Exercise 2.10 - Chapter 2 Numbers and Sequences 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 2.10$
Multiple choice questions
Question 1.
Euclid's division lemma states that for positive integers a and $b$, there exist unique integers $q$ and $r$ such that $\mathrm{a}=\mathrm{bq}+\mathrm{r}$, where $r$ must satisfy.
(1) $1 (2) $0 (3) $0 \leq r (4) $0 Answer:
(3) $0 \leq r  

Question 2.
Using Euclid's division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are
(1) $0,1,8$
(2) $1,4,8$
(3) $0,1,3$
(4) $1,3,5$
Answer:
(1) $0,1,8$
Hint:
Cube of any +ve integers $1^{3}, 2^{3}, 3^{3}, 4^{3}, \ldots$
$1,8,27,64,125,216 \ldots$
Remainders when $27,64,125$ are divided by 9 .

Question $3 .$
If the H.C.F of 65 and 117 is expressible in the form of $65 \mathrm{~m}-117$, then the value of $\mathrm{m}$ is
(1) 4
(2) 2

(3) 1
(4) 3
Answer:
(2) 2

Question $4 .$
The sum of the exponents of the prime factors in the prime factorization of 1729 is
(1) 1
(2) 2
(3) 3
(4) 4
Answer.
(3) 3

Question $5 .$
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(1) 2005
(2) 5220
(3) 5025
(4) 2520
Answer:
(4) 2520

Question $6 .$
$7^{4 \mathrm{k}} \equiv$ $(\bmod 100)$
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(1) 1
Hint:
$7^{4 \mathrm{k}} \equiv \ldots .(\bmod 100)$
$7^{4 \mathrm{k}}=\left(7^{4}\right)^{\mathrm{k}} \equiv \ldots \ldots \ldots(\bmod 100)\left(7^{4}-2401\right)$
The value is 1 .

Question 7.
Given $F_{1}=1, F_{2}=3$ and $F n=F_{n-1}+F_{n-2}$ then
(1) 3
(2) 5
(3) 8
(4) 11
Answer:
(4) 11

Question $8 .$
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
(1) 4551
(2) 10091
(3) 7881
(4) 13531
Answer:
(3) 7881
Hint:
Here $a=1, d=4$
It is not a term of A.P.
(ii) $4 \mathrm{n}-3=10091 \Rightarrow 4 \mathrm{n}=10091+3=10094$
$\mathrm{n}=\frac{10094}{4}=2523.5$ it is a term of A.P.
(iii) $4 \mathrm{n}-3=7881 \Rightarrow 4 \mathrm{n}=7881+3$
$\mathrm{n}=\frac{7884}{4}=1971$.
$\therefore 7881$ is a term of the A.P.

Question $9 .$
If 6 times of $6^{\text {th }}$ term of an A.P is equal to 7 times the 7 th term, then the 13 th term of the A.P. is
(1) 0
(2) 6
(3) 7
(4) 13
Answer:
(1) 0

Question $10 .$
An A.P consists of 31 terms. If its $16^{\text {th }}$ term is $m$, then the sum of all the terms of this A.P. is
(1) $16 \mathrm{~m}$
(2) $62 \mathrm{~m}$
(3) $31 \mathrm{~m}$
(4) $\frac{31}{2} \mathrm{~m}$
Answer:
(3) $31 \mathrm{~m}$
Hint:
$\mathrm{M}=31$
$\mathrm{t}_{16}=\mathrm{m} \Rightarrow \mathrm{a}+15 \mathrm{~d}=\mathrm{m}$
$\mathrm{S}_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]$
$\mathrm{S}_{\mathrm{n}}=\frac{31}{2}[2 \mathrm{a}+30 \mathrm{~d}]=\frac{31}{2} \times 2[\mathrm{a}+15 \mathrm{~d}]$
$=31(\mathrm{~m})=31 \mathrm{~m}$

Question 11.
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 ?
(1) 6
(2) 7
(3) 8
(4) 9
Answer:
(3) 8

Question $12 .$
If $\mathrm{A}=2^{65}$ and $\mathrm{B}=2^{64}+2^{63}+2^{62}++2^{0}$ which of the following is true?
(1) $\mathrm{B}$ is 264 more than $A$
(2) $A$ and $B$ are equal
(3) $\mathrm{B}$ is larger than $\mathrm{A}$ by 1
(4) $\mathrm{A}$ is larger than $\mathrm{B}$ by 1
Answer:
(4) $\mathrm{A}$ is larger than $\mathrm{B}$ by $\mathrm{l}$

Question $13 .$
The next term of the sequence $\frac{3}{16}, \frac{1}{8}, \frac{1}{12}, \frac{1}{18}, \ldots$
(1) $\frac{1}{24}$
(2) $\frac{1}{27}$
(3) $\frac{2}{3}$
(4) $\frac{1}{81}$
Answer:
(2) $\frac{1}{27}$

Question $14 .$
If the sequence $t_{1}, t_{2}, t_{3}, \ldots \ldots . \ldots .$ are in A.P. then the sequence $t_{6}, t_{12}, t_{18}, \ldots .$ is
(1) a Geometric progression
(2) an Arithmetic progression
(3) neither an Arithmetic progression nor a Geometric progression
(4) a constant sequence
Answer:
(2) an Arithmetic progression

Question $15 .$
The value of $\left(1^{3}+2^{3}+3^{3}+\ldots+15^{3}\right)-(1+2+3+\ldots+15)$ is
(1) 14400
(2) 14200
(3) 14280
(4) 14520
Answer:
(3) 14280