CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set D

Read and download free pdf of CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set D. Download printable Mathematics Class 10 Worksheets in pdf format, CBSE Class 10 Mathematics Chapter 5 Arithmetic Progression Worksheet has been prepared as per the latest syllabus and exam pattern issued by CBSE, NCERT and KVS. Also download free pdf Mathematics Class 10 Assignments and practice them daily to get better marks in tests and exams for Class 10. Free chapter wise worksheets with answers have been designed by Class 10 teachers as per latest examination pattern

Chapter 5 Arithmetic Progression Mathematics Worksheet for Class 10

Class 10 Mathematics students should refer to the following printable worksheet in Pdf in Class 10. This test paper with questions and solutions for Class 10 Mathematics will be very useful for tests and exams and help you to score better marks

Class 10 Mathematics Chapter 5 Arithmetic Progression Worksheet Pdf

Question. The 19th term from the end of the series 2 + 6 + 10 + …. + 86 is:
a. 6
b. 18
c. 14
d. 10
Answer : C

Question. In a certain A.P., 5 times the 5th term is equal to 8 times the 8th term, then its 13th term is:
a. 0
b. – 1
c. – 12
d. – 13
Answer : A

Question. The nth term of the series 3 + 10 + 17 + ….. and 63 + 65 + 67 + …… are equal, then the value of n is:
a. 11
b. 12
c. 13
d. 15
Answer : C

Question. The sum of the first and third term of an A.P. is 12 and the product of first and second term is 24, the first term is:
a. 1
b. 8
c. 4
d. 6
Answer : C

Question. If Sr denotes the sum of the first r terms of an A.P., then S3r - Sr-1/S2r - S2r-1 is equal to:
a. 2r – 1
b. 2r + 1
c. 4r + 1
d. 2r + 3
Answer : B

Question. If the sum of the first 2n terms of 2, 5, 8…. is equal to the sum of the first n terms of 57, 59, 61…., then n is equal to:
a. 10
b. 12
c. 11
d. 13
Answer : C

Question. The sum of all odd numbers of two digits is:
a. 2475
b. 2530
c. 4905
d. 5049
Answer : A

Question. If sum of n terms of an A.P. is 3n2 + 5n and 164 Tm = ,then m =?
a. 26
b. 27
c. 28
d. 29
Answer : B

Question. The sum of n terms of the series 1/1+√3 + 1/√3+v5

""CBSE-Class-10-Mathematics-Arithmetic-Progressions-Worksheet-Set-D

Answer : D

Question. If a1, a2 , ,.., an+1 + are in A.P., then 1a1a2 + 1/a2a3 + ..... + 1/anan+1
a. n - 1 / a1an+1
b. 1 / a1an+1
c. n + 1 / a1an+1
d. n / a1an+1

Answer : D

Question. The number of terms in the series 101+ 99 + 97 +.....+ 47 is:
a. 25
b. 28
c. 30
d. 20
Answer : B

Question. The numbers (2 +1),1, (2 −1) will be in:
a. A.P.
b. G.P.
c. H.P.
d. None of these
Answer : B

Question. If x, 2x + 2,3x + 3 are in G.P., then the fourth term is:
a. 27
b. – 27
c. 13.5
d. – 13.5
Answer : D

Question. The first term of an infinite geometric progression is x and its sum is 5. Then:
a. 0 ≤ x ≤ 10
b. 0 < x < 10
c. −10 < x < 0
d. x > 10
Answer : B

Question.

""CBSE-Class-10-Mathematics-Arithmetic-Progressions-Worksheet-Set-D-1

a. e + 1
b. e – 1
c. 1 – e
d. e
Answer : B

Question. The G.M. of the numbers 2 3 3,3 , 3 ......3n is:
a. 32/n
b. 3n+1/2
c. 3n/2
d. 3n−1/2
Answer : B

Question. The two geometric mean between the number 1 and 64 are:
a. 1 and 64
b. 4 and 16
c. 2 and 16
d. 8 and 16
Answer : B

Question. The 4th term of a H.P. is 3/5 and 8th term is 1/3 then its 6th term is:
a. 1/6
b. 3/7
c. 1/7
d. 3/5
Answer : B

Question. The harmonic mean of the roots of the equation 2 (5 + √2)x − (4 +√3) x +8 + 2 √3 = 0 is:
a. 2
b. 4
c. 6
d. 8
Answer : B

Question. If a,b,c are in H.P., then the value of (1/b + 1/c - 1/a) (1/c + 1/a - 1/b)
a. 2/bc + 1b2
b. 3/c2 + 2/ca
c. 3/b2 - 2b ab
d. None o]f these
Answer : C

Question. 1 + 3 /22  5/22  ∞ is equal to:
a. 3
b. 6
c. 9
d. 12
Answer : B

Question. Sum of the series 2 3 99 1+ 2.2 + 3.2 + 4.2 +....+100.2 is:
a. 100.2100+1
b. 100 99.100 +1
c. 100 99.100 −1
d. 100 100.100 −1
Answer : B

Question. The sum to n terms of the series 1 + 3 + 7 + 15 + 31 +...is:
a. 2n+1 − n
b. n n + − −
c. 2n - n − 2
d. None of theseSummation of Series
Answer : B

Question. If the A.M., G.M. and H.M. between two positive numbers a and b are equal, then:
a. a = b
b. ab = 1
c. a > b
d. a < b
Answer : A

Question. If 5 7 9 x = log 3+ log 5 + log 7 then:
a.x > 3/2
b .x > 3/2
c. > 1/13 
d.1/√3
Answer : C

Question. If the A.M. of two numbers is greater than G.M. of the numbers by 2 and the ratio of the numbers is 4 : 1, then the numbers are:
a. 4, 1
b. 12, 3
c. 16, 4
d. None of these
Answer : C

Question. If the ratio of H.M. and G.M. of two quantities is 12 : 13, then the ratio of the numbers is:
a. 1: 2
b. 2: 3
c. 3: 4
d. None of these
Answer : D

Question. Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation:
a. 2 x −18x −16 = 0
b. 2 x −18x +16 = 0
c. 2 x +18x −16 = 0
d. 2 x +18x +16 = 0
Answer : B

Question. If the arithmetic, geometric and harmonic means between two positive real numbers be A, G and H , then:
a. A2 = GH
b. H2 = AG
c. G = AH
d. G2 = AH
Answer : D

Question. If log , log , log a b c x x x be in H.P., then a, b, c are in:
a. A.P.
b. H.P.
c. G.P.
d. None of these
Answer : C

Question. If a, b, c are in A.P. as well as in G.P., then:
a. a = b ≠ c
b. a ≠ b = c
c. a ≠ b ≠ c
d. a = b = c
Answer : D

Question. If a, b, c are in G.P., a − b, c − a, b − c are in H.P., then a + 4b + c is equal to:
a. 0
b. 1
c. −1
d. None of these
Answer : A

Question. If A1, A2 are the two A.M.'s between two numbers a and b and  G1 ,G2 be two G.M.'s between same two numbers, then A1 + A2 / G1,G= ?
a. a+b/ab
b. 2+b/2ab
c. 2ab/a+b
d. ab/a+b
Answer : A

Question. If G.M. = 18 and A.M. = 27, then H.M. is:
a. 1/18
b. 1/12
c. 12
d. 9√6
Answer : C

Question. If 9 A.M.'s and H.M.'s are inserted between the 2 and 3 and if the harmonic mean H is corresponding to arithmetic mean A , then A + 6/H = ?
a.1
b.3
c.5
d.6
Answer : C

Question. If | x |<1, then the sum of the series 1+ 2x + 3x3 + 4x3 + ...∞ will be:
a. 1/1 − x
b. 1/1 + x
c. 1/(1 + x)2
d. 1/(1- x)2
Answer : D

Question. The sum of infinite terms of the following series 1 + 4 / 5 + 7 / 52 + 10 / 53
a. 3/16
b. 35/8
c. 35/4
d. 35/16
Answer : D

Question. 1+ 3+ 7 +15 + 31+.......... to n terms = ?
a. 2n+1 − n
b. 2n+1 − n − 2
c. 2n − n − 2
d. None of these
Answer : B

Question. Sum of n terms of series 12 + 16 + 24 + 40 + ..... will be:
a. 2(2n − 1) + 8n
b. 2(2n − 1) + 6n
c. 3(2n − 1) + 8n
d. 4(2n − 1) + 8n
Answer : D

Question. The sum of the series 1 + 1.3 / 6 + 1.3.5 / 6.8 + ∞ is:
a. 1
b. 0
c. ∞
d. 4
Answer : D

Question. The sum of the series 1 + 2x + 3x2 + 4x3 +......... upto n terms is:

""CBSE-Class-10-Mathematics-Arithmetic-Progressions-Worksheet-Set-D-2

Answer : A

Question. If the set of natural numbers is partitioned into subsets S1 = {1}, S2 = {2, 3}, S3 = {4, 5, 6} and so on. Then the sum of the terms in S50 is:
a. 62525
b. 25625
c. 62500
d. None of these
Answer : A

Question. The sum first n odd natural number is
(a) 2n-1
(b) n2
(c) 2n+1
(d) n2-1
Answer : B

Question. The next term of an AP √7 , √28 , √63 , ………..
(a) √84
(b) √70
(c) √97
(d) √112
Answer : A

Question. For the A.P. 3/2 , 1/2 , -1/2 , -3/2 write common difference
(a) 1
(b) 14
(c) -1
(d) 3/2
Answer : C

Question. What is common difference of an AP . in which a18 – a14 = 32
(a) 8
(b) -8
(c) -4
(d) 4
Answer : A

Question. If 6/5, a , 4 are in AP then value of a is
(a) 13/3
(b) 13/6
(c) 13/5
(d) 13/7
Answer : C

 
Short Answer Type Question
 
Question. Find the sum of first 30 terms of an A.P. whose second term is 2 and seventh term is 22.
Answer : ​​​​Let a be the first term and d be the common difference of the given A.P. Then,
a2 = 2 and a7 = 22
=> a + d = 2 and a + 6d = 22
Solving these two equations, we get
a = – 2 and d = 4.
Sn = n/2
[2a + (n – 1) d]
 S30 = 30/2
[2 × (–2) + (30 – 1) × 4]
=> 15 (–4 + 116) = 15 × 112
= 1680
Hence, the sum of first 30 terms is 1680.
 
Question.  Find the sum of all natural numbers between 250 and 1000 which are exactly divisible by 3.
Answer : Clearly, the numbers between 250 and 1000 which are divisible by 3 are 252, 255, 258,...., 999.
This is an A.P. with first term a = 252, common difference = 3 and last term = 999. Let there be n terms in this A.P. Then,
=> an = 999
=> a + (n – 1)d = 999
=> 252 + (n – 1) × 3 = 999 => n = 250
 Required sum = Sn = n/2 [a + l]
= 250/2 [252 + 999] = 156375
 
Question. How many terms of the series 54, 51, 48, ....be taken so that their sum is 513 ? Explain the double answer.
Answer : a = 54, d = – 3 and Sn = 513
=> n/2 [2a + (n – 1) d] = 513
 n/2[108 + (n – 1) × – 3] = 513
 => n2 – 37n + 342 = 0
=> (n – 18) (n – 19) = 0   =>n = 18 or 19
Here, the common difference is negative, So,
19th term is a19 = 54 + (19 – 1) × – 3 = 0.
Thus, the sum of 18 terms as well as that of 19 terms is 513.
 
CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set D

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