CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set F

Question. The 10^th term of an A.P. is 52 and 16^th term is 82. Find the 32nd term and the general term.
Solution. a₃₂ = 162, a_n = 5n + 2

Question. between 50 and 500 which are divisible by 3 and 5.
Solution. 8325

Question. In an A.P., the sum of the first n terms is 3n²/2 + 13/n , n. Find its 25^th term.
Solution. 80

Question. Find the common difference of an A.P. whose first term is 1 and the sum of the first four terms is one-third the sum of the next four terms.
Solution. 2

Question. Find the 20^th term from the end of the A.P. 3, 8, 13, ....., 253.
Solution. 158

Question. The sum of the first 9 terms of an A.P. is 171 and that of the first 24 terms is 996, find the first term and the common difference.
Solution. a = 7, d = 3

Question. Salman Khan buys a shop for Rs. 1, 20,000. He pays half of the amount in cash and agrees to pay the balance in 12 annual instalments of Rs. 5000 each. If the rate of interest is 12% and he pays with the instalment the interest due on the unpaid amount, find the total cost of the shop.
Solution. Rs. 1,66,800

Question. Find the term of A.P. 9, 12, 15, 18, .... which is 39 more than its 36th term.
Solution. 49^th

Question. The sum of three numbers in A.P is 21and their product is 231 find the numbers.
Solution. 2475

Question. If the sum of the first 20 terms of an A.P. is 400 and the sum of the first 40 terms is 1600, find the sum of its first 10 terms.
Solution. 100

Question. Find A.P which fifth terms are 5 and common difference is –3.
Solution. 17

Question. The nth term of an A.P is 3n+5 find its common difference.
Solution. 3

Question. The 7^th term of an A.P. is – 4 and its 13^th term is –16. Find the A.P.
Solution. 8, 6, 4, .....

Question. Find the sum of first 25 terms of an A.P. whose nth term is given by an = 2 – 3n.
Solution. –925

Question. Find the sum of first 24 term of AP 5,8,11,14………..
Solution. 948

Question. If the sum of 7 terms of an A.P. is 49 and that of 17 terms is 289, find the sum of n terms.
Solution. n²

Question. Find whether 0 is a term of the A.P: 40,37,34,31………….
Solution. no

Question. Determine the 10thterm from the end of the A.P:4,9,14……..254
Solution. 209

Question. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the sum of first 20 terms of the A.P.
Solution. 690

Question. Find the sum of all two digit odd positive numbers
Solution. 2, 26

Question. A man saved Rs. 32 during the first year, Rs. 36 in the second year and in this way be increases his savings by Rs. 4 every year. Find in what time his savings will be Rs. 200.
Solution. 5 years.

Question. Pranshi saved Rs. 5 in the first week of the year and then increased her weekly savings by Rs. 1.75 each week. In what week will her weekly savings be Rs. 20.75?
Solution. 10^th week

Question. Find the sum of first 15 multiples of 8.
Solution. 0

Question. Find the middle term of A.P 6,13,20,………..216
Solution. 111

Question. Find the sum of 25^th term of an AP which nth term is given by tn=(7-3n)
Solution. 44550

Question. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of first sixteen terms of the A.P.
Solution. 20 or 76

Question. Find the number of integers between 50 and 500 which are divisible by 7.
Solution. 64

Question. (a) Find the 12^th term from the end of the A.P. 3, 8, 13, ..... , 253.
(b) How many three digit numbers are divisible by 7?
Solution. (a) 198 (b) 128

Question. The 6t hterm of an A.P is -10 and its 10thterm is -26.Determine the15thterm of an A.P
Solution. -46

Question. Write the next term of an A.P. √2 , √18
Solution. 5√2

Question. Find the sum of three digits numbers which are divisible by 11
Solution. 6,12

Question. Write the value of x for which x+2,2x,2x+3 are three consecutive terms of an A.P
Solution. 5

Question. If 4/5 ,a,2 three consecutive term of an A.P then find ’a’.
Solution. 7/5

Question. Write fourth term of an AP if its nth term is 3n+2.
Solution. 14

Question. If the sum of the first 7 terms of an A.P. is 49 and that of its first 17 terms is 288, find the sum of first n terms of the A.P. 
Solution. Let the first term of an A.P be 'a’ and its common difference be 'd’.
given: S₇ = 49
and S₁₇ = 289
Then, S₇ = 7/2[2a + (7 – 1) × d]
[∵ S_n = n/2[2a + (n -1 )d]
⇒ 49 = 72[2a + 6d]
⇒ 7 = a + 3d ...(i)
and S₁₇ = 17/2[2a + (17 – 1) × d]
⇒ 289 = 172× [2a + 16 × d]
⇒ 17 = a + 8d ...(ii)
On subtracting equation (i) from equation (ii), we get:
a + 8d = 17 89
If we put the value of ‘d’ in equation (i), we get
a + 3 × (2) = 7
⇒ a = 7 – 6 = 1
Sum of the first 'n’ terms,
S_n =n²[2a + (n – 1)d]
=n²[2 + (n – 1) × 2]
=n²[2 + 2n – 2]
= n²
Hence, the sum of the first 'n’ terms is n².

Question. Find the:
(A) sum of those integers between 1 and 500 which are multiples of 2 as well as of 5. 
(B) sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
(C) sum of those integers from 1 to 500 which are multiples of 2 or 5.
[Hint (iii): These numbers will be: multiple of
2 + multiple of 5 – multiples of 2 as well as of 5]
Solution. (A) Multiples of 2 and 5 will be multiples of
LCM of 2 and 5. LCM of ( 2, 5) = 10
Multiples of 2 as well as 5 between 1 and
500 is 10, 20, 30, 40, ...490
this forms an AP with first term, a= 10
Common difference, d = 20 - 10 = 10
Last term, L = 490
We know that sum of n terms between 1 and 500 is
S_n = n/2[a + L]
Also, L = a + (n – 1) d
⇒ 490 = 10 + (n – 1)10
⇒ 480 = (n – 1)10
⇒ (n – 1) = 48
⇒ n = 49
Putting this value in equation (i), we get
S₄₉ =49/2[10 + 490]
=49/2× 500 = 49/2× 250= 12250
⇒ S₄₉ = 12250
(B) Here, multiples of 2 as well as 5 from 1 to
500 are 10, 20, 30, ...500
Here, first term, a = 10
common difference, d = 20 - 10 = 10
Last term, L = 500
We know that an = a + (n – 1)d
L = a + (n – 1)d
[Where, n is total no. of terms]
500 = 10 + (n – 1)d
490 = (n – 1)10
⇒ (n – 1) = 49 ⇒ n = 50
Also we know that Sn =n/2(a + L)
⇒ S₅₀ = 50/2(10 + 500)
= 25 × 510 = 12750
Hence, S50 = 12750
(C) Multiples of 2 or 5 = Multiples of 2 +
multiples of 5 – [Multiples of 2 and 5] ...(i)
Multiples of 2 [2, 4, 6, ...500]
Multiples of 5 [5, 10, 15, ...500]
Multiples 2 and of 5 [10, 20, 30, ...500]
1st list of multiples of 2 [2, 4, 6, ...500]
Here, first term, a1 = 2
and common difference d₁ = 2
Let number of terms be n₁
Then, last term, L = a + (n₁ – 1)d
500 = 2 + (n₁ – 1)(2)
498 = (n₁ – 1)2
⇒ (n₁ – 1) = 249 ⇒ n₁ = 250
Sum of [2, 4, 6, ... 500]
S_n,=n₁/2[a + l]
= 250/2[2 + 500]
S_n, = 225 × 502 = 62750
2nd list of multiples of 5 [5, 10, 15, ...500]
Here, first term, a’ = 5,
common difference, d’ = 5
Last term, L’ = 500
Let n² be the number of terms of second
list. We know that an = a + (n – 1)d
L’ = a’ + (n₂ – 1) d’
500 = 5 + (n₂ – 1)5
⇒ 495 = (n2 – 1)5
⇒ n₂ – 1 = 99
⇒ n₂ = 100
Sum of 2nd List,
S_n =n/2(a + L)
S_n2 =n₂/2(5 + 500)
= 100/2×505= 25250
3rd list of multiples of 2 as well as 5
[10, 20, 30, ...500 ]
Here, first term, a’’ = 10
Common difference, d’’ = 10
Last term , L’’ = 500
Let n3 be the number of terms, then
L’’ = a’’ + (n3 – 1) d’’
500 = 10 + (n3 – 1)(10)
490 = (n3 – 1)10
⇒ n₃ – 1 = 49
⇒ n₃ = 50
Sum of 3rd List,
Sn³ = n₃/2(10 + 500)
= 50/2×510 =12750
⇒ Sum of multiples of 2 or 5
= Sn₁+ Sn₂ – Sn₃
= 62750 + 25250 – 12750
= 88000 – 12750 = 75250

Question. Find the sum of the following series :
5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 + .... + (– 5) + 81 + (– 3)
Solution. Given: series is:
5 + (– 41) + 9 + (– 39) + 13 + (– 37) + 17 +
(– 5) + 81 + (– 3)
= [5 + 9 + 13 + ... + 81) + [(– 41) + (– 39) +
(– 37) + (– 5) + (– 3)]
Here are the two series:
1. 5 + 9 + 13 + ...... + 81
2. (– 41) + (– 39) + (– 37) + .... + (– 3)
For the first series :
first term a = 5, common difference, d = 9 - 5
= 13 – 9 = 4
last term, l(an) = 81
Then, an = a + (n – 1)d [where ‘n’ is the number of terms
81 = 5 + (n – 1) × 4
⇒ (n – 1) = 76/4 = 19
⇒ n = 20
Sum of the first series, S20
S₂₀= 20/2[2 × 5 + (20 − 1) × 4 ]
= 10(10 + 19 × 4)
= 10(10 + 76)
= 860
For the second series :
first term, a′ = – 41,
common difference, d′ = (– 39) – (– 41)
= (– 37) – (– 39)
= 2
last term, l′(an) = – 3
Then, an′ = a′ + (n′ – 1)d′
[where n′ are the number of terms]
⇒ – 3 = – 41 + (n′ – 1) × 2
⇒ (n′ – 1) = 19
⇒ n′ = 20
Sum of the second series, S′20 =n'[2d' + (n' - 1)d']
=20/2[2 × (− 4 1) + (19) × 2]
= 10[– 82 + 38]
= 10 × (– 44)
= – 440
Total sum = S₂₀ + S_′20
= 860 – 440 = 420
Hence, the total sum of series is 420.

Question. An AP consists of 37 terms. The sum of the 3 middle most terms is 225 and the sum of the last 3 is 429. Find the AP.
Solution. It is given that
Total number of terms, n = 37
Since n is odd, therefore
middle most term =(n + 1)/2 th term = 19th term 3 middle most terms = 18th, 19th and 20th term by given condition
a₁₈ + a₁₉ +a₂₀ = 225
We know that an = a + (n – 1) d
⇒ (a + 17d) + (a + 18d) + (a + 19d) = 225
⇒ 3a + 54d = 225
⇒ a + 18d = 75 ...(i)
Also, it is given that sum of last 3 terms = 429
⇒ a₃₅ + a₃₆ + a₃₇ = 429
⇒ (a + 34d) + (a + 35d) + (a + 36d) = 429
⇒ 3a + 105d = 429
⇒ a + 35d = 143 ...(ii)
Subtracting equation (i) from equation (ii), we get
(a + 35d) – (a + 18d) = 143 – 75
a + 35d – a – 18d = 68
⇒ 17d = 68 ⇒ d = 4
Putting value of d in equation (i), we get
a + 18(4) = 75
⇒ a + 72 = 75
⇒ a = 3, d = 4
Required AP is
a, a + d, a + 2d, a + 3d, ...
3, 3 + 4, 3 + 2(4), 3 + 3(4), ...
3, 7, 11, 15, ...

Question. Which term of the Arithmetic Progression – 7, – 12, – 17, – 22, ... will be –82? Is –100 any term of the A.P.? Give reason for your answer.
Solution. The first term of the A.P., a = – 7
Common difference,
d = (– 12) – (– 7) = (–12 + 7)
= – 5
Let the nth term of A.P. be – 82.
Then, an = – 82
a + (n – 1) d = – 82
⇒ – 7 + (n – 1) (– 5) = – 82
⇒ – 7 – 5n + 5 = – 82
⇒ – 5n = – 80
⇒ n = 16
Let, mth term of the given A.P. be – 100
am = – 100
a + (m – 1) d = – 100
⇒ – 7 + (m – 1) (– 5) = – 100
⇒ (m – 1) 5 = 93
⇒ m =93/5+ 1=98/5∉ N
Therefore, – 100 is not any term of the given A.P.
Hence, – 82 is the 16th term of the A.P. and – 100 is not a term of the given A.P.

Question. How many terms of the arithmetic progression 45, 39, 33, ... must be taken so that their sum is 180? Explain the double answer.
Solution. Given: The arithmetic progression is 45, 39, 33, ......
Here, a = 45
d = 39 – 45 = 33 – 39 = ........
= – 6
and the sum the of nth term, Sn = 180
Sn =n/2[2a + (n – 1)d]
where, n is the number of terms.
180 =n/2[2 × 45 + (n – 1) (– 6)]
⇒ 360 = 96n – 6n2
⇒ 6n2 – 96n + 360 = 0
⇒ 6[n2 − 16n + 60] = 0
⇒ 6(n2 − 10n − 􀇩n + 60] = 0
⇒ 6(n(n(n − 10) − 6(n − 10)] = 0
⇒ 6[(n – 6) (n – 10)] = 0
⇒ n = 6, 10
Sum of a7, a8, a9 and a10 terms = 0
Hence, on either adding 6 terms or 10 terms we get a total of 180.

Question. Find the sum of the integers between 100 and 200 that are:
(A) divisible by 9
(B) not divisible by 9
Solution. (A) Integers between 100 and 200 that are divisible by 9 are 108, 117, 126, ...198
Let n be the number of terms between 100 and 200
Here, first term, a = 108
Common difference, d = 117 – 108 = 9
Last term, L = 198
We know that an = a + (n – 1)d
L = a + (n – 1)d
⇒ 198 = 108 + (n – 1)9
⇒ 198 – 108 = (n – 1)9
⇒ 90 = (n – 1)9
⇒ (n – 1) = 10 ⇒ n = 11
Sum of n terms, Sn =n/2[2a + (n – 1)d]
∴ Sum of 11 terms between 100 and 200 which is divisible by 9 is
S_n =n/2[2a + (n – 1)d]
⇒ S₁₁ =11/2[2(108) + (11 – 1)(9)]
=11/2[2(108) + 10 × 9]
= 11 [108 + 5 × 9]
= 11 [108 + 45]
= 11× 153 = 1683
Hence, the required sum of integers is 1683.
(B) Sum of integers between 100 and 200 which is not divisible by 9 = sum of total no. between 100 and 200 –
sum of no. between 100 and 200 which are divisible by 9 ...(i)
From part (i), we know that sum of no.
between 100 and 200 divisible by 9 = 1683
Total numbers between 100 and 200 is
101, 102, 103, ... 199.
Here, first term, a = 101
Common difference, d = 102 – 101 = 1
Last term, L = 199
Let n be total no. of terms.
We know that an = a + (n – 1)d
⇒ L = a + (n – 1)d
199 = 101 + (n – 1)(1)
199 – 101 = (n – 1)
n = 99
∴ Sum of 99 terms between 100 and 200
S_n =n/2[2a + (n – 1)d]
S₉₉ =99/2[2(101) + (99 – 1)(1)]
=99/2[202 + 98
=99/2× 300
S₉₉ = 99 × 150 = 14850
Putting this value in equation (i), we get
sum of no. between 100 and 200 not divisible by 9
= 14850 – 1683
= 13167
Hence, the required sum is 13167.

Question. If the ratio of the 11th term of an AP to its 18th term is 2 : 3, find the ratio of the sum of the first five terms to the sum of its first 10 terms. 
Solution. Let, the first term of A.P. be 'a’ and its common
difference be 'd’.
Then, 11th term of A.P., a11 = a + 10d
18th term of A.P., a18 = a + 17d
a +10d /a + 17d 2/3
⇒ 3(a + 10d) = 2(a + 7d)
⇒ 3a + 30d = 2a + 34d
⇒ a = 4d ...(i)
Now, sum of first five terms,
S5 = n/2[2a + (n – 1)d]
=5/2[2 × 4d + 4 × d] [from (i)]
=5/2× 12d = 30d
Sum of first 10 terms,
S10 = n/2[2a + (n – 1)d]
=10/2[2 × 4d + 9 × d]
= 5(8d + 9d)
= 85d
S5/S10 = 30d/85d = 6/17
Hence, the required ratio is 6 : 17.