Q1: If a, b, form an A.P. with common difference d. then the value of a 2bc is equal to
(a) 2a + 4d
(b) 0
(c) 2a 4d
(d) 2a  3d [2023, 1 Mark]
Ans: (c)
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
Q3: If k + 2, 4k  6. and 3k  2 are three consecutive terms of an A.P. then the value of k is
(a) 3
(b) 3
(c) 4
(d) 4 [2023, 1 Mark]
Ans: (a)
Since, k + 2, 4k  6 and 3k  2 are three consecutive terms of A.P.
a_{2}  a_{1} = a_{3}  a_{2}
⇒ (4k  6) (k + 2) = (3k  2)  (4k  6)
⇒ 4k 6  k  2= 3k  2  4k + 6
⇒ 3k  8 = k + 4
⇒ 4k = 12
⇒ k = 3
Q4: 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, 3 Marks]
► View AnswerAns: We have
First term, a_{1} =  14
Fifth term, a_{5} = 2
Last term, a_{n} = 62
Let d be the common difference and n be the number of terms.
∵ a_{5} = 2
⇒ 14 +(5  1)d = 2
⇒ 4d = 16
⇒ d =4
Now, a_{n }= 62
⇒ 14 + (n  1)4 = 62
⇒ 4n  4 = 76
⇒ 4n = 80
⇒ n= 20
There are 20 terms in A.P.
Q5: Which term of the A.P. : 65, 61, 57, 53, _____ is the first negative term ? [2023, 3 Marks]
► View AnswerAns: Given, A.P. is 65, 61, 57, 53,.....
Here, first term a = 65 and common difference, d = 4
Let the n^{th} term is negative.
Last term, a_{n} = a + (n  1) = 65 + (n  1)(4)
= 65  4n + 4
= 69  4n, which will be negative when n = 18
So, 18^{th} term is the first negative term.
Q6: Assertion: a, b, c are in AP if and only if 2b = a + c
Reason: The sum of first n odd natural numbers is n^{2}.
(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. [2023, 1 Mark]
Ans: (b)
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 =
Hence, assertion and reason are true but reason is not the correct expiation of assertion.
Q7: The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20^{th} term. [2023, 3 Marks]
► View AnswerAns: Here, a = 15 and S_{15} = 750
⇒ 15(15 + 7d) = 750
⇒ 15 + 7d = 50
⇒ 7d = 35
⇒ d = 5
Now, 20^{th} term = a + (n  1)d
= 15 + (20  1] 5
= 15 + 95
= 110
Q8: 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 30^{th} instalment? What amount of loan has he paid after 30^{th} instalment? [2023, 3 Marks]
► View AnswerAns: 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 30^{th} 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
n^{th} term of an A.P. a_{n}= a + (n  1)d
For 30^{th} instalment, a_{30} = a + (30  1)d
= 1000 + (29) 100 = 1000 + 2900 = 3900
So Rs. 3900 will be paid by Rohan in the 30^{th} instalment.
Now, we have a = 1000, last term (l)= 3900
⇒ S_{30} = 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 30^{th} instalment.
Q9: The ratio of the 11^{th} term to 17^{th} term of an A.P. is 3 : 4. Find the ratio of 5^{th} term to 21^{st} term of the same A.P. Also, find the ratio of the sum of first 5 terms to that of first 21 terms. [2023, 4,5,6 Marks]
► View AnswerAns: Given,
Required ratio =
Q10: 250 logs are stacked in the following manner:
22 logs in the bot tom 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? [2023, 4,5,6 Marks]
Ans: Number of logs in 1^{st} row = 22
Number of logs in 2^{nd} row = 21
Number of logs in 3^{rd} row = 20
The number of Logs i.e., 22, 21, 20, __ forms an A.P.. where
a = 22 ,d = a_{2}  a_{1} ,= 21 22 =  1
Let the number of rows be n.
Now,
∴ Number of logs in the 20^{th} (top) row is 3.
Q1: Find a and b so that the numbers a, 7, b, 23 are in A.P. [2022, 2 Marks]
► 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.
Q2: For the A.P. ; a_{1}, a_{2}, a_{3}, .....a_{4}/a_{7} = 2/3, then find a_{6}/a_{8}. [2022, 2 Marks]
► View AnswerAns: Let a be the first term and d be the common difference of given A.P.
We have, a_{4}/a_{7} = 2/3
Q3: Find the number of terms of the A.P. : 293, 235, 27,....., 53 [2022, 2 Marks]
► View AnswerAns: Given, 293, 285, 277..... 53 be an A.P.
a = 293, d = 285  293 = 8
We know. a_{n} = a + (n  1 )d
⇒ 53 = 293 = (n  1)(8)
⇒ 53  293 = (n  1) (8)
⇒ 240 = (n  1) (8)
⇒ 30 = n  1
⇒ n = 31
Q4: For what value of n are the n^{th} terms of the A.P.'s : 9, 7, 5, .... and 15, 12, 9.... the same? [2022, 2 Marks]
► View AnswerAns:
Q5: Which term of the A .P. [2022, 2 Marks]
► View AnswerAns: Let n^{th} term of the given A.P. be 49/2.
Here a = 11/2
⇒ a_{1} = 11/2 and a_{2} = 3
Q6: Determine the A.P. whose third term is 5 and seventh term is 9. [2022, 2 Marks]
► View AnswerAns: Let the first term and common difference of an A.P. be a and d, respectively.
Given a_{3}= 5 and a_{7} = 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, .....
Q7: In Mathematics, relations can be expressed in various ways. The matchstick patterns are based on linear relations. Different strategies can be used to calculate the number of matchsticks used in different figures. One such pattern is shown below. Observe the pattern and answer the following question using Arithmetic Progression:
(a) Write the AP for the number of triangles used in the figures. Also, write the nth term of this AP.
(b)Which figure has 61 matchsticks. [2022, 2 Marks]
Ans: (a): 20
(b) Numbers of matchsticks used in figure 1 = 12
Number of matchsticks used in figure 2 = 19
Number of matchsticks used in 3 = 26
Thus Required A.P. be 12 , 19 , 26 .....
Hence, figure 8 has 61 matchsticks.
Q8: In an AP if S_{n} = n (4n + 1), then find the AP. [2022, 2 Marks]
► View AnswerAns: Given S_{n} = n (4n + 1)
S_{1 }=1(4 + 1) = 5 = a_{1}
_{S}_{2 }= 2(8 + 1) = 2(9) = 18
S_{3 }= 3(4(3)+1)= 3(13) = 39
S_{4} = 4(4(4) + 1) =4(17) = 68
We know that a_{n} = S_{n}  S_{n1}
Q9: Find the common difference 'd' of an A.P. whose first term is 10 and sum of the first 14 terms is 1505. [2022, 2 Marks]
► View AnswerAns:
Q11: Find the sum of first 20 terms of an A.P whose n^{th }term is given as a_{n} = 5  2n. [2022, 2 Marks]
Ans: Given, n^{th} term of the A.P. series is
So, the series becomes 3, 1, 1,  3 .....
Here, a = 3 and d = a_{2}  a_{1} , = 1  3 =  2
We know that, sum of n terms of an A.P. is
∴ Sum of first 20 terms of an A.P. is
Hence, the required sum of 20 terms of given A.P. is 320
Ans: (b)
Given, 5/7, a, 2 are in A.P. therefore common difference is same.
∴ a_{2}  a_{1} = a_{3}  a_{2}
Q2: Which of the following is not an A.P? [2020, 1 Mark]
(a) 1.2, 0.8.2.8, ....
(b) 3, 3+√2, 3+2√2,3 + 3√2,...
(c)
(d)
Ans: (c)
In option (c), We have
As a_{2}  a_{1 ≠ } a_{3} a_{2} the given list of numbers does not form an A.P.
Q3: The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an A.P, is
(a) 6
(b) 6
(c) 18
(d) 18 [2020, 1 Mark]
Ans: (a)
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
Q4: Show that (a  b)^{2}, (a^{2} + b^{2}) and (a + b)^{2} are in A.P. [2020, 2 Marks]
► View AnswerAns: Let a_{1} = (a  b)^{2}, a_{2} = (a^{2} + b^{2}) and a_{3}= (a + b)^{2}
Now. a_{2}  a_{1} = (a^{2} + b^{2})  (a  b)^{2}
= a^{2} + b^{2}  (a^{2} + b^{2}  2ab)
= a^{2} + b^{2} a^{2} b^{2} + 2ab = 2ab
Again a_{3 } a_{2} = (a + b)^{2}  (a^{2} + b^{2})
= a^{2} + b^{2} + 2ab  a^{2}  b^{2} = 2ab
∴ a_{2}  a_{1} = a_{3}  a_{2}
So, (a  b)^{2}, (a^{2} + b^{2}) and (a + b)^{2} are in A.P.
Q6: 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, 4,5,6 Marks]
► View AnswerAns: 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.
Q7: Find the sum of the first 100 natural numbers. [2020, 1 Mark]
► View AnswerAns: First 100 natural numbers are 1 , 2 , 3 ...... 100 which form an A.P. with a = 1, d = 1.
Sum of n terms =
Q8: Find the sum of first 16 terms of an Arithmetic Progression whose 4^{th} and 9^{th} terms are  15 and  30 respectively. [2020, 3 Marks]
► View AnswerAns: Given, a_{4} = 15 and a_{9} = 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,
Q9: in an A.P. given that the first term (a) = 54. the common difference (d) = 3 and the n^{th} term (a_{n}) = 0. find n and the sum of first n terms (S_{n}) of the A.P. [2020, 3 Marks]
► View AnswerAns: Given, d =  3, a = 54 and a_{n}= 0
Since a_{n} = a + (n 1)d
∴ 0 = 54 + (n  1)(3)
⇒ 0 = 54  3n + 3
⇒ 3n = 57
⇒ n = 19
Now,
Q10: Find the sum: (  5 ) + (  8 ) + ( 11)+...+ ( 230) [2020, 3 Marks]
► View AnswerAns: Let S_{n} = ( 5) + ( 8) + ( 11) + ...+ (230)
Q11: For an A.P, it is given that the first term (a)  5, common difference (d) = 3, and the n^{th} term (a_{n}) = 50. Find n and sum of first n terms (S„) of the A.P [2020, 3 Marks]
Ans: Given, a = 5, d = 3 and a_{n}= 50
Since, a_{n} = a + (n 1)d
Q1: Find the common difference of the Arithmetic Progression (A.P.) [2019, 1 Mark]
► View AnswerAns: Given A.P. is
So. common difference
Q2: Write the common difference of A.P. [2019, 1 Mark]
√3, √12, √27, √48, ......
Ans: Give A.P. is
√3, √12, √27, √48, ......
or √3, 2√3, 3√3,4√3, .......
∴ d = common difference = 2√3  √3 = √3
Q3: If the n^{th} term of an A.P. is pn + q. find its common difference. [2019, 1 Mark]
► View AnswerAns: Given, a_{n} = pn + q
⇒ d + (n  1)d = pn + q
⇒ (n  1)d = pn + q  a
⇒
Q4: Which term of the A.P. 10, 7, 4, ...is 41 ? [2019, 1 Mark]
► View AnswerAns: Let n^{th} term of A.P. 10, 7, 4, ...is 41
∴ a_{n} = a + (n  1)d
∴ 18^{th} term of given A.P. is  41
Q5: If in an A.P. a = 15, d =  3 and a_{n} = 0, then find the value of n. [2019, 1 Mark]
► View AnswerAns: Given, a = 15, d =  3 and a_{n} = 0
∴ a_{n} = a + (n  1)d
⇒ 15 + (n  1 )( 3)= 0
⇒ 15  3n +3 = 0
⇒ 18  3n = 0
⇒  3n = 18
⇒ n = 6
Q6: How many two digit numbers are divisible by 3? [2019, 1 Mark]
► View AnswerAns: Twodigit 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 n^{th} term = 99
a + (n  1)d = 99
⇒ 12 + (n  1)3 = 99
⇒ 3n = 99  9
⇒ n = 90/3
⇒ n = 30
Q7: If the 9^{th} term of an AR is zero, then show that its 29^{th} term is double of its 19^{th} term. [2019, 2 Marks]
► View AnswerAns: Given, a_{9} = 0 . we have to show that a_{2}_{9 =} 2a_{19}
a + 8d = 0
⇒ a =  8 d
Now, a_{19 = }a + 18d = 8d + 18d = 10d
a_{29} = a + 28d = 8d + 28d = 20d = 2(10d ) = 2a_{19}
Hence, a_{2}_{9 =} 2a_{19}
Q8: Which term of the A.P. 3, 15, 27, 39, ... will be 120 more than its 21^{st} term? [2019, 2 Marks]
► View AnswerAns: We have, first term, a = 3, common difference, d = 15  3 =12
n^{th} term of an A.P. is given by a_{n} = a + (n  1)d
∴ a_{21} = 3 + (20) x 12
= 3 + 240
= 243
Let the r^{th} term of the AP. be 120 more than the 21^{st} term.
Q9: If the 17^{th} term of an A.P. exceeds its 10^{th }term by 7, find the common difference. [2019, 2 Marks]
► View AnswerAns: According to question, a_{17}  a_{10} = 7
i.e. a + 16d (a + 9d) = 7
where a = first term d = common difference
⇒ 7d = 7
∴ d = 1
Q10: 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. [2019, 4,5,6 Marks]
► View AnswerAns: Given, Arithmetic Progression 7, 12, 17, 22..
and n^{th} term of given AP. is 82.
∴ a_{n} = a + (n  1)d
Hence, 100 is not any term of given A.P.
Q11: If S_{n} the sum of first it terms of an A.P. is given by S_{n} = 3n^{2}  4n, find the n^{th} term. [2019, 2 Marks]
► View AnswerAns: Given, S_{n} = 3n^{2}  4n
We know that S_{n}  S_{n1} = a_{n}
Q12: If S_{n} the sum of first it terms of an A.P. is given by S_{n} = 2n^{2} + n, find the n^{th} term. [2019, 2 Marks]
► View AnswerAns: We have S_{n} = 2n^{2} + n
Q13: If m^{th} term of an A.P. is 1/n and n^{th} term is 1/m, then find the sum of its first mn terms. [2019, 3 Marks]
► View AnswerAns: Let a be first term and d be the common difference of an A.P.
Since, we have,
Subtracting (ii) from (i), we get
On Substituting (iii) in (i), we get
Sum of first mn terms is
=
126 videos477 docs105 tests

1. What is an arithmetic progression (AP)? 
2. How do you find the nth term of an arithmetic progression? 
3. What is 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 negative terms? 

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