Arithmetic Progression Review

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Definitions/Properties/Formulas/Methods/Tips

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A sequence is a collection of numbers arranged in a definite order according to some definite rule.

Examples: (1) n

1 t:rule ...,

3

1,

2

1,1 n

(2) 1n 15 t: Rule...,5,5,5,5 n

(3)P = 100, R = 10 p. c. p. a. A = Amount in N years 110, 120, 130, … Rule:

100

PNRP

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?11111

12343211111

12321111

12111

2

2

2

2

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?10001

10020011001

10201101

12111

2

2

2

2

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?6667

444889667

448967

497

2

2

2

2

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1 2

L1

1 2 3 4

L1

L2

1 2 3 4 5 6 7

8

L1

L2

L3

No of lines 1 2 3 4 No of parts 21 = 2 22 = 4 23 = 8 ?

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A progression is a special type of sequence, in which the relationship between any two consecutive terms is the same. Examples:

1. 20, 10, 0, -10, … Relation: 101 nn tt

2. 3, 12, 48, 192

Relation: 41

n

n

t

t

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Arithmetic Progression is a sequence in which difference between two

consecutive terms is constant.

Example: The arrangement of chairs in an auditorium.

Row No 1 2 3 … n No of chairs 30 40 50 … 20+10n

Screen

For an Arithmetic Progression

nth term tn = a + (n-1)d

where a = first term and d= common difference

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For an Arithmetic Progression sum

of the first n terms sn = dna

n12

2 or

sn =

21 ntt

n

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2

1

...321

numbers natural nfirst of Sum

1

nn

nrnr

r

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1

2...6422

numbers even nfirst of Sum

1

nn

nrnr

r

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2

1

12...53112

numbers odd nfirst of Sum

n

nrnr

r

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For an A. P. with first term ‘a’ and the common difference ‘d’, if any term k is added to each term of an A. P. then the new sequence is also A. P. with first term ‘a + k’ and the same common difference ‘d’.

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For an A. P. with first term ‘a’ and the common difference ‘d’, if each term of this A. P. is multiplied by any real number k, then the new sequence is also A. P. with first term ‘ak’ & the common difference ‘dk’.

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It is convenient to consider three consecutive terms of

an A. P. as a - d, a, a + d

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It is convenient to consider four consecutive terms of an A. P. as a-2d, a-d, a+d, a+2d

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It is convenient to consider five consecutive terms of an AP as a-2d, a-d, a, a+d, a+2d

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The first five terms of A. P. with first term a = 2 and

the common difference d= 3 are

14311

1138

835

532

2

5

4

3

2

1

t

t

t

t

t

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Solution: Here a = 10 and d = -3 As first four terms of General A. P.

are a, a + d, a + 2d, a + 3d, First four terms of given A. P. are

10, 10-3, 10-6, 10-9 i. e. 10, 7, 4, 1

Find the first four terms of A. P. with first term 10 & common difference -3

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The first four terms of an A. P. with first term a = 5 and the common difference d= 0

are 5, 5, 5, 5 5

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Solution Here a = 1 and d = 7-1 = 6 For an A. P. tn = a + d(n-1)

t11 = 1 + 6(11-1) The 11th term of given AP is 61

What is the 11th term an A. P. 1, 7, 13, …?

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Solution: Let a be the first term & d be the common difference of given AP.

For an AP tn = a + d(n-1) …(1) t3 = a + 2d and t4 = a + 3d

a + 2d = 12 …(2) and a + 3d = 17 …(3) By solving (2) & (3) d = 5 & a = 2

By substituting a =2 and d =5 in (1) tn = 2 + 5(n-1) i.e. tn = 5n - 3

Find tn for an AP where t3 = 12 & t4 = 17

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Find tn for an AP where t1 = 3 and t7 = 15

Solution: Here a = 3

For an AP tn = a + d(n-1) …(1) t7 = 3 + 6d

3 + 6d = 15 d = 2 By substituting a=3 and d=2 in (1)

tn = 3 + 2(n-1) tn = 2n + 1

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Find tn for an AP where t11=33 & t21=63 Solution: Let a be the first term & d be

the common difference of given AP. For an AP tn = a + d(n-1) …(1) t11 = a + 10d and t21 = a + 20d a + 10d = 33 …………..(2) & a + 20d = 63 …………..(3)

By solving (2) & (3) d = 3 & a = 3 By substituting a =3 and d =3 in (1)

tn = 3n

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How many two digit natural numbers are divisible by 3?

Solution: The smallest and the largest two digit natural numbers divisible by 3 are 12

and 99 respectively. The sequence of two digit natural numbers divisible by 3 is

12, 15, 18, … 99. This sequence is an AP with a=12 & d=3 For an AP tn = a +d(n-1), 99 = 12 + 3(n-1)

n = 30

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How many two digit natural numbers are divisible by 7?

The sequence of two digit natural numbers divisible by 7 is

14, 28, 42, …,98 This sequence is an AP with the first

term a = 14 & common diff d = 7 tn =14 + 7(n-1) 98 =14 + 7(n-1)

n = 13

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Solution: Here a = 4 & d = -1 - 4 = -5 For an AP, tn = a + d(n-1)

T15 = 4 + -5x14 = -66 -66 is the fifteenth term of

given A. P.

Find the fifteenth term of an A. P. 4, -1, -6, . . .

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The 11th and 21st term of an AP are 47 & 87 respectively then find the common

difference. Let a be the first term & d be the common difference of given A. P.

For an AP, tn = a + d(n-1) t11 = a + 10d & t21 = a + 20d

a + 10d = 47 …(1) & a + 20d = 87 …(2) By subtracting (1) from (2), 10d = 40

4 is the common difference.

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Solution: The sequence of natural numbers is 1, 2, 3, …n This sequence is an AP with a =1 & d =1 For an AP, Sn = 12

2 nda

n

Sn = 11122

nn = 2

)1( nn

Find the sum of first n natural numbers

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Solution:

For an AP, Sn = 122

ndan

S5 = 154122

5

S5 = 45

Find the sum of first 5 terms of an AP if a = 1 and d = 4

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Find the sum of first 5 terms of an AP if a = 1 and d = 4

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Find the sum of all odd numbers from 1 to 20

Solution: The sequence of odd numbers from 1 to 10 is 1, 3, 5, …, 19

This sequence is an AP with a=1 & d=2. For an AP, tn = a + d(n-1)

19 = 1 + 2(n-1) n = 10

For an AP Sn = 122

ndan

S10 = 1102122

10 = 100

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The sum of first 11 terms of an AP is 297. Find the 6th term. Solution: For an AP, Sn = 12

2 nda

n

da 1022

11297

a + 5d = 27 ------------------------(1) As tn = a + d(n-1), t6 = a + 5d ----(2) From (1) & (2) t6 = 27 27 is the 6th term of given A. P.

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tm = a + (m-1)d, tn = a + (n-1)d and dna

nSn 12

2 , find S(m+n-1)

Solution: tm + tn = a + (m-1)d + a + (n-1)d tm + tn = 2a + d(m + n -2) .......(1) As dna

nSn 12

2 ,

11221

nmdanm

S nm ..........(2) From (1) and (2) nmnm tt

nmS

21

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For an A. P. tn = a + d(n-1) a + 4d = 19 …..(1) & a + 15d = 63…..(2) By adding (1) and (2) 2a + 19d = 82 S20 = da 192

2

20 = 10x82 = 820

t5 = 19 & t16 = 63 then find S20

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The sum of three consecutive terms of an A. P. is 48 find the middle term

Solution: Let the three consecutive terms of A. P. be a - d, a, a + d. Then according to the given condition a – d + a + a + d = 48 3a = 48 a = 16. But a is the middle term. 16 is the middle term of given A. P.

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The sum of four consecutive terms of an A. P. is 50, find the sum of first term and last term Solution: Let the four consecutive terms of AP be a - 2d, a - d, a + d, a + 2d. Then according to the given condition a – 2d + a – d + a + d + a + 2d = 50 4a = 50 2a = 25 ......... (1) Now t1 + t4 = a-2d+a+2d = 2a ......(2) 25 is the sum of first & last term of AP

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Solution: Let the five consecutive terms of AP be a-2d, a-d, a, a+d, a+2d Then according to given condition a - 2d + a – d + a + a + d + a + 2d = 60 5a = 60. a = 12 But a is the middle term of given A. P. 12 is the middle term of given A. P.

The sum of five consecutive terms of an A. P. is 60, find the middle term