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Published on: 29/10/2022
QB365 provides a detailed and simple solution for every Possible Book Back Questions in Class 12 Maths Subject -Numbers and Sequences , English Medium. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
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Questions + Answers key
Take MCQ Maths Test1.
The value of (13 + 23 + 33 +...+153) - (1 + 2 + 3 +...+ 15)is
14400
14200
14280
14520
2.
If the sequence t1, t2, t3... are in A.P. then the sequence t6, t12, t18,.... is
a Geometric Progression
an Arithmetic Progression
neither an Arithmetic Progression nor a Geometric Progression
a constant sequence
3.
The next term of the sequence \(\frac { 3 }{ 16 } ,\frac { 1 }{ 8 } ,\frac { 1 }{ 12 } ,\frac { 1 }{ 18 } \), ..... is
\(\frac { 1 }{ 24 } \)
\(\frac { 1 }{ 27 } \)
\(\frac { 2 }{ 3 } \)
\(\frac { 1 }{ 81 } \)
4.
If A = 265 and B = 264 + 263 + 262 +...+ 20 Which of the following is true?
B is 264 more than A
A and B are equal
B is larger than A by 1
A is larger than B by 1
5.
In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P. must be taken for their sum to be equal to 120?
6
7
8
9
6.
An A.P. consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is
16 m
62 m
31 m
\(\frac { 31 }{ 2 } \) m
7.
8.
The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P.
4551
10091
7881
13531
9.
Given F1 = 1, F2 = 3 and Fn = Fn-1 + Fn-2 then F5 is
3
5
8
11
10.
74k \(\equiv \) ________ (mod 100)
1
2
3
4
11.
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
2025
5220
5025
2520
12.
The sum of the exponents of the prime factors in the prime factorization of 1729 is
1
2
3
4
13.
If the HCF of 65 and 117 is expressible in the form of 65m - 117 , then the value of m is
4
2
1
3
14.
Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are
0, 1, 8
1, 4, 8
0, 1, 3
0, 1, 3
15.
1.
(c)
14280
2.
(b)
an Arithmetic Progression
3.
(b)
\(\frac { 1 }{ 27 } \)
4.
(i) To check if h is one – one, we assume that h(b1) = h(b2)
Then we get 2.47b1 + 54.10 = 2.47 b2 + 54.10
2.47b1 = 2.47b2 ⇒ b1 = b2
Thus, h(b1) = h(b2) ⇒ b1 = b2, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 = 177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) =147.96 and so the length of the thigh bone is given by 2.47b + 54.10 = 147.96.
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cms.
5.
(c)
8
6.
(i) To check if h is one – one, we assume that h(b1) = h(b2)
Then we get 2.47 b1 + 54.10 = 2.47 b2 + 54.10
2.47b1 = 2.47b2 ⇒ b1 = b2
Thus, h(b1) = h(b2) ⇒ b1 = b2, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 = 177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) = 147.96 and so the length of the thigh bone is given by 2.47b + 54.10 = 147.96.
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cms.
7.
(a)
8.
(c)
7881
9.
(d)
11
10.
(a)
1
11.
(d)
2520
12.
(c)
3
13.
(b)
2
14.
(a)
0, 1, 8
15.
(c)
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Tamilnadu Stateboard 10th Standard Subjects
Tamilnadu Stateboard Standards
