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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 15 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 15
Find the sum of $$2,4,6,8,,,........2n$$
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$$n(n+1)$$
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$$n^2$$
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$$n(n-1)$$
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$$n^2+n$$
Explanation
The given sequence is $$2,4,6,8,..... 2n$$
$$\therefore a= 2, d = 2$$ & $$a_k= 2n$$
$$\therefore 2n = a+(k-1)d $$
$$\Rightarrow 2n = 2+(k-1)2$$
$$\Rightarrow k = n$$
Sum of $$n$$ terms $$=\dfrac{k}{2}(2a+(k-1)d)$$
$$= \dfrac {n}{2} \times [2(2)+(n-1)(2)] $$
$$=n(n+1)$$
The $$27^{th}$$ term of AP $$7,9,11,13,15,17,19,\ldots$$ is
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$$59$$
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$$61$$
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$$63$$
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None of these
Explanation
Given series is $$7,9,11,13,15,17,19,\dots$$
$$a=7$$
$$d=9-7$$
$$=2$$
Formula for $$n^{th}$$ term of an AP whose first term is $$a$$ and common difference is $$d$$ is given by,
$$a_n=a+(n-1)d$$
So,
$$27^{th} $$ term is given as,
$$a_{27}=7 + (27-1)2$$
$$=7+26\times2$$
$$=7+52$$
$$=59$$
How many terms are there in the sequence 3, 6, 9, 12, ..., 111 ?
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$$32$$
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$$37$$
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$$42$$
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$$49$$
Explanation
The given sequence is an A.P. with first term $$a=3$$ and common difference $$d=3$$. Let there be n terms in the given sequence. Then,
$$nth \ term=111$$
$$a+(n-1)d=111$$
$$3+(n-1) \times 3=111$$
$$n=37$$
Thus, the given sequence contains 37 terms.
Which of the following pair of terms do you prefer if the sum of two consecutive terms of an $$AP$$ is given whose first term is $$a$$ and common difference is $$d$$?
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$$a$$ and $$a+d$$
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$$a$$ and $$a-d$$
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$$a-d$$ and $$a+d$$
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None of these
State true/false:
$$3,4,5,6,7,......$$ forms a progression. As they are formed by the rule $$n+1$$
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True
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False
It is given that the sum of four terms of an $$AP$$ is $$2$$ and their product is zero, then what will be the terms?
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$$-1, 0, 1, 2$$
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$$-2, 0, 2, 4$$
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$$-2, -1, 0, 2$$
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None of these
Find the four consecutive terms of an $$AP$$ whose sum is $$12$$ and sum of $$2nd$$ and $$4th$$ term is $$8$$.
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$$0, 2, 4, 6$$
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$$0, 3, 5, 7$$
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$$2, 4, 6, 7$$
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None of these
In an arithmetic progression, if $${ S }_{ n }=n(5+3n)\quad and\quad { t }_{ n }=32$$, then the value of n is
[Note: $${ S }_{ n }$$ and $${ t }_{ n }$$ denote the sum of first n terms and $${ n }^{ th }$$ term of arithmetic progression respectively.]
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4
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5
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6
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7
Find the sum of all odd integers between 2 and 100 divisible by 3.
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452
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754
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867
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765
Explanation
$$\textbf{Step-1: Finding the value of n}$$
$$\text{The}$$ $$n^{th}$$ $$\text{term in an A.P. is given by}$$
$$t_n=a+(n-1)d$$
$$\text{We know that the}$$ $$n^{th}$$ $$\text{term is 99}$$
$$99=3+(n-1)6$$
$$96=6(n-1)$$
$$n-1=16$$
$$n=17$$
$$\textbf{Step-2: Finding the required sum}$$
$$\text{Sum of n terms in an A.P. is}$$
$$S_n=\dfrac{n}{2}(a+t_n)$$
$$S_n=\dfrac{17}{2}(3+99)=867$$
$$\textbf{Hence,The correct option is (C)}$$
Choose the correct answer from the given four options in the following question:
In an A.P., if $$d = -4, n = 7, a_n = 4,$$ then $$a$$ is
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6
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7
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20
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28
Explanation
Hint :
$$n^{th }$$ term of an A.P. is $$a_n=a+(n-1)d$$ where $$a$$ is first term and $$d$$ is common difference.
Given:
common difference $$=d=-4$$
$$n=7$$
$$a_n=4$$ [$$n^{th}$$ term]
As we know, that
general term of an A.P. is given by
$$a_n=a+(n-1)d$$
putting the given values in the above formula
$$\Rightarrow 4=a+(7-1)(-4)$$
$$\Rightarrow 4=a+(6)(-4)$$
$$\Rightarrow 4=a-24\\ \Rightarrow 4+24=a$$
$$\Rightarrow 28=a$$
Hence, value of first term$$,a$$ is $$28$$
Choose the correct answer from the followings for the sequence $$-10, -6, -2, 2, ...$$
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an A.P. with $$d = -16$$
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an A.P. with $$d = 4$$
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an A.P. with $$d = -4$$
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not an A.P.
Explanation
Hint :
In an A.P. difference between two consecutive terms is always same
Step 1 :
find out the difference between two consecutive terms
for the given sequence $$-10,\,-6,\,-2,\,2,...$$
second term - first term $$=-6-(-10)=4$$
third term - second term $$=-2-(-6)=4$$
fourth term - third term $$=2-(-2)=4$$
Step 2 : Comparison of all the difference quantities and conclusion
Thus, difference between two consecutive terms is always $$4$$
$$\therefore \,$$common difference $$=d=4$$
Final step :
Hence the given sequence is an A.P. with $$d=4$$.
Choose the correct answer from the given four options in the following question:
The first four terms of an A.P., whose first term is $$-2$$ and the common difference is $$-2$$, are
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$$-2, 0, 2, 4$$
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$$-2, 4, -8, 16$$
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$$-2, -4, -6, -8$$
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$$-2, -4, -8, -16$$
Explanation
Hint:
use the formula of $$n^{th}$$ term of an A.P.
$$a_n=a+(n-1)d$$
Given:-
first term $$=a=a_1=-2$$
common difference $$=d=-2$$
Step 1 :
find the second term by replacing $$n$$ with $$2$$ in the formula of $$n^{th}$$ term of the A.P
$$a_n=a+(n-1)d$$
$$\Rightarrow a_2=(-2)+(2-1)(-2)$$
$$\Rightarrow a_2=-2+(1)(-2)$$
$$\Rightarrow a_2=-2-2$$
$$\therefore a_2=-4$$
Step 2 :
find the third term by replacing $$n$$ with $$3$$ in the formula of
$$n^{th}$$ term of the A.P
$$a_n=a+(n-1)d$$
$$\Rightarrow a_3=(-2)+(3-1)(-2)$$
$$\Rightarrow a_3=-2+(2)(-2)$$
$$\Rightarrow a_3=-2-4$$
$$\therefore a_3=-6$$
Step 3 : find the fourth term by replacing $$n$$ with $$4$$ in the formula of $$n^{th}$$ term of the A.P
$$a_n=a+(n-1)d$$
$$\Rightarrow a_4=(-2)+(4-1)(-2)$$
$$\Rightarrow a_4=-2+(3)(-2)$$
$$\Rightarrow a_4=-2-6$$
$$\therefore a_4=-8$$
Final step:
write down the first four terms of A.P.
first four terms of A.P. are $$a_1,\,a_2,\,a_3,\,a_4$$
i.e, $$-2,\,-4,\,-6,\,-8$$
Next three consecutive numbers in the pattern are
11, 8, 5, 2...
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$$0, – 3, – 6$$
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$$– 1, – 5, – 8$$
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$$– 2, – 5, – 8$$
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$$– 1, – 4, – 7$$
Explanation
Option (d) is correct; any two numbers have the common difference of 3,
$$\begin{array}{l}11-3=8,8-3=5, \ldots \\\text { Similarly, }(2-3)=-1,(-1-3)=-4,(-4-3)=-7\end{array}$$
Hence, three consecutive terms will be -1, -4, and -7.
What is the common difference of an A.P. in which $$a_{18}-a_{14} = 32$$?
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$$8$$
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$$-8$$
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$$-4$$
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$$4$$
Explanation
Hint:
The general term formula of an $$A.P$$ is $$\{a_n=a+(n-1)d\}$$ will be used along with the related terms.
Given:
Let $$ \text {a=first term (Let)}$$
$$ \text {d= common difference (Let)}$$
$$a_{18}-a_{14}=32$$
Step 1: Finding the general term expression of $$a_{18}$$
We know,
$$a_n=a+(n-1)d$$
For $$n=18$$
$$ a_{18}=a+(18-1)d=a+17d.....(i)$$
Step 2: Finding the general term expression of $$a_{14}$$
We know,
$$a_n=a+(n-1)d$$
For $$n=14$$
$$ a_{14}=a+(14-1)d=a+13d.....(ii)$$
Step 3: Finding the difference $$a_{18}-a_{14}$$ by subtracting $$eq (i)$$ from $$eq. (ii)$$
$$eq.(ii)-eq. (i)$$
$$a_{18}-a_{14}=a+17d-(a+13d)=a+17d-a-13d=4d.....(iii)$$
Step 4: Comparing and equating the value of $$a_{18}-a_{14}$$ from the question and $$eq(iii)$$
$$a_{18}-a_{14}=4d=32$$
$$\Rightarrow 4d=32 \Rightarrow d=8$$
Hence, the value of
'd'
is $$8$$
Final step:
The value of $$ \text {common difference, d=8}$$
Two APs have the same common difference and the 1st term of one AP is -1 and first term of other AP is -Then the difference between their 4th terms is
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-1
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-8
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7
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-9
Explanation
Hint:
use the formula of $$n^{th}$$ term of an A.P.
Given:-
first term of one A.P. $$=a=-1$$
first term of other A.P. $$=A=-8$$
same common difference of both the A.Ps $$=d$$
Step 1 :
find the $$4^{th}$$ term of first A.P.
$$\Rightarrow a_n=a+(n-1)d$$
$$\Rightarrow a_4=-1+(4-1)d$$
$$\therefore a_4=-1+3d$$
Step 2 : find the $$4^{th}$$ term of other A.P.
$$\Rightarrow A_n=A+(n-1)d$$
$$\Rightarrow A_4=-8+(4-1)d$$
$$\therefore A_4=-8+3d$$
Step 3 :
find the difference between $$4^{th}$$ terms of both A.P.s
$$|a_4-A_4|=|(-1+3d)-(-8+3d)|=|-1+3d+8-3d|=|-1+8|=|7|=7$$
Final step:
Hence, difference between their fourth terms is $$7.$$
In an A.P $$1^{st}$$ term is $$1$$ and the last term is $$20$$. The sum of all terms is $$=399$$ then $$n=$$....
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$$42$$
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$$38$$
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$$21$$
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$$19$$
Explanation
The is given that,
First term $$(a)=1$$
Last term $$(t_n)=20$$
Sum of terms $$(S_n)=399$$
We know that,
$$t_n=a+(n-1)d$$
$$S_n=\dfrac{n}{2}(2a+(n-1)d)\\$$
$$S_{n}=\dfrac{n}{2} (a+(a+(n-1)d))\\$$
$$\Rightarrow S_n=\dfrac{n}{2}(a+t_n)\\$$
$$\Rightarrow 399=\dfrac{n}{2}(1+20)\\$$
$$\Rightarrow 399=\dfrac{21n}{2}\\$$
$$\Rightarrow 21n=399 \times 2\\$$
$$\Rightarrow n=\dfrac{798}{21}\\$$
$$\Rightarrow n=38$$
If $$a$$ is constant then $$a + 2a + 3a + ..... + na$$ is
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$$\dfrac{an(n + 1)}{4}$$
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$$\dfrac{an(n + 1)}{2}$$
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$$\dfrac{n(n + 1)}{2}$$
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$$\dfrac{an(n + 1)}{6}$$
Explanation
For the terms $$a,\ 2a,\ 3a,\ 4a...,\ na$$
$${ a }_{ 2 }-{ a }_{ 1 }=2a-a$$
$$=a$$
$${ a }_{ 3 }-{ a }_{ 2 }=3a-2a$$
$$=a$$
Thus, the difference between consecutive terms of A.P. is the same.
Sum of first n terms of A.P. is given by
$${ S }_{ n }=\dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] $$
$$\therefore { S }_{ n }=\dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) a \right] $$
$$=\dfrac { n }{ 2 } \left[ 2a+na-a \right] $$
$$=\dfrac { n }{ 2 } \left[ a\left( n+2 \right) -a \right] $$
$$=\dfrac { n }{ 2 } \left[ a\left( n+2-1 \right) \right] $$
$$=\dfrac { n }{ 2 } \left[ a\left( n+1 \right) \right] $$
$$=\dfrac { an\left( n+1 \right) }{ 2 } $$
In an A.P, if $$a_{4} = 8 \,\& \,a = 2,$$ then its common difference is
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$$6$$
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$$4$$
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$$2$$
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$$10$$
Explanation
nth term of the A.P. is given by,
$${ a }_{ n }=a+\left( n-1 \right) d$$
$$\therefore { a }_{ 4 }=a+\left( 4-1 \right) d$$
$$\therefore 8=2+3d$$ (Given)
$$\therefore 3d=6$$
$$\therefore d=2$$
Thus, common difference is 2.
If sum of $$n$$ term of $$A.P.$$ is $$S_{n}$$ and $$S_{2n}=3S_{n}$$, then $$S_{3n}: S_{n}$$ will be:
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$$10$$
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$$11$$
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$$6$$
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$$4$$
Explanation
$$S_{2n}=3 S_{n}$$
$$\dfrac{2n}{2}[2a(2n-1)d]=\dfrac{3n}{2}[2a+(n-1)d]$$
$$\Rightarrow 4a+4nd-2d=6a+3nd-3d$$
$$\Rightarrow nd+d=2a$$
Now, $$S_{3n}:S_{n}$$
$$\dfrac{S_{3n}}{S_{n}}=\dfrac{\dfrac{3n}{2}[2a+(3n-1)d]}{\dfrac{n}{2}[2a+(n-1)d]}$$
$$\dfrac{S_{3n}}{S_{n}}=\dfrac{3[nd+d+3nd-d]}{[nd+d+nd-d]}$$
$$[\because 2a=nd+d]$$
$$=\dfrac{3\times 4nd}{2\ nd}$$
$$=\dfrac{12}{2}$$
$$=6$$
Hence, option $$(C)$$ is correct.
If sum of $$n$$ terms of $$A.P$$ is $$3n^2+5n$$, then its which term is $$164$$:
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$$12^{th}$$
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$$15^{th}$$
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$$27^{th}$$
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$$20^{th}$$
Explanation
Given: $$S_{n}=3n^{2}+5n$$
$$S_{1}=3(1)^{2}+5(1)=8$$
$$S_{2}=3(2)^{2}+5(2)=22$$
$$S_{3}=3(3)^{2}+5(3)=42$$
$$S_{4}=3(4)^{2}+5(4)=68$$
$$\therefore a_{1}=S_{1}=8$$
$$a_{2}=S_{2}-S_{1}\Rightarrow 22-8\Rightarrow 14$$
$$a_{3}=S_{2}-S_{1}\Rightarrow 42-22\Rightarrow 20$$
$$a_{4}=S_{2}-S_{1}\Rightarrow 68-42\Rightarrow 26$$
Thus $$A.P.$$ will be $$8,14,20,26,......164$$
$$a=8$$,
$$d=14-8=6$$ and $$a_{n}=164$$
$$\therefore 164=a+(n-1)d$$
$$164=8+(n-1)$$
$$(n-1)=156/6=26$$
$$\therefore n=26+1=27$$
Hence, option $$(C)$$ is correct.
A manufacture company made $$15000$$ mobile phones in its $$17^{th}$$ year. The production has been increasing by $$500$$ every year since the first year.
How many mobile phones did the company make in second year?
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$$7000$$
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$$7500$$
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$$8000$$
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$$8500$$
In the A.P. $$2, -2, -6, -10........$$ common difference (d) is:
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$$-4$$
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$$2$$
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$$-2$$
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$$4$$
If we divide $$18$$ into three parts that are in $$AP$$ and whose product is $$120,$$ then formed $$AP$$ is given by
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$$2, 7, 9$$
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$$2, 5, 11$$
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$$2, 6, 9$$
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$$2, 4, 10$$
Dishan started reading a novel which had $$240$$ pages. At the end of the first day, she was on page number $$45$$. She slowed after that. She read $$39$$ pages every day after the first day. How long did it take Dishan to complete that novel?
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$$5$$
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$$6$$
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$$7$$
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$$8$$
A roll of thread $$264$$ $$cm$$ long is cut into four parts to make four circles. The radii of the circles increases by $$1$$ $$cm$$ consecutively. Find the radius of smallest circle?
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$$9$$
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$$11$$
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$$12$$
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$$13$$
M.C.M college had enrollment of $$1625$$ students in the year of $$2014$$ which increases by $$100$$ students per year. What was the enrollment in the year $$2020?$$
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$$2125$$
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$$2225$$
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$$2235$$
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$$2215$$
Find an $$AP$$ whose general term is given by $$3n+1$$.
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$$4, 7, 10, 13, .....$$
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$$3, 6, 9, 12, .....$$
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$$3, 7, 12, 15, .....$$
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$$2, 5, 9, 13, .....$$
In a quiz competition, if the first prize is worth $$Rs. 200$$. Each other prize is $$Rs.15$$ less than its proceeding prize, then the worth of $$9th$$ prize is ____.
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$$60$$
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$$70$$
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$$80$$
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$$90$$
Which of the following $$AP$$ can be formed if the third and fifth terms are $$7$$ and $$11$$ respectively?
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$$3, 10,7, 13, 11, .......$$
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$$3, 5, 7, 9, 11, ......$$
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$$1, 2, 7, 9, 11, .......$$
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Nine of these
The arithmetic progression whose $$2nd$$ and $$5th$$ terms are $$13$$ and $$19$$ is
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$$11, 13, 15, 17, ......$$
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$$11, 13, 14, 17, ......$$
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$$11, 13, 14, 16, ......$$
0%
None of these
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