. In (i), each term is 1 more than the term preceding it. In (ii), each term is 30 less than the term preceding it. In (iii), each term is obtained by adding 1 to the term preceding it. In (iv), all the terms in the list are 3 , i.e., each term is obtained by adding (or subtracting) 0 to the term preceding it. In (v), each term is obtained by adding – 0.5 to (i.e., subtracting 0.5 from) the term preceding it. General Form of an AP: Let us denote the first term of an AP by a1, second term by a2, . . ., nth term by an and the common difference by d. Then the AP becomes a1, a2, a3, . . ., an. So, a2 – a1 = a3 – a2 = . . . = an – an – 1 = d. So, AP can also be written as: a, a+d, a+2d, a+3d …………………….. Finite AP: The AP which is having a last term is called as finite AP. Infinite AP: The AP which is not having a last term is called as infinite AP.

Example: For the AP :

3 1 1 3 , ,- ,2 2 2 2

write the first term a and the common difference d. Solution: Here, a =

3 2

And d= 1 Example: Which of the following list of numbers does form an AP? If they form an AP, write the next two terms: (i) 4, 10, 16, 22, . . . (ii) 1, – 1, – 3, – 5, . . .

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Finish Line & Beyond (iii) – 2, 2, – 2, 2, – 2, . . . (iv) 1, 1, 1, 2, 2, 2, 3, 3, 3, . . . Solution: (i) We have a2 – a1 = 10 – 4 = 6 a3 – a2 = 16 – 10 = 6 a4 – a3 = 22 – 16 = 6

a

a

a

a

i.e., k + 1 – k is the same every time. So, the given list of numbers forms an AP with the common difference d = 6. The next two terms are: 22 + 6 = 28 and 28 + 6 = 34. (ii) a2 – a1 = – 1 – 1 = – 2 a3 – a2 = – 3 – ( –1 ) = – 3 + 1 = – 2 a4 – a3 = – 5 – ( –3 ) = – 5 + 3 = – 2 i.e., k + 1 – k is the same every time. So, the given list of numbers forms an AP with the common difference d = – 2. The next two terms are: – 5 + (– 2 ) = – 7 and – 7 + (– 2 ) = – 9 (iii) a2 – a1 = 2 – (– 2) = 2 + 2 = 4 a3 – a2 = – 2 – 2 = – 4 As a2 – a1 a3 – a2 , the given list of numbers does not form an AP. ARITHMETIC PROGRESSIONS 99 (iv) a2 – a1 = 1 – 1 = 0 a3 – a2 = 1 – 1 = 0 a4 – a3 = 2 – 1 = 1 Here, a2 – a1 = a3 – a2 ≠ a4 – a3. So, the given list of numbers does not form an AP.

nth Term of an AP: The nth term an of the AP with first term a and common difference d is given by an = a + (n – 1) d.

Example: Reena applied for a job and got selected. She has been offered the job with a starting monthly salary of Rs 8000, with an annual increment of Rs 500. What would be her monthly salary for the fifth year? Solution: a = 8000 d = 500 n = 5 years So, a5 = a + (5 – 1) 500 = 8000 + 2000 = 10000

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Finish Line & Beyond Example: Find the 10th term of the AP : 2, 7, 12, . . . Solution: Here, a = 2, d = 7 – 2 = 5 and n = 10. We have an = a + (n – 1) d So, a10 = 2 + (10 – 1) × 5 = 2 + 45 = 47 Therefore, the 10th term of the given AP is 47. Example: Which term of the AP : 21, 18, 15, . . . is – 81? Also, is any term 0? Give reason for your answer. Solution: Here, a = 21, d = 18 – 21 = – 3 and an = – 81, and we have to find n. As an = a + ( n – 1) d, we have – 81 = 21 + (n – 1)(– 3) – 81 = 24 – 3n – 105 = – 3n So, n = 35 Therefore, the 35th term of the given AP is – 81. Next, we want to know if there is any n for which an = 0. If such an n is there, then 21 + (n – 1) (–3) = 0, i.e., 3(n – 1) = 21 i.e., n = 8 So, the eighth term is 0. Example: Determine the AP whose 3rd term is 5 and the 7th term is 9. Solution: We have a3 = a + (3 – 1) d = a + 2d = 5 ………………………………………………… (1) and a7 = a + (7 – 1) d = a + 6d = 9 …………………………………………. (2) From equation (1) and (2) a = 5 – 2d and a = 9 – 6d ∴ 5 – 2d = 9 – 6d Or, 4d = 4 ∴ d = 1 and a = 3 Hence, the required AP is 3, 4, 5, 6, 7, . . . Example: Check whether 301 is a term of the list of numbers 5, 11, 17, 23, . . . Solution: We have : a2 – a1 = 11 – 5 = 6, a3 – a2 = 17 – 11 = 6, a4 – a3 = 23 – 17 = 6 Now, a = 5 and d = 6. Let 301 be a term, say, the nth term of the this AP. As we know, an = a + (n – 1) d So, 301 = 5 + (n – 1) × 6 Or, 301 = 6n – 1

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Finish Line & Beyond

Or, n = 302 ÷ 6 = 50

1 3

But n should be a positive integer as position of a particular term can be an integer only. So, 301 is not a term of the given list of numbers. Example: How many two-digit numbers are divisible by 3? Solution: The smallest two digit number divisible by 3 is 12 and the greatest two digit number divisible by 3 is 99. So, a = 12, d = 3, an = 99. Using the formula, an = a + (n – 1) d, we have 99 = 12 + (n – 1) × 3 Or, 87 = (n – 1) × 3 Or, n – 1 = 87 ÷ 3 = 29 Or, n = 29 + 1 = 30 So, there are 30 two-digit numbers divisible by 3. Example: Find the 11th term from the last term (towards the first term) of the AP : 10, 7, 4, . . ., – 62. Solution: Here, a = 10, d = 7 – 10 = – 3, l = – 62, where l = a + (n – 1) d To find the 11th term from the last term, we need to find the total number of terms in the AP. So, – 62 = 10 + (n – 1)(–3) Or, – 72 = (n – 1)(–3) Or, n – 1 = 24 Or, n = 25 So, there are 25 terms in the given AP. The 11th term from the last term will be the 15th term So, a15 = 10 + (15 – 1)(–3) = 10 – 42 = – 32 i.e., the 11th term from the last term is – 32. Alternative Solution: If we write the given AP in the reverse order, then a = – 62 and d = 3 So, the question now becomes finding the 11th term with these a and d. So, a11 = – 62 + (11 – 1) × 3 = – 62 + 30 = – 32 So, the 11th term, which is now the required term, is – 32. Example: A sum of Rs 1000 is invested at 8% simple interest per year. Calculate the interest at the end of each year. Do these interests form an AP? If so, find the interest at the end of 30 years making use of this fact. Solution: Using the formula of Simple Interest =

P× R × T 100

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Finish Line & Beyond

So, the interest at the end of the 1st year =

The interest at the end of the 2nd year =

1000 × 8 × 1 = 80 100

1000 × 8 × 2 = 160 100

Similarly for every additional year Rs. 80 will be added as interest So, a = 80, d = 80 Using the formula an = a + (n-1)d We get, a30 = 80 +(30-1)80 = 2400, which is the interest at the end of 30 years.

Sum of First n Terms of an AP: The sum of the first n terms of an AP is given by S=

n [2a + (n – 1) d ] 2

We can also write this as S =

n (a + an) 2

This form of the result is useful when the first and the last terms of an AP are given and the common difference is not given.

Example: Find the sum of the first 22 terms of the AP : 8, 3, –2, . . . Solution: Here, a = 8, d = 3 – 8 = –5, n = 22. Using the formula we get, S=

22 (2 × 8 + 21 × -5) 2

= 11(-89) = -979 So, the sum of the first 22 terms of the AP is – 979. Example: If the sum of the first 14 terms of an AP is 1050 and its first term is 10, find the 20th term. Solution: Here, S14 = 1050, n = 14, a = 10. Or, 1050 = 7(20 + 13d) Or, 1050 = 140 + 91d Or, 91d = 910 Or, d = 10 ∴ n20 = 10 + 19 × 10 = 200

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Finish Line & Beyond Example: How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78? Solution: Here, a = 24, d = 21 – 24 = –3, Sn = 78. We need to find n. Using the formula we get, 78 =

n (48+(n-1)-3) 2

Or, 156 = n(48-3n +3) Or, 156 = 51n-3n² Or, 3n²-51n+156 = 0 Or, n²-17n+52 = 0 Or, n²-4n-13n+52 = 0 Or, n(n-4)-13(n-4) = 0 Or, (n-4)(n-13) = 0 Or, n= 4 and n = 13 Both values of n are admissible. So, the number of terms is either 4 or 13. Remarks: 1. In this case, the sum of the first 4 terms = the sum of the first 13 terms = 78. 2. Two answers are possible because the sum of the terms from 5th to 13th will be zero. This is because a is positive and d is negative, so that some terms will be positive and some others negative, and will cancel out each other.

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