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Previous Year Questions 2024 | |
Previous Year Questions 2023 | |
Previous Year Questions 2022 | |
Previous Year Questions 2020 | |
Previous Year Questions 2019 |
(a) 110º
(b) 115º
(c) 120º
(d) 125º
Ans: (d)
∠OTP = 90º [Line from centre is ⊥ to tangent at point of contact]
∠x = ∠TPO + ∠OTP [Exterior Angle Prop.]
x = 35º + 90º
= 125º
Q2: The ratio of the 10th term to its 30th term of an A.P. is 1 : 3 and the sum of its first six terms is 42. FInd the first term and the common difference of A.P. (2024)
Ans:
Let the AP be a, a + d, a + 2d,.............
⇒ 3a + 27d= a + 29d
⇒ 2a – 2d = 0
⇒ 2a + 2d
∴ a = d ...(i)
Now, S6 = 42
⇒ 3[2a + 5d] = 42
⇒ 2a + 5d =14
⇒ 2a + 5a = 14 [From eqn (i)]
⇒ 7a = 14
⇒ a = 14/7
∴ a = 2
So First term = 2
Common difference = 2.
Q3: If the sum of first 7 terms of an A.P. is 49 and that of first 17 terms is 289, find the sum of its first 20 terms. (2024)
Ans:
S7 = 49
⇒
⇒ 2a + 6d = 14
⇒ a + 3d = 7 ...(i)
S17 = 289
⇒ 2(a + 8d) = 34
⇒ a + 8d = 17 ...(ii)
From eqn (i) and (ii).
–5d = –10
∴ d =2
Put the value of d in eqn.(i),
∴ a + 3 × 2 =7
⇒ a + 6 = 7
⇒ a = 7 - 6
⇒ a = 1
∴
Q4: If a, b, form an A.P. with common difference d. then the value of a- 2b-c is equal to (2023)
(a) 2a + 4d
(b) 0
(c) -2a- 4d
(d) -2a - 3d
Ans: (c)
Sol: We have, a, b, c are in A.P.
b = a + d, and c = a + 2d
Now, a - 2b - c = a - 2(a + d) - (a + 2d)
= a - 2a - 2d - a - 2d
= -2a- 4d
Q5: If k + 2, 4k - 6. and 3k - 2 are three consecutive terms of an A.P. then the value of k is (2023)
(a) 3
(b) -3
(c) 4
(d) -4
Ans: (a)
Sol: Since, k + 2, 4k - 6 and 3k - 2 are three consecutive terms of A.P.
a2 - a1 = a3 - a2
⇒ (4k - 6)- (k + 2) = (3k - 2) - (4k - 6)
⇒ 4k -6 - k - 2= 3k - 2 - 4k + 6
⇒ 3k - 8 = -k + 4
⇒ 4k = 12
⇒ k = 3
Q6: How many terms are there in A.P. whose first and fifth term are -14 and 2, respectively and the last term is 62. (2023)
Ans: We have
First term, a1 = - 14
Fifth term, a5 = 2
Last term, an = 62
Let d be the common difference and n be the number of terms.
∵ a5 = 2
⇒ -14 +(5 - 1)d = 2
⇒ 4d = 16
⇒ d =4
Now, an = 62
⇒ -14 + (n - 1)4 = 62
⇒ 4n - 4 = 76
⇒ 4n = 80
⇒ n= 20
There are 20 terms in A.P.
Q7: Which term of the A.P. : 65, 61, 57, 53, _____ is the first negative term ? (2023)
Ans: Given, A.P. is 65, 61, 57, 53,.....
Here, first term a = 65 and common difference, d = -4
Let the nth term is negative.
Last term, an = a + (n - 1) = 65 + (n - 1)(-4)
= 65 - 4n + 4
= 69 - 4n, which will be negative when n = 18
So, 18th term is the first negative term.
Q8: Assertion: a, b, c are in AP if and only if 2b = a + c
Reason: The sum of first n odd natural numbers is n2. (2023)
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Ans: (b)
Sol: Since, a, b ,c are in A.P. then b - a = c -b
⇒ 2b = a + c
First n odd natural number be 1, 3, 5 ..... (2n - 1).
which form an A.P. with a = 1 and d = 2
Sum of first n odd natural number = n/2[2a + (n -1)d]
= n/2 [2 + (n - 1)2] = n2
Hence, assertion and reason are true but reason is not the correct expiation of assertion.
Q9: The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term. (2023)
Ans: Here, a = 15 and S15 = 750
∵ Sn = n/2[2a + (n -1)d]
∴ S15 = 15/2 [2 x 15 + (15 -1)d] = 750
⇒ 15(15 + 7d) = 750
⇒ 15 + 7d = 50
⇒ 7d = 35
⇒ d = 5
Now, 20th term = a + (n - 1)d
= 15 + (20 - 1) 5
= 15 + 95
= 110
Q10: Rohan repays his total loan of Rs. 1,18,000 by paying every month starting with the first instalment of Rs. 1,000. If he increase the instalment by Rs. 100 every month. what amount will be paid by him in the 30th instalment? What amount of loan has he paid after 30th instalment? (2023)
Ans: Total amount of loan Rohan takes = Rs. 1,18,000
First instalment paid by Rohan = Rs. 1000
Second instalment paid by Rohan = 1000 + 100 = Rs. 1100
Third instalment paid by Rohan = 1100 + 100 = Rs. 1200 and so on.
Let its 30th instalment be n.
Thus, we have 1000,1100,1200, which forms an A.P. with first term (a) = 1000
and common difference (d) = 1100 - 1000 = 100
nth term of an A.P. an= a + (n - 1)d
For 30th instalment, a30 = a + (30 - 1)d
= 1000 + (29) 100 = 1000 + 2900 = 3900
So Rs. 3900 will be paid by Rohan in the 30th instalment.
Now, we have a = 1000, last term (l)= 3900
Sum of 30 instalment, S30 = 30/2[a + 1]
⇒ S30 = 15(1000 + 3900) = Rs. 73500
Total amount he still have to pay after the 30th instalment = (Amount of loan) - (Sum of 30 instalments)
= Rs. 1,18,000 - Rs. 73,500 = Rs. 44,500
Hence, Rs. 44,500 still have to pay after the 30th instalment.
Q11: The ratio of the 11th term to the 18th term of an AP is 2:3. Find the ratio of the 5th term to the 21st term and also the ratio of the sum of the first five terms to the sum of the first 21 terms. (2023)
Ans: Let a and d be the first term and common difference of an AP.
Given that, a11 : a18 = 2 : 3
⇒ 3a + 30d = 2a + 34d
⇒ a = 4d ...(i)
Now, a5 = a + 4d = 4d + 4d = 8d [from Eq.(i)]
And a21 = a + 20d = 4d + 20d = 24d [from Eq. (i)]
a5 : a21 = 8d : 24d = 1 : 3
Now, sum of the first five terms, S5 = 5/2 [2a + (5−1)d]
= 5/2 [2(4d) + 4d] [from Eq.(i)]
= 5/2 (8d + 4d) = 5/2 × 12d = 30d
And, sum of the first 21 terms, S21 = 21/2 [2a + (21−1)d]
= 21/2 [2(4d) + 20d]= 21/2 × 28 d = 294 d from Eq..(i)]
So, ratio of the sum of the first five terms to the sum of the first 21 terms is,
S5 : S21 = 30d : 294d = 5 : 49
Q12: 250 logs are stacked in the following manner: 22 logs in the bottom row, 21 in the next row, 20 in the row next to it and so on (as shown by an example). In how many rows, are the 250 logs placed and how many logs are there in the top row?
Ans: Let there be n rows to pile of 250 logs
Here, the bottom row has 22 logs and in next row, 1 log reduces
It means, we get an AP 22, 21, 20, 19, ..................... n with first term or a = 22 and d = -1
Now, we know that total logs are 250 or we can say that,
Sn =250
Since sum of n terms of an A.P. Sn = n/2 (2a + (n-1) d)
= 250 Therefore, n/2 (2 x 22 + (n-1) x (-1))
or 500 = n (44 - (n-1))
500 = n (45- n)
n2 - 45 n + 500 = 0
By solving this, we get (n-20) (n-25) = 0
Since, there are 22 logs in first row and in next row, 1 log reduces, then we can not have more than 22 terms
∴ n ≠ 25
and n = 20
Means, 20th row is the top row of the pile
Now let's find out number of logs in 20th row
We know that value of nth term of an A.P. = a + (n-1) d
N20 = [22 + (20-1) (-1)]
=(22-19) = 3
Therefore, there are 3 logs in the top row.
Q13: Find a and b so that the numbers a, 7, b, 23 are in A.P. (2022)
View AnswerAns: Since a, 7, b, 23 are in A.P.
∴ Common difference is same.
∴ 7 - a = b -7 = 23 - b
Taking second and third terms, we get
b - 7 = 23 - b
⇒ 2b = 30
⇒ b = 15
Taking first and second terms, we get
⇒ 7 - a = b - 7
⇒ 7 - a = 15 - 7
⇒ 7 - a = 8
⇒ a = -1
Hence, a = -1, b = 15.
Q14: Find the number of terms of the A.P. : 293, 235, 27,....., 53 (2022)
Ans: Given, 293, 285, 277..... 53 be an A.P.
a = 293, d = 285 - 293 = -8
We know. an = a + (n - 1 )d
⇒ 53 = 293 = (n - 1)(-8)
⇒ 53 - 293 = (n - 1) (-8)
⇒ -240 = (n - 1) (-8)
⇒ 30 = n - 1
⇒ n = 31
Q15: Determine the A.P. whose third term is 5 and seventh term is 9. (2022)
Ans: Let the first term and common difference of an A.P. be a and d, respectively.
Given a3= 5 and a7 = 9
a + (3 - 1 ) d = 5 and a + (7 - 1)d = 9
a + 2d = 5 --------------(i)
and a + 6d = 9--------------(ii)
On subtracting (i) from (ii), we get
⇒ 4d = 4
⇒ d = 1
From (i), a + 2(1) = 5 ⇒ a + 2 = 5 ⇒ a = 3
So. required A.P. is a, a + d, a + 2d, a + 3d......
i.e. 3, 3 +1, 3 + 2(1), 3 + 3(1), ....., i.e.. 3, 4, 5, 6, .....
Ans: (b)
Sol: Given, -5/7, a, 2 are in A.P. therefore common difference is same.
∴ a2 - a1 = a3 - a2
Q17: Which of the following is not an A.P? (2020)
(a) -1.2, 0.8.2.8, ....
(b) 3, 3+√2, 3+2√2,3 + 3√2,...
(c) 4/3, 7/3, 9/3, 12/3, ...
(d) -1/5, -2/5, -3/5,..
Ans: (c)
Sol: In option (c), We have
As a2 - a1 ≠ a3- a2 the given list of numbers does not form an A.P.
Q18: The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P, is (2020)
(a) 6
(b) -6
(c) 18
(d) -18
Ans: (a)
Sol: Given, 2x, (x + 10) and (3x + 2) are in A.P.
(x + 10) - 2x = (3x + 2) -(x + 10)
⇒ -x + 10= 2x - 8
⇒ - 3x = -18
⇒ x = 6
Q19: Show that (a - b)2, (a2 + b2) and (a + b)2 are in A.P. (2020)
Ans: Let a1 = (a - b)2, a2 = (a2 + b2) and a3= (a + b)2
Now. a2 - a1 = (a2 + b2) - (a - b)2
= a2 + b2 - (a2 + b2 - 2ab)
= a2 + b2- a2- b2 + 2ab = 2ab
Again a3 - a2 = (a + b)2 - (a2 + b2)
= a2 + b2 + 2ab - a2 - b2 = 2ab
∴ a2 - a1 = a3 - a2
So, (a - b)2, (a2 + b2) and (a + b)2 are in A.P.
Q20: The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the numbers. (2020)
Ans: Let the four consecutive numbers be (a - 3d), (a - d), (a + d), (a + 3d).
Sum of four numbers = 32 (Given)
⇒ (a - 3d) + (a - d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32 ⇒ a = 8
If d = 2. then the numbers are (8 - 6), (8 - 2), (8 + 2) and (8 + 6) i.e., 2,6, 10, 14 .
If d = -2. then the numbers are (8 + 6), (8 + 2), (8 - 2). (8 - 6) i.e.,14 ,10 ,6 ,2 .
Hence, the numbers are 2, 6, 10, 14 or 14, 10, 6, 2.
Q21: Find the sum of the first 100 natural numbers. (2020)
Ans: First 100 natural numbers are 1 , 2 , 3 ...... 100 which form an A.P. with a = 1, d = 1.
Sum of n terms =Sn = n/2 [2a + (n - 1)d]
= 100/2 [2 x 1 + (100 - 1) x 1] = 50 [2 + 99] = 50 x 101 = 5050
Q22: Find the sum of first 16 terms of an Arithmetic Progression whose 4th and 9th terms are - 15 and - 30 respectively. (2020)
Ans: Given, a4 = -15 and a9 = -30
a + 3d = - 15 (i)
a + 8d = -30 (ii)
On subtracting (ii)from (i), we have
-5d = 15
⇒ d = - 3
Put d = - 3 in (i), we have
a + 3(-3)= - 15
⇒ a - 9 = - 15
⇒ a = - 6
Now, Sn = n/2 [2a + (n - 1)d]
⇒ S16 = 16/2 [2(-6) + (16 - 1) (-3)]
= 8 [2(-6) + (15) (-3)] = 8 [-12 - 45] = -456
Q23: in an A.P. given that the first term (a) = 54. the common difference (d) = -3 and the nth term (an) = 0. find n and the sum of first n terms (Sn) of the A.P. (2020)
Ans: Given, d = - 3, a = 54 and an= 0
Since an = a + (n -1)d
∴ 0 = 54 + (n - 1)(-3)
⇒ 0 = 54 - 3n + 3
⇒ 3n = 57
⇒ n = 19
Now,
Sn = n/2 [2a+(n-1)d]
= 19/2 [2 × 54 +(19 - 1)(-3)]
= 19/2 [108 - 54] = 19/2 × 54 = 513
Q24: Find the Sum (−5) + (−8)+ (−11) + ... + (−230). (2020)
Ans: (−5) + (−8)+ (−11) + ... + (−230) .
Common difference of the A.P. (d) = a2 - a1
= -8-(-5)
= -8+5
= -3
So here,
First term (a) = −5
Last term (l) = −230
Common difference (d) = −3
So, here the first step is to find the total number of terms. Let us take the number of terms as n.
Now, as we know,
an = a + (n-1) d
So, for the last term,
- 230 = -5 + ( n-1) (-3)
- 230 = -5-3n + 3
-23 +2 = -3n
- 228/-3 = n
n = 76
Now, using the formula for the sum of n terms, we get
Sn = 76/2 [2(-5) + (76-1)(-3)]
= 38 [-10 + (75)(-3)]
=38 (-10-225)
= 38(-235)
= -8930
Therefore, the sum of the A.P is Sn = -8930
Ans: Give A.P. is
√3, √12, √27, √48, ......
or √3, 2√3, 3√3,4√3, .......
∴ d = common difference = 2√3 - √3 = √3
Q26: Which term of the A.P. 10, 7, 4, ...is -41 ? (2019)
Ans: Let nth term of A.P. 10, 7, 4, ...is -41
∴ an = a + (n - 1)d
⇒ - 41 = 10+(n-1)(-3) [∵ d = 7 - 10 = -3]
⇒ - 41 = 10 - 3n + 3
⇒ -41 = 13 - 3n
⇒ 3n = 54=n= 18
∴ 18th term of given A.P. is - 41
Q27: If in an A.P. a = 15, d = - 3 and an = 0, then find the value of n. (2019)
Ans: Given, a = 15, d = - 3 and an = 0
∴ an = a + (n - 1)d
⇒ 15 + (n - 1 )(- 3)= 0
⇒ 15 - 3n +3 = 0
⇒ 18 - 3n = 0
⇒ - 3n = -18
⇒ n = 6
Q28: How many two digit numbers are divisible by 3? (2019)
Ans: Two-digit numbers which are divisible by 3 are 12, 15, 13..... 99. which forms an A.P. with first term (a) = 12, common difference (d) = 15 - 12 = 3 and last term (l) or nth term = 99
a + (n - 1)d = 99
⇒ 12 + (n - 1)3 = 99
⇒ 3n = 99 - 9
⇒ n = 90/3
⇒ n = 30
Q29: If the 9th term of an AR is zero, then show that its 29th term is double of its 19th term. (2019, 2 Marks)
Ans: Given, a9 = 0 . we have to show that a29 = 2a19
a + 8d = 0
⇒ a = - 8 d
Now, a19 = a + 18d = -8d + 18d = 10d
a29 = a + 28d = -8d + 28d = 20d = 2(10d ) = 2a19
Hence, a29 = 2a19
Q30: Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21st term? (2019)
Ans: We have, first term, a = 3, common difference, d = 15 - 3 =12
nth term of an A.P. is given by an = a + (n - 1)d
∴ a21 = 3 + (20) x 12
= 3 + 240
= 243
Let the rth term of the AP. be 120 more than the 21st term.
⇒ a + (r - 1) d = 243 + 120
⇒ 3 + (r - 1) 12 = 363
⇒ (r - 1) 12 = 360 ⇒ r - 1= 30 ⇒ r = 31
Q31: If the 17th term of an A.P. exceeds its 10th term by 7, find the common difference. (2019)
Ans: According to question, a17 - a10 = 7
i.e. a + 16d- (a + 9d) = 7
where a = first term d = common difference
⇒ 7d = 7
∴ d = 1
124 videos|457 docs|77 tests
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1. What is an arithmetic progression (AP) and how is it defined? |
2. How can I find the nth term of an arithmetic progression? |
3. What is the formula for the sum of the first n terms of an arithmetic progression? |
4. How do you determine if a given sequence is an arithmetic progression? |
5. Can an arithmetic progression have a negative common difference? |
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