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Arithmetic progression : An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number d to the preceding term except the first term.
This fixed number d is called the common difference of the AP. It can be positive, negative or zero.
Common difference →Positive → Increasing AP.
Common difference →Negative → Decreasing AP.
Common difference →Zero → Constant AP.
Types of an Arithmetic Progression
1. Finite A.P. : The Arithmetic Progression which has only a finite number of terms called a finite A.P. Such AP has a last term. For example multiples of 5 till 50.
2. Infinite A.P. : The Arithmetic Progression which has not finite number of terms called infinite arithmetic Progression. Such AP do not has a last term. For example all multiples of 5.
The general form of an AP : a, a + d, a + 2d, a + 3d, . . . . .and so on. Where a is the first term and d is the common difference.
If the first term of an A.P. is a₁, second term is a₂, third term is a₃, . . . and n th term is aₙ then
A.P → a₁ a₂, a₃, . . . . . . . aₙ.
Common difference is d → a₂ – a₁ = a₃ – a₂ = . . . . . . . = aₙ – aₙ₋₁
nth Term of an AP
aₙ = a + (n – 1) d.
a = first term
n= Number of term
d = common difference
aₙ = nth Term
aₙ is also called the general term of the AP. If there are m terms in the AP, then
aₘ represents the last term which is sometimes also denoted by l
The sum of the first n terms of an AP
When first tern (a) and common difference (d) are know
This fixed number d is called the common difference of the AP. It can be positive, negative or zero.
Common difference →Positive → Increasing AP.
Common difference →Negative → Decreasing AP.
Common difference →Zero → Constant AP.
Types of an Arithmetic Progression
1. Finite A.P. : The Arithmetic Progression which has only a finite number of terms called a finite A.P. Such AP has a last term. For example multiples of 5 till 50.
2. Infinite A.P. : The Arithmetic Progression which has not finite number of terms called infinite arithmetic Progression. Such AP do not has a last term. For example all multiples of 5.
The general form of an AP : a, a + d, a + 2d, a + 3d, . . . . .and so on. Where a is the first term and d is the common difference.
If the first term of an A.P. is a₁, second term is a₂, third term is a₃, . . . and n th term is aₙ then
A.P → a₁ a₂, a₃, . . . . . . . aₙ.
Common difference is d → a₂ – a₁ = a₃ – a₂ = . . . . . . . = aₙ – aₙ₋₁
nth Term of an AP
aₙ = a + (n – 1) d.
a = first term
n= Number of term
d = common difference
aₙ = nth Term
aₙ is also called the general term of the AP. If there are m terms in the AP, then
aₘ represents the last term which is sometimes also denoted by l
The sum of the first n terms of an AP
When first tern (a) and common difference (d) are know
or 
When first tern (a) and last term are known
The sum of n positive integers is given by
When first tern (a) and last term are known
The sum of n positive integers is given by
EXERCISE 5.1
- In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
Taxi fare for 1st km (a₁)= Rs 15
Taxi fare after 2 km ( a₂)= 15 + 8 = Rs 23
Taxi fare after 3 km (a₃)= 23 + 8 = Rs 31
Taxi fare after 4 km (a₄)= 31 + 8 = Rs 39
Therefore, the sequence is 15, 23, 31, 39…
d₁ = a₂ – a₁ = 23-15 = 8
d₂ = a₃ – a₂ = 31-23 = 8
d₃ = a₄ – a₃ = 39-31 = 8
It is an arithmetic progression because difference between any two consecutive terms is equal which is 8.
(ii) The amount of air present in a cylinder when a vacuum pump removes ¼ of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
The cost of digging the well after 1 meter (a1)= Rs. 150
The cost of digging the well after 2 meters (a2)= Rs. 150 + 50 = Rs. 200
The cost of digging the well after 3 meters(a3) = Rs. 200+50 = Rs. 250
d₁ = a₂ - a₁ = 200 – 150 = Rs.50
d₂ = a₃ - a₂ = 250 – 200 = Rs.50
d₃ = a₄ – a₃= 300 – 250 = Rs.50
d₁ = d₂ = d₃ = 50Therefore, the given list of numbers froms an AP
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8 % per annum.
- Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
The general form of an AP is a, a + d, a + 2d, a + 3d, . . .
First term= a =10
Second term= a + d =10+10 = 20
Third term= a + 2d =10+ 20 = 30
Fourth term= a +3d =10+30 = 40
The first four terms of AP are 10, 20, 30, and 40.
(ii) a = –2, d = 0
The general form of an AP is a, a + d, a + 2d, a + 3d, . . .
First term a = −2
Second term a + d = −2+ 0 = −2
Third term a + 2d = −2+0 = −2
Fourth term a +3d = −2+0 = −2
The first four terms of AP are -2, -2, -2, and -2.
(iii) a = 4, d = – 3
The general form of an AP is a, a + d, a + 2d, a + 3d, . . .
First term a = 4
Second term a + d = 4 + (−3) =1
Third term a + 2d = 4 – 6 = −2
Fourth term a +3d = 4 – 9 = −5
The first four terms of AP are 4, 1, -2, -5.
(iv) a = – 1, d =½
The general form of an AP is a, a + d, a + 2d, a + 3d, . . .
First term= a = -1 + ½ = -½
Second term= a + d = - ½ + ½ = 0
Third term= a + 2d = 0 + ½ = ½
Fourth term= a +3d = ½ + ½ =1
The first four terms of AP are -1, -½ ,0, ½,
(v) a = – 1.25, d = – 0.25
The general form of an AP is a, a + d, a + 2d, a + 3d, . . .
First term a = -1.25
Second term= a + d = -1.25 +(-.25) = -1.50
Third term =a + 2d = -1.50 + (-.25) = -1.75
Fourth term= a +3d = -1.75 + (-.25) = -2.00
The first four terms of AP are -1.25, -1.5, -1.75, and - 2.00. - For the following APs, write the first term and the common difference:
(i) 3, 1, – 1, – 3, . . .
First term a = 3
Common difference d = a₂ - a₁ = 1-3 =-2
(ii) – 5, – 1, 3, 7, . . .
First term a = -5
Common difference d = a₂ - a₁ = -1 - (-5) = 4
(iv) 0.6, 1.7, 2.8, 3.9, . . .
The AP is 0.6, 1.7, 2.8, 3.9
First term a = 0.6
Common difference is d = a₂ - a₁
= 1.7 – 0.6
= 1.1 - Which of the following are APs ? If they form an AP, find the common difference d and write three more terms.
(i) 2, 4, 8, 16, . . .
The given numbers are 2, 4,8,16.........
First term (a₁)= 2
Common difference d₁ = a₂ - a₁ =4 − 2 = 2
Common difference d₂ = a₃ - a₂ = 8− 4 = 4
d₁ ≠ d₂
Therefore, the given terms 2, 4, 8, 16 are not in AP
(iii) – 1.2, – 3.2, – 5.2, – 7.2, . . .
The given numbers are – 1.2, – 3.2, – 5.2, – 7.2, . . .
First term (a1)= -1.2
Common difference d₁ = a₂ - a1 = -3.2 − (-1.2) = -2
Common difference d₂ = a₃ - a₂ = -5.2 − (-3.2) = -2
Common difference d₃ = a₄ - a₃ = -7.2 − (-5.2) = -2
d₁ = d₂ = d₃ = -2Therefore, the given terms – 1.2, – 3.2, – 5.2, – 7.2, . . . are in AP
Common difference d = a₂ - a₁ = -3.2 − (-1.2) = -2
Fifth term= a + 4d = -1.2 + 4 (-2) = 9.2
Six term= a + 5d = -1.2 + 5 (-2) = 11.2
Seventh term= a +6d = -1.2 + 46 (-2) = 13.2
(iv) – 10, – 6, – 2, 2, . . .
The given numbers are – 10, – 6, – 2, 2, . . .
First term (a₁)= -10
Common difference d₁ = a₂ - a₁ = -6 − (-10) = 4
Common difference d₂ = a₃ - a₂ = -2− (-6) = 4
Common difference d₃ = a₄ - a₃ = 2 − (-2) = 4
d₁ = d₂ = d₃ = 4
Therefore, the given terms – 10, – 6, – 2, 2, . . .are in AP
Common difference d₁ = a₂ - a1 = -6 − (-10) = 4
Fifth term= a +4 d = -10 + 4(4) = 6
Six term= a + 5d = -10 + 5(4) = 10
Seventh term= a +6d = -10 + 6(4) = 14
(v) 3, 3 + √2 , 3 + 2 √2 , 3 + 3 √2 , . . .
The given numbers are 3, 3 + √2 , 3 + 2 √2 , 3 + 3 √2 , . . .
First term (a₁)= -3
Common difference d₁ = a₂ - a₁ = 3 + √2 − 3 = √2
Common difference d₂ = a₃ - a₂ = 3 + 2 √2 - (3 + √2) = √2
Common difference d₃ = a₄ - a₃ = 3 + 3 √2 - (3 + 2 √2) = √2
d₁ = d₂ = d₃ = √2
Therefore, the given terms 3, 3 + √2 , 3 + 2 √2 , 3 + 3 √2 , . . . are in AP
Common difference d₁ = a₂ - a₁ = 3 + √2 − 3 = √2
Fifth term= a +4 d = 3 + 4(√2)
Six term= a + 5d = 3 + 5(√2)
Seventh term= a +6d = 3 + 6(√2)
(vi) 0.2, 0.22, 0.222, 0.2222, . . .
First term (a₁)= 0.2
Common difference d₁ = a₂ - a₁ = 0.22-0.2 = 0.02
Common difference d₂ = a₃ - a₂ = 0.222-0.22 = 0.002
d1₁ ≠ d₂
Therefore, the given terms 2, 4, 8, 16 are not in AP
(vii) 0, – 4, – 8, –12, . . .
The given numbers are 0, – 4, – 8, –12, . . .
First term (a1)= 0
Common difference d₁ = a₂ - a₁ = (-4) - 0 = -4
Common difference d₂ = a₃ - a₂ = (-8) - (-4) = -4
Common difference d₃ = a₄ - a₃ = (-12) - (-8) = -4
d₁ = d₂ = d₃ = -4Therefore, the given terms 0, – 4, – 8, –12, . . . are in AP
Common difference d = a₂ - a₁ = (-4) - 0 = -4
Fifth term= a + 4d = 0 + 4(-4) -16
Six term= a + 5d = 0 + 5(-4) = -20
Seventh term= a +6d = 0 + 6(-4) = -24
(viii) –½, -½, -½, -½, . . . . . .
The given numbers are –½, -½, -½, -½, . . . . . .
First term (a₁)= –½,
Common difference d₁ = a₂ - a₁ = -½, - ( -½) = 0
Common difference d₂ = a₃ - a₂ = -½, - ( -½) = 0
Common difference d₃ = a₄ - a₃ = -½, - ( -½) = 0
d₁ = d₂ = d₃ = 0Therefore, the given terms –½, -½, -½, -½, . . . . . . are in AP
Common difference d = a₂ - a₁ = = (-4) - 0 = -4
Fifth term= a + 4d = -½ + 4(0) =-½
Six term= a + 5d = -½ + 5(0) = -½Seventh term= a +6d = -½ + 6(0) = -½
(ix) 1, 3, 9, 27, . . .
First term (a₁)= 1
Common difference d₁ = a₂ - a₁ =3 - 1 = 2
Common difference d₂ = a₃ - a₂ = 9 - 3 = 6
d1₁ ≠ d₂
Therefore, the given terms 1, 3, 9, 27, . . . are not in AP
(x) a, 2a, 3a, 4a, . . .
The given numbers are a, 2a, 3a, 4a, . . .
First term (a1)= 0
Common difference d₁ = a₂ - a₁ = 2a – a = a
Common difference d₂ = a₃ - a₂ = 3a - 2a = a
Common difference d₃ = a₄ - a₃ = 4a - 3a = a
d₁ = d₂ = d₃ = a Therefore, the given terms a, 2a, 3a, 4a, . . . are in AP
Common difference d = a₂ - a₁ = 2a – a = a
Fifth term= a + 4d = a + 4(a) = 5a
Six term= a + 5d = a + 5(a) = 6a
Seventh term= a +6d = a + 6(a) = 7a
(xi) a, a², a³, a⁴, . . .
First term (a₁)= a
Common difference d₁ = a₂ - a₁ =a²– a = a(a - 1)
Common difference d₂ = a₃ - a₂ = a³ – a² = a²(a - 1)
d1₁ ≠ d₂
Therefore, the given terms a, a², a³, a⁴, . . . are not in AP
(xii) √2, √8, √18, √32 …
The given numbers are √2, √8, √18, √32 …
First term (a₁)= √2,
Common difference d₁ = a₂ - a₁ = √8-√2 = 2√2-√2 = √2
Common difference d₂ = a₃ - a₂ =√18-√8 = 3√2-2√2 = √2
Common difference d₃ = a₄ - a₃ = 4√2-3√2 = √2
d₁ = d₂ = d₃ = √2Therefore, the given terms √2, √8, √18, √32 … are in AP
Common difference d = a₂ - a₁ = √8-√2 = 2√2-√2 = √2
Fifth term= a + 4d = √2 + 4√
= 5√2
= √25 X 2
= √50
Six term= a + 5d = √2 + 5√2
=6√2
= √36 X 2
= √72
Seventh term= √2 +6d = √2 + 6√2
= 7√2
=√49 X 2
= √98
(xiii) √3, √6, √9, √12 …
First term (a₁) = √3
Common difference d₁ = a₂ - a₁ =√6-√3 = √3×√2-√3 = √3(√2-1)
Common difference d₂ = a₃ - a₂ = √9-√6 = 3-√6 = √3(√3-√2)
d1₁ ≠ d₂
Therefore, the given terms 1, 3, 9, 27, . . . are not in AP
(xiv) 1², 3², 5², 7² …
Or, 1, 9, 25, 49 …..
First term (a₁) = 1
Common difference d₁ = a₂ - a₁ = 9−1 = 8
Common difference d₂ = a₃ - a₂ =25−9 = 16
d1₁ ≠ d₂
Therefore, the given terms 1, 3, 9, 27, . . . are not in AP
(xv) 1² , 5² , 7² , 73 …
Or 1, 25, 49, 73 …
The given numbers are 1² , 5² , 7² , 73 …
First term (a₁)= 1
Common difference d₁ = a₂ - a₁ = 25−1 = 24
Common difference d₂ = a₃ - a₂ = 49−25 = 24
Common difference d₃ = a₄ - a₃ = 73−49 = 24
d₁ = d₂ = d₃ = 24Therefore, the given terms 1² , 5² , 7² , 73 … are in AP
Common difference d = a₂ - a₁ = 25−1 = 24
Fifth term= a + 4d = 1 + 4(24) = 97
Six term= a + 5d = 1 + 5(24) = 121
Seventh term= a +6d = 1 + 6(24) = 145
EXERCISE 5.2
- Fill in the blanks in the following table, given that a is the first term, d the common difference and aₙ the nth term of the AP:a d n aₙ
(i) 7 3 8
(ii) –18 10 0
(iii) –3 18 –5
(iv) – 18.9 2.5 3.6
(v) 3.5 0 105
(i) Here, a = 7, d = 3, n= 8, aₙ =?
We know that the nth term of an A.P. is,
aₙ = a + (n – 1) d.
= 7+(8 −1) 3
= 7+(7) 3
= 7+21
aₙ = 28
Therefore, the aₙ th term of the given A.P. is 28
(ii) Here, a = -18, d = ? n = 10 aₙ = 0
We know that the nth term of an A.P. is,
aₙ = a + (n – 1) d.
0 = − 18 +(10−1)d
18 = 9d
d = 18/9
d = 2
Therefore, the common difference of the given A.P. is 2
(iii) Here, a = ? d = -3 n = 18 aₙ = -5
We know that the nth term of an A.P. is,
aₙ = a + (n – 1) d.
−5 = a+(18−1) (−3)
−5 = a+(17) (−3)
−5 = a−51
a = 51−5
a = 46
Therefore, the first term of the given A.P. is 46
(iv) Here, a = -18.9, d = 2.5, n =?, aₙ = 3.6
We know that the nth term of an A.P. is,
aₙ = a + (n – 1) d.
3.6 = − 18.9 + (n −1)2.5
3.6 + 18.9 = (n −1)2.5
22.5 = (n −1)2.5
(n – 1) = 22.5/2.5
n – 1 = 9
n = 10
Therefore, the 10th term of the given AP is 3.6
(v) Here, a = 3.5, d = 0, n = 105, aₙ =?
We know that the nth term of an A.P. is,
aₙ= a + (n – 1) d.
aₙ= 3.5 + (105 − 1) 0
aₙ= 3.5 + 104 × 0
aₙ= 3.5
Therefore, the aₙ th term of the given AP is 3.5 - Choose the correct choice in the following and justify:
(i) 30th term of the A.P: 10,7, 4, …, is
(A) 97 (B) 77 (C) −77 (D) −87 [C]
Here, a₁ = 10, d = a₂ − a₁ = 7−10 = −3
We know that the nth term of an A.P. is,
aₙ = a + (n – 1) d.
a₃₀ = 10 + (30−1) (−3)
a₃₀ = 10 + (29) (−3)
a₃₀ = 10 − 87 = −77
(ii) 11th term of the A.P. -3, -1/2, ,2 …. is
(A) 28 (B) 22 (C) – 38 (D)– 48½ [B ]
A.P. = -3, -1/2, ,2 …
n =11
First term a₁ = – 3
Common difference, d = a₂ − a₁ = (-1/2) -(-3)
⇒ (-1/2) + 3 = 5/2
As we know,
aₙ = a + (n − 1) d
a₁₁ = -3 + (11 - 1) (5/2)
a₁₁ = -3 + (10) (5/2)
a₁₁ = -3 + 25
a₁₁ = 22 - In the following APs find the missing term in the boxes.
(i) 2, , 26
Here ,a₁ = 2, a₃ = 26
As we know,
aₙ = a + (n −1) d
a₃ = 2 + (3 - 1) d
26 = 2 + 2d
24 = 2d
d = 12
a₂ = 2 + (2 - 1)12
a₂ = 14
Therefore, the missing terms is 14
(ii) , 13, ,3
Here, a₂ = 13,a₄ = 3
As we know
aₙ = a + (n − 1) d
a₂ = a + (2 - 1)d
13 = a + d ………………. (i)
a₄ = a + ( 4- 1) d
3 = a + 3d ………….. (ii)
On subtracting equation (i) from (ii)
– 10 = 2d
d = – 5
Putting the value of d in equation (ii)
13 = a + (-5)
a = 18
a₃ = 18 + (3 - 1) ( -5)
a₃ = 18 + 2 (-5)
a₃ = 18 - 10
a₃ = 8
Therefore, the missing terms are 18 and 8, respectively.
(iii) 5, , , 9½
Here, a = 5 and a₄ = 9½
As we know
aₙ = a + (n−1)d
a₄ = a + (4-1)d
9½ = 5 + 3d
(9½) – 5 = 3d
3d = 9½
d = 3/2
a₂ = a + (2 - 1) d
a₂= 5 + (1) (3/2)
a₂ = 13/2
a₃ = a + (3 - 1)d
a₃ = 5 + 2×(3/2)
a₃ = 8
Therefore, the missing terms are 13/2 and 8, respectively.
(iv) -4, , , , 6
Here, a = −4, a₆ = 6
As we know
aₙ = a + (n−1) d
a₆ = a + (6 − 1)d
6 = − 4 + 5d
10 = 5d
d = 2
a₂ = a + d = − 4 + 2 = −2
a₃ = a + 2d = − 4 + 2(2) = 0
a₄ = a + 3d = − 4+ 3(2) = 2
a₅ = a + 4d = − 4 + 4(2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
(v) , 38, , ,-22
Here, a₂ = 38, a₆ = −22
As we know
aₙ = a + (n − 1)d
a₂ = a + (2 − 1)d
38 = a + d ……………………. (i)
a₆ = a + (6 − 1)d
−22 = a + 5d …………………. (ii)
On subtracting equation (i) from (ii), we get
− 22 − 38 = 4d
−60 = 4d
d = −15
a = a₂ − d = 38 − (−15) = 53
a₃ = a + 2d = 53 + 2 (−15) = 23
a₄ = a + 3d = 53 + 3 (−15) = 8
a₅ = a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7, respectively. - Which term of the A.P. 3, 8, 13, 18, … is 78?
Given A.P. series is 3, 8, 13, 18, …
First term, a = 3
Common difference, d = a₂ − a₁ = 8 − 3 = 5
aₙ =78
aₙ = a + (n−1)d
78 = 3 + (n −1)5
75 = (n−1)5
(n−1) = 15
n = 16
Hence, 16th term of this A.P. is 78. - Find the number of terms in each of the following A.P.
(i) 7, 13, 19, , 205
First term, a = 7
Common difference, d = a₂ − a₁ = 13 − 7 = 6
Let there are n terms in this A.P.
aₙ = 205
aₙ = a + (n − 1) d
205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n = 34
Therefore, the number of terms in this A.P. is 34
(ii) 18,15½, 13, . . . , – 47
First term, a = 18
Common difference, d = a₂ -a₁ = (15½ -18) = -5/2
Let there are n terms in this A.P.
aₙ = -47
aₙ = a+(n − 1)d-47 = 18 + (n - 1) (-5/2)
-47 - 18 = (n - 1) (-5/2)
-65 = (n - 1) (-5/2
(n - 1) = -130/-5
(n-1) = 26
n = 27
Therefore, the number of terms in this A.P. is 34 - Check whether -150 is a term of the A.P. 11, 8, 5, 2,
Given A.P. - 11, 8, 5, 2,
First term, a = 11
Common difference, d = a₂ −a₁ = 8−11 = −3
Let −150 be the nth term of this A.P.
aₙ = a+(n − 1)d
-150 = 11 + (n - 1)(-3)
-150 = 11 - 3n + 3
164 = -3n
n = 164/3
Clearly, n is not an integer but a fraction.
Therefore, – 150 is not a term of this A.P. - Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Given : 1th term, a₁₁ = 38 and 16th term, a₁₆ = 73aₙ = a + (n−1)d
a₁₁ = a + (11−1)d
38 = a + 10d ………………………………. (i)
In the same way,
a₁₆ = a + (16−1)d
73 = a + 15d ………………………………………… (ii)
On subtracting equation (i) from (ii), we get
35 = 5d
d = 7
Putting the value of d in equation (i)
38 = a + 10×(7)
38 − 70 = a
a = −32
a₃₁ = a + (31 − 1) d
= − 32 + 30 (7)
= − 32 + 210a₃₁ = 178
Hence, 31st term is 178. - An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Given : 3rd term, a₃ = 12 and 50th term, a₅₀ = 106
aₙ = a + (n − 1)d
a₃ = a + (3 − 1)d
12 = a + 2d ……………………………. (i)
In the same way,
a₅₀ = a + (50 − 1)d
106 = a + 49d …………………………. (ii)
On subtracting equation (i) from (ii)
94 = 47d
d = 2
Putting the value of d in equation (i)
12 = a + 2(2)
a = 12 − 4 = 8
a₂₉ = a + (29−1) d
a₂₉ = 8 + (28)2
a₂₉ = 8 + 56 = 64
Therefore, 29th term is 64. - If the 3rd and the 9th terms of an A.P. are 4 and − 8, respectively. Which term of this A.P. is zero?
Given
3rd term, a₃ = 4
9th term, a₉ = −8
aₙ = a + (n − 1)d
a₃ = a + (3 − 1)d
4 = a + 2d ……………………………………… (i)
a₉ = a + (9 − 1)d
−8 = a + 8d ………………………………………………… (ii)
On subtracting equation (i) from (ii)
−12 = 6d
d = −2
Putting the value of d in equation (ii)
4 = a + 2(−2)
4 = a − 4
a = 8
Let nth term of this A.P. be zero.
aₙ = a + (n −1)d
0 = 8 + (n − 1)(−2)
0 = 8−2n + 2
2n = 10
n = 5
Hence, 5th term of this A.P. is 0. - If 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Let the first term of A.P. be a and common difference be d
aₙ = a+ (n − 1)
a₁₇ = a+ (17−1)d
a₁₇ = a + 16d ...............(1)
In the same way,
a₁₀ = a + (10 - 1)d
a₁₀ = a + 9d .............(2)
According to question
a₁₇ − a₁₀ = 7
(a + 16d) − (a + 9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1. - Which term of the A.P. 3, 15, 27, 39,.. will be 132 more than its 54th term?
Given A.P. is 3, 15, 27, 39, …
first term, a = 3
Common difference, d = a₂ − a₁ = 15 − 3 = 12
aₙ = a + (n − 1)d
a₅₄ = a + (54 − 1)d
= 3 + (53) (12)
= 3 + 636
a₅₄ = 639
Let nth term be 132 more than its 54th term
aₙ = a₅₄ + 132
aₙ = 639 + 132
aₙ = 771
aₙ = a + (n − 1)d
771 = 3 + (n − 1)12
768 = (n − 1)12
(n −1) = 64
n = 65
Therefore, 65th term was 132 more than 54th term. - Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms?
Let, the first term of two APs be a₁ and a₂ respectively. and the common difference of these APs be d.
For the first A.P.
aₙ = a + (n − 1)d
a₁₀₀ = a₁ + (100 − 1)d
= a₁ + 99d
a₁₀₀₀ = a₁ + (1000 − 1)d
a₁₀₀₀ = a₁ + 999d
For second A.P.
aₙ = a + (n −1)d
a₁₀₀ = a₂ + (100 − 1)d
= a₂ + 99d
a₁₀₀₀ = a₂ +(1000 −1)d
= a₂ + 999d
Given that, difference between 100th term of the two APs = 100
Therefore, (a₁ + 99d) − (a₂ + 99d) = 100
a₁ −a₂ = 100…………….. (i)
Difference between 1000th terms of the two APs
(a₁ + 999d) − (a₂ + 999d) = a₁ −a₂
From equation (i),
This difference, a₁ −a₂ = 100
Hence, the difference between 1000th terms of the two A.P. will be 100. - How many three digit numbers are divisible by 7?
First three-digit number that is divisible by 7 are;
First number = 105
Second number = 105+7 = 112
Third number = 112+7 =119
maximum number = 994
Therefore, A.P. 105, 112, 119, …....994
Let 994 be the nth term of this A.P.
first term, a = 105
common difference, d = 7
aₙ = 994
n = ?
aₙ = a + (n − 1)d
994 = 105 + (n − 1)7
889 = (n − 1)7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7. - How many multiples of 4 lie between 10 and 250T
The numbers between 10 and 250 which are multiple of 4 are 12,16,20,...,248.
Therefore, the series formed as;
12, 16, 20, …...,248.
Let 248 be the nth term of this A.P.
First term, a = 12
common difference, d = 4
aₙ = 248
aₙ = a+(n − 1)d
248 = 12 + (n - 1) × 4
236/4 = n - 1
59 = n - 1
n = 60
Therefore, there are 60 multiples of 4 between 10 and 250. - For what value of n, are the nth terms of two APs 63, 65, 67, and 3, 10, 17, … equal?
Given first AP 63, 65, 67,…
First term, a = 63
Common difference, d = a₂ −a₁ = 65 − 63 = 2
nth term of first A.P. aₙ = a + (n − 1)d
aₙ = 63 + (n − 1)2 = 63 + 2n − 2
aₙ = 61 + 2n …………………. (i)
Given second AP 3, 10, 17, …
First term, a = 3
Common difference, d = a₂ − a₁ = 10 − 3 = 7
nth term of second A.P. aₙ = a + (n − 1)d
aₙ = 3 + (n − 1)7
aₙ = 3 + 7n − 7
aₙ = 7n − 4 …………………….. (ii)
According to question nth term of these A.P.s are equal to each other.
61 + 2n = 7n − 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13th terms of both these A.P.s are equal to each other. - Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
We know that term of an A.P. is
aₙ =a + (n − 1)d
Third term of A.P.
a₃ = a + (3 − 1)d
16 =a + 2d [ Given a₃ =16]
a + 2d = 16……………………. (i)
Fifth term of A.P.,
a₅ = a + (5 − 1)d
Seventh term of A.P.
a₇ = a + (7 − 1)d
According to question
a₇ − a₅ = 12
[a + (7 − 1)d] − [a + (5 − 1)d] = 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
Putting the value of d in equation (i)
a + 2 (6) = 16
a+12 = 16
a = 4
First term = 4
Scond term = 4 + 6 =10
Third term =16 [Given ]
Forth term = 16 + 6 =22
Therefore, A.P. will be 4, 10, 16, 22, … - Find the 20th term from the last term of the A.P. 3, 8, 13, …, 253.
Given A.P. is 3, 8, 13, …, 253
Common difference, d= 8-3 = 5.
Therefore, we can write the given AP in reverse order as;
253, 248, 243, …, 13, 8, 5
Now first term, a = 253
Common difference, d = 248 − 253 = −5
n = 20
The nth term of an A.P. is
aₙ = a + (n - 1) d
a₂₀ = a + (20 − 1)d
a₂₀ = 253 + (19)(−5)
a₂₀ = 253 − 95
a = 158
Therefore, 20th term from the last term of the AP 3, 8, 13, …, 253 is 158. - The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.
We know that, the nth term of the AP is;
aₙ = a + (n − 1)d
a₄ = a + (4 − 1)d
a₄ = a + 3d
In the same way
a₈ = a + (7 - 1)d
a₈ = a + 7d
According to question
a₄ + a₈ = 24
a + 3d + a + 7d = 24
2a + 10d = 24
A + 5d = 12 ……………………… (i)
Now in the same way
a₆ = a + (6 -1)d
a₆ = a + 5d
a₁₀ = a + (10 -1)d
a₁₀ = a + 9d
According to question
a₆ + a₁₀ = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a+ 7d = 22 …………………….. (ii)
On subtracting equation (i) from (ii)
2d = 22 − 12
2d = 10
d = 5
Putting the value of d in equation (i)
a + 5d = 12
a + 5(5) = 12
a + 25 = 12
a = −13
a₂ = a + d = − 13 + 5 = −8
a₃ = a₂ + d = − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3. - Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?
The salary of Subba Rao in 1995 =5000
The salary of Subba Rao in 1996 = 5000 + 200 = 5200
The salary of Subba Rao in 1997 = 5200 + 200 = 5400
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
First term, a = 5000
Common difference, d = 200
Let after nth year, his salary be Rs 7000.
The nth term formula of AP,
aₙ = a+(n − 1) d
7000 = 5000 + (n − 1)200
200(n − 1)= 2000
(n − 1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000. - Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.
Ramkali saved in first week =5
Ramkali saved in second week =5 + 1.75 = 6.75
Ramkali saved in Third week = 6.75 + 1.75 = 8.5
Therefore, it is an AP of the form: 5, 6.75, 8.5...
First term, a = 5
Common difference, d = 1.75
Ramkali saved in nth week aₙ = 20.75
Find, n = ?
The nth term of A.P. aₙ = a + (n − 1)d
20.75 = 5 + (n - 1) ×1.75
15.75 = (n - 1) ×1.75
(n - 1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n - 1 = 9
n = 10
Hence, n is 10.
EXERCISE 5.3
- Find the sum of the following APs:
(i) 2, 7, 12, . . ., to 10 terms.
Here, a = 2, d = 7 – 2 = 5, n =10
We know that S = [2a + (n – 1) d ]
Sₙ = n/2 [2a +(n - 1)d]
S₁₀ = 10/2 [2(2)+(10 - 1)×5]
= 5[4+(9)×(5)]
= 5 × 49 = 245
Therefore, the sum of the first 10 terms of the AP is 245.
(ii) –37, –33, –29, . . ., to 12 terms.
Here, a = -37, d = -33– (-37) = 4, n =1n = 12
We know that
S = [2a + (n – 1) d ]
Sn = n/2 [2a + (n - 1)d]
S12 = 12/2 [2( -37) + (12 - 1)×4]
= 6[ -74 + (11 × 4)]
= 6[-74 + 44]
= 6(-30) = -180
Therefore, the sum of the first 12 terms of the AP is -180
(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.
Here, a = 0.6, d = 1.7 – 0.6 = 1.1, n =100
We know that
Sn = n/2 [2a +(n-1)d]
S100 = 100/2 [2(0.6) + (100 - 1)×1.1]
= 50/2 [1.2 + (99)×1.1]
= 50[1.2 + 108.9]
= 50[110.1]
= 5505
Therefore, the sum of the first 100 terms of the AP is 5505
(iv) 1/15, 1/12, 1/10, …… , to 11 terms
Here, a = 1/15, d = 1/12 – 1/15 =1/60, n =11
We know that
Sₙ = n/2 [2a +(n-1)d]
S₁₁= 11/2[2(1/15) + (11 - 1)1/60]
= 11/2 [2/15 + 10/60)
= 11/2 (18/60)
= 11/2 (9/30)
= 99/60
= 33/20 - Find the sums given below :
(i) 7 +101/2 + 14 + . . . + 84
Here, a = 7, d =101/2 – 7 = 7/2, l = 84
We know that
aₙ = a(n-1)d
84 = 7+(n – 1)×7/2
77 = (n-1)×7/2
22 = n−1
n = 23
We know that the sum of n term is;
Sₙ = n/2 (a + l) ,
S₂₃ = 23/2 (7+84)
S₂₃ = (23×91/2) = 2093/2
Therefore, the sum of the first 23 terms of the AP is 2093/2
(ii) 34 + 32 + 30 + . . . + 10
Here, a = 34, d = 32 – 34 = -2, l = 10
we know that,
aₙ= a +(n−1)d
10 = 34+(n−1)(−2)
−24 = (n −1)(−2)
12 = n −1
n = 13
We know that the sum of n terms is;
Sₙ = n/2 (a +l) ,
= 13/2 (34 + 10)
= (13×44/2) = 13 × 22
= 286
Therefore, the sum of the first 13 terms of the AP is 286
(iii) –5 + (–8) + (–11) + . . . + (–230)
Here, a = -5, d = -8 – (-5) = - 3, l = -230
we know that,
aₙ= a+(n−1)d
−230 = − 5+(n−1)(−3)
−225 = (n−1)(−3)
(n−1) = 75
n = 76
We know that the sum of n terms is;
Sₙ = n/2 (a + l)
= 76/2 [(-5) + (-230)]
= 38(-235)
= -8930
Therefore, the sum of the first 76 terms of the AP is -8930 - In an AP:
(i) given a = 5, d = 3, an = 50, find n and Sₙ
Here, a = 5, d = 3, an = 50
We know that,
an = a +(n −1)d,
⇒ 50 = 5+(n -1)×3
⇒ 3(n -1) = 45
⇒ n -1 = 15
⇒ n = 16
Now, sum of n terms,
Sₙ = n/2 (a +an)
Sₙ = 16/2 (5 + 50) = 440
Therefore, the sum of the first 50 terms of the AP is 440
(ii) given a = 7, a₁₃ = 35, find d and S₁₃.
Here, a = 7, d = ?, a₁₃ = 35
We know that,
aₙ = a+(n−1)d,
35 = 7+(13-1)d
12d = 28
d = 28/12 = 2.33
Now, sum of n terms,
Sₙ = n/2 (a+aₙ )
S₁₃ = 13/2 (7+35)
S₁₃ = 273
Therefore, the sum of the first 13 terms of the AP is 273
(iii) given a₁₂ = 37, d = 3, find a and S₁₂ .
Here, a = ?, a₁₂ = 37, d = 3, S₁₂ .
As we know,
aₙ = a + (n − 1)d,
a₁₂ = a + (12 − 1)3
37 = a + 33
a = 4
Now, sum of nth term,
Sₙ = n/2 (a + aₙ )
Sₙ = 12/2 (4+37)
= 246
Therefore, the sum of the first 12 terms of the AP is 246
(iv) given a3 = 15, S10 = 125, find d and a10.
As we know, from the formula of the nth term in an AP,
aₙ = a +(n−1)d,
Therefore, putting the given values, we get,
a3 = a+(3−1)d
15 = a+2d ………………………….. (i)
Sum of the nth term,
Sₙ = n/2 [2a+(n-1)d]
S₁₀ = 10/2 [2a+(10-1)d]
125 = 5(2a+9d)
25 = 2a+9d ……………………….. (ii)
On multiplying equation (i) by (ii), we will get;
30 = 2a+4d ………………………………. (iii)
By subtracting equation (iii) from (ii), we get,
−5 = 5d
d = −1
From equation (i),
15 = a+2(−1)
15 = a−2
a = 17 = First term
a₁₀ = a+(10−1)d
a₁₀ = 17+(9)(−1)
a₁₀ = 17−9 = 8
(v) given d = 5, S9 = 75, find a and a9.
As, sum of n terms in AP is,
Sₙ = n/2 [2a +(n -1)d]
Therefore, the sum of first nine terms are;
S₉ = 9/2 [2a +(9-1)5]
25 = 3(a+20)
25 = 3a+60
3a = 25−60
a = -35/3
As we know, the nth term can be written as;
aₙ = a+(n−1)d
a₉ = a+(9−1)(5)
= -35/3+8(5)
= -35/3+40
= (35+120/3) = 85/3
(vi) given a = 2, d = 8, Sn = 90, find n and an.
As, sum of n terms in an AP is,
Sₙ = n/2 [2a +(n -1)d]
90 = n/2 [2a +(n -1)d]
180 = n(4+8n -8) = n(8n-4) = 8n2-4n
8n²-4n –180 = 0
2n²–n-45 = 0
2n²-10n+9n-45 = 0
2n(n -5)+9(n -5) = 0
(n-5)(2n+9) = 0
So, n = 5 (as n only is a positive integer)
∴ a₅ = 8+5×4 = 34
(vii) given a = 8, an = 62, Sn = 210, find n and d.
As, the sum of n terms in an AP is,
Sₙ = n/2 (a + an)
210 = n/2 (8 +62)
35n = 210
n = 210/35 = 6
Now, 62 = 8+5d
5d = 62-8 = 54
d = 54/5 = 10.8
(viii) given an = 4, d = 2, Sₙ = –14, find n and a.
As we know, from the formula of the nth term in an AP,
aₙ = a+(n −1)d,
Therefore, putting the given values, we get,
4 = a+(n −1)2
4 = a+2n−2
a + 2n = 6
a = 6 − 2n …………………………………………. (i)
As we know, the sum of n terms is;
Sₙ = n/2 (a+an)
-14 = n/2 (a+4)
−28 = n (a+4)
−28 = n (6 −2n +4) {From equation (i)}
−28 = n (− 2n +10)
−28 = − 2n²+10n
2n² −10n − 28 = 0
n² −5n −14 = 0
n² −7n+2n −14 = 0
n (n−7)+2(n −7) = 0
(n −7)(n +2) = 0
Either n − 7 = 0 or n + 2 = 0
n = 7 or n = −2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we get
a = 6−2n
a = 6−2(7)
= 6−14
= −8
(ix) given a = 3, n = 8, S = 192, find d.
Here, a = 3, n = 8, Sn = 192
As we know,
Sₙ = n/2 [2a+(n -1)d]
192 = 8/2 [2×3+(8 -1)d]
192 = 4[6 +7d]
48 = 6+7d
42 = 7d
d = 6
(x) given l = 28, Sₙ = 144, and there are total 9 terms. Find a.
Here, a =?, aₙ/l =28, Sₙ = 144, n =9
Sum of n terms formula,
Sₙ = n/2 (a + l)
144 = 9/2(a + 28)
(16) × (2) = a + 28
32 = a + 28
a = 4 - How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636?
Here, a = 9, d= 17-9 = 8, Sₙ =636, n = ?
We know that sum of n terms of an AP
Sₙ = n/2 [2a + (n - 1) d]
636 = n/2 [2 × 9 + (n - 1) 8]
636 = n/2 [18 + 8n - 8]
636 = n/2 [10 + 8n]
636 = n[5 + 4n]
636 = 5n + 4n²
4n² + 5n - 636 = 0
4n² + 53n - 48n - 636 = 0
n (4n + 53) - 12 (4n + 53) = 0
(4n + 53)(n - 12) = 0
Either 4n + 53 = 0 or n - 12 = 0
n = - 53/4 or n = 12
n cannot be -53/4 because the number of terms can neither be negative nor fractional, therefore, n = 12 - The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Here, a = 5, aₙ /l = 45, Sₙ = 400
As we know, the sum of AP formula is;
Sₙ = n/2 (a+l)
400 = n/2(5+45)
400 = n/2(50)
Number of terms, n =16
As we know, the last term of AP series can be written as;
aₙ /l = a+(n −1)d
45 = 5 +(16 −1)d
40 = 15d
d = 40/15 = 8/3 - The first and the last term of an AP are 17 and 350, respectively. If the common difference is 9, how many terms are there and what is their sum?
Here, a = 17, aₙ /l = 350, d = 9
Let there be n terms in the A.P., thus the formula for last term can be written as;
aₙ /l = a+(n −1)d
350 = 17+(n −1)9
333 = (n−1)9
(n−1) = 37
n = 38
Sₙ = n/2 (a+l)
S38 = 38/2 (17+350)
= 19×367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973. - Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
Solution:
Here, d = 7, a₂₂ = 149, S₂₂ = ?
By the formula of nth term,
aₙ = a+(n−1)d
a₂₂ = a + (22 − 1)d
149 = a + 21×7
149 = a + 147
a = 2
Sum of n terms,
Sₙ = n/2(a+ aₙ)
S22 = 22/2 (2+149)
= 11×151
= 1661 - Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18, respectively.
Here, Second term, a₂ = 14, Third term, a₃ = 18, d = 18−14 = 4
We know that,
a₂ = a+d
14 = a+4
a = 10
Sum of n terms;
Sₙ = n/2 [2a + (n – 1)d]
S₅₁ = 51/2 [2×10 (51 - 1) 4]
= 51/2 [20+(50)×4]
= 51 × 220/2
= 51 × 110
= 5610 - If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
Here, S₇ = 49, S₁₇ = 289 Sₙ= ?
We know, sum of n terms;
Sₙ = n/2 [2a + (n – 1)d]
S₇= 7/2 [2a +(n -1)d]
49= 7/2 [2a + (7 -1)d]
49 = 7/2 [2a + 6d]
7 = (a + 3d)
a + 3d = 7 …………………………………. (i)
In the same way,
S₁₇ = 17/2 [2a + (17 - 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 ………………………………. (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
Putting the value of d in equation (i)
a + 3(2) = 7
a + 6 = 7
a = 1
Sum of first n terms.
Sₙ = n/2[2a + (n - 1)d]
= n/2[2(1)+(n – 1)×2]
= n/2(2 + 2n - 2)
= n/2(2n)
= n² - Show that a1, a2 … , an , … form an AP where an is defined as below
(i) aₙ = 3 + 4n
(ii) aₙ = 9 − 5n
Also, find the sum of the first 15 terms in each case.
(i) aₙ = 3 + 4n
a₁ = 3 + 4(1) = 7
a₂ = 3 + 4(2) = 3 + 8 = 11
a₃ = 3 + 4(3) = 3 + 12 = 15
a₄ = 3 + 4(4) = 3 + 16 = 19
Common difference
a₁₂ − a1 = 11−7 = 4
a₃− a₂ = 15−11 = 4
a₄ − a₃ = 19−15 = 4
We can see here, the common difference between the terms are same. Therefore, this is an AP with common difference as 4 and first term as 7.
sum of the first 15 terms
Sₙ = n/2[2a+(n -1)d]
S15 = 15/2[2(7)+(15-1)×4]
= 15/2[(14)+56]
= 15/2(70)
= 15×35
= 525
(ii) aₙ = 9−5n
a₁ = 9−5×1 = 9−5 = 4
a2 = 9−5×2 = 9−10 = −1
a3 = 9−5×3 = 9−15 = −6
a4 = 9−5×4 = 9−20 = −11
common difference
a2 − a₁ = −1−4 = −5
a3 − a2 = −6−(−1) = −5
a4 − a3 = −11−(−6) = −5
We can see here, the common difference between the terms are same. Therefore, this is an A.P. with common difference as −5 and first term as 4.
sum of the first 15 terms
Sₙ= n/2 [2a +(n-1)d]
S₁₅ = 15/2[2(4) + (15 - 1)(-5)]
= 15/2[8 + 14(-5)]
= 15/2(8-70)
= 15/2(-62)
= 15(-31)
= -465 - If the sum of the first n terms of an AP is 4n − n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.
Given that,
Sₙ = 4n−n²
(i) first term/ Sum of first terms (n=1)
= 4(1) − (1)² = 4 − 1 = 3
S₁ = 3
(ii) Sum of first two terms[ n= 2]
S₂= 4(2)−(2)² = 8−4 = 4
(iii) Second term, a₂ = S₂ − S₁ = 4−3 = 1
(iv) Third term: a= 3, d = 1-3 =-2
nth term, an = a+(n−1)d
a3 = 3+(3 −1)(−2)
a3 = 3 + (-4)
a3 = 3 – 4
a3 = −1
(v) 10th term: a= 3, d = 1-3 =-2
nth term, an = a+(n−1)d
a₁₀ = 3 +(10-1)(-2)
a₁₀ = 3 + (-18)
a₁₀ = −15
(vi) nth terms a= 3, d = 1-3 =-2
nth term, an = a+(n−1)d
= [3+ (n -1)(-2)]
= 3+(-2n +2)
= 5 -2n - Find the sum of the first 40 positive integers divisible by 6.
The positive integers that are divisible by 6 are 6, 12, 18, 24 ….
We can see here, that this series forms an A.P.
Here, a = 6, d = 6, S40 = ?
By the formula of sum of n terms, we know,
Sₙ = n/2 [2a +(n – 1)d]
Therefore, putting n = 40, we get,
S₄₀ = 40/2 [2(6)+(40-1)6]
= 20[12+(39)(6)]
= 20(12+234)
= 20×246
= 4920 - Find the sum of first 15 multiples of 8.
The multiples of 8 are 8, 16, 24, 32…
The series is in the form of AP,.
Here, a = 8, d = 8, S₁₅ = ?
By the formula of sum of nth term, we know,
Sₙ = n/2 [2a+(n-1)d]
S₁₅ = 15/2 [2(8) + (15-1)8]
= 15/2[16 +(14)(8)]
= 15/2[16 +112]
= 15(128)/2
= 15 × 64
= 960 - Find the sum of the odd numbers between 0 and 50.
The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.
Therefore, we can see that these odd numbers are in the form of A.P.
Here, a = 1, d = 2, l = 49, n =?, Sₙ = ?
By the formula of last term, we know,
l = a+(n−1) d
49 = 1+(n−1)2
48 = 2(n − 1)
n − 1 = 24
n = 25 = Number of terms
By the formula of sum of nth term, we know,
Sₙ = n/2(a +l)
S₂₅ = 25/2 (1+49)
= 25(50)/2
=(25)(25)
= 625 - A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money does the contractor have to pay as penalty, if he has delayed the work by 30 days.
We can see that the given penalties are in the form of A.P.
200, 250, 300, ……..
Here, a = 200, d = 50, n = 30, S₃₀ = ?
By the formula of sum of nth term, we know,
Sₙ = n/2[2a+(n -1)d]
S₃₀= 30/2[2(200)+(30 – 1)50]
= 15[400+1450]
= 15(1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty. - A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Let the cost of 1st prize be Rs. a.
Cost of 2nd prize = Rs. a − 20
And cost of 3rd prize = Rs. a − 40
We can see that the cost of these prizes are in the form of A.P.
Here, a =?, d = -20. S₇ =700, n = 7
Thus, a = P and d = −20
By the formula of sum of nth term, we know,
Sₙ = n/2 [2a + (n – 1)d]
700 = 7/2 [2a + (7 – 1)(-20)]
7/2 [2a -120)] = 700
2a - 120 = 200
2a = 100 +120
2a = 320
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40. - In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant, will be the same as the class, in which they are studying. E.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?
Total numbers plats planted by class I = 3 X 1 = 3
Total numbers plats planted by class II = 3 X 2 = 6
Total numbers plats planted by class III = 3 X 3 =9
Total numbers plats planted by class XII = 3 X 12 = 36
The number of trees planted by the students are in an AP.
3, 6, 9,……………36
Here, a = 1, d = 6−3 = 2
Sₙ = n/2 [2a +(n-1)d]
We know that sum of n terms of AP,
Sₙ = n/2 [2a + (n - 1) d]
S₁₂ = 12/2 [2 × 3 + (12 - 1) × 3]
= 6 [6 + 11 × 3]
= 6 [6 + 33]
= 6 × 39
= 234
Therefore, 234 trees will be planted by the students. - A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in the figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take π = 22/7)Circumference of a semi-circle = πr
Therefore,
Circumference of first semi-circle = π(0.5) = π/2 cm
Circumference of second semi-circle = π(1) = π cm
Circumference of third semi-circle = π(1.5) = 3π/2 cm
Hence we got a series here, as,
π/2, π, 3π/2, 2π, ….
Here, a= π/2, d = π – π/2 = π/2
By the sum of n term formula, we know,
Sₙ = n/2 [2a + (n – 1)d]
S₁₃ = 13/2 [2(π/2) + (13 – 1)π/2]
= 13/2 [π + 6π]
=13/2 (7π)
= 13/2 × 7 × 22/7
= 143 cm - 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?We can see that the numbers of logs in rows are in the form of an A.P. 20, 19, 18…
Here, a = 20, d = 19−20 = −1, , Sₙ = 200, n = ?
The sum of nth term :
Sₙ = n/2 [2a +(n -1)d]
Sₙ = n/2 [2(20)+(n -1)(-1)]
200 = n/2 (40−n+1)
200 = n/2 (41-n)
400 = 41n−n2
n²−41n + 400 = 0
n²−16n−25n + 400 = 0
n(n −16)−25(n −16) = 0
(n −16)(n −25) = 0
Either (n −16) = 0 or n−25 = 0
n = 16 or n = 25
By the nth term formula,
aₙ = a+(n−1)d
a₁₆ = 20+(16−1)(−1)
a₁₆ = 20−15
a₁₆ = 5
The number of logs in the 16th row is 5
Similarly, the 25th term could be written as;
a₂₅ = 20+(25−1)(−1)
a₂₅ = 20−24
= −4
The number of logs in the 25th row is -4 It is impossible because the numbers of logs cannot be negative.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5. - In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run?
The distance of the potatoes from the bucket are 5, 8, 11, 14…, which is in the form of AP.
The distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept. Therefore, the distance to be run, could be written as;
2[5, 8, 11, 14…, ] = 10, 16, 22, 28, 34,……….
Here, a = 10, d = 16−10 = 6 S10 =?
The sum of n terms:
Sₙ = n/2 [2a + (n – 1)d]
S₁₀ = 10/2 [2(10) + (10 - 1)(6)]
= 5[20+54]
= 5(74)
= 370
Therefore, the competitor will run a total distance of 370 m.
