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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 16 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 16
Check whether $$46$$ is a term of an $$AP$$ $$2, 5, 8, 11, ..... 65$$ or not.
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Yes
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No
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Can't say anything
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None of these
State true/false:
The first negative term of an $$AP$$ $$35, 30, 25, 20, ..... $$ is $$9th$$ term.
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True
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False
Which of the following arithmetic progression has $$0$$ as its term?
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$$9, 7, 5, 3, ......$$
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$$15, 11, 7, ....$$
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$$12, 7, 2, .....$$
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$$6, 4, 2, .....$$
Which of the following is the term of given $$AP$$
$$5, 8, 11, 14, 17, ......$$?
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$$19$$
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$$20$$
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$$21$$
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$$22$$
The first three terms of an $$AP$$ are $$57, 53$$ and $$49$$. Which of the following term will be its first negative term?
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$$14th$$
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$$15th$$
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$$16th$$
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None of these
The middle most term of an $$AP$$ $$-7, -3, 1, ......, 49$$ is
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$$a_7 = 17$$
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$$a_8 = 17$$
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$$a_7 = 21$$
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$$a_8 = 21$$
Check whether $$6$$ is a term of an $$AP$$ $$34, 30, 26, 22, 18, ...$$ or not?
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Yes
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No
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Can't say anything
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Nine of these
Which of the following term of the arithmetic progression $$36, 33, 30, 27, ....$$ is $$0$$?
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$$12th$$ term
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$$13th$$ term
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$$14th$$ term
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$$15th$$ term
The arithmetic progression whose common difference is $$5$$ and $$6th$$ term is $$30,$$ is
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$$5, 10, 15, 20, 25, 30, ......$$
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$$4, 10, 14, 19, 23, 30, .......$$
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$$3, 8, 13, 19, 23, 30, ......$$
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None of these
Is $$132$$ a term of an $$AP$$ $$7, 10, 13, 16, 19, ......$$?
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Yes
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No
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Can't say anything
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None of these
For an $$AP,$$ if the sum of four consecutive terms is $$42,$$ and common difference is $$1.$$ Then its first term is
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$$9$$
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$$10$$
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$$11$$
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$$12$$
5,5,5,... forms a ______
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Series
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Sequence
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Progression
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None of the above
The sum of three terms in an $$AP$$ is $$9$$ and sum of their squares is $$45.$$ Find the terms.
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$$0, 3, 6$$
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$$0, 4, 5$$
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$$0, 2, 7$$
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None of these
If the sum and product of 3 consecutive terms of an $$AP$$ is $$12$$ and $$0$$ respectively, then what will be the terms?
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$$2, 3, 7$$
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$$1, 5, 7$$
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$$0,4,8$$
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None of these
Whole numbers from a sequence/progression, as they are formed by the fixed rule of adding 1 to previous whole number.
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True
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False
If $$a_1, a_2, a_3, ....$$ is an A.P. such that $$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$$ then $$a_1+a_2+a_3+ ... +a_{23}+a_{24}$$ is equal to-
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$$909$$
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$$75$$
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$$750$$
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$$900$$
Explanation
Here we use the following formula to calculate the sum of first n terms of an AP:
$$S_{n}=\frac {n}{2}(2a+(n-1)d)$$
Since $$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$$
$$\Rightarrow 6a+69d=225$$
$$\Rightarrow 2a+23d=75$$
Now $$ S_{24} = a_1+a_2+a_3+ ... +a_{23}+a_{24}$$
$$S_{24} =12(2a+23d)$$
$$=900$$
The sum of digits of all numbers from 1 to 300 is equal to
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3000
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3003
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3033
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none of these
Let $$(1+x)^n=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r } } $$. Then $$(1+\frac {a_1}{a_0})(+\frac {a_2}{a_1})....(1+\frac {a_n}{a_{n-1}})$$ is equal to:
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$$\frac {(n+1)^{n+1}}{n!}$$
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$$\frac {(n+1)^{n}}{n!}$$
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$$\frac {(n)^{n-1}}{(n-1)!}$$
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$$\frac {(n+1)^{n+1}}{(n-1)!}$$
If $${ S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 },.......,{ S }_{ r }$$ are the sums of first n terms of r arithmetic progression whose first terms are $$1,2,3......$$ and whose common differences are $$1,3,5,.......$$ respectively, then the value of $${ S }_{ 1 }+{ S }_{ 2 }+{ S }_{ 3 }+.......+{ S }_{ r }$$ is
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$$\frac { (nr-1)(nr+1) }{ 2 } $$
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$$\frac { (nr+1)nr }{ 2 } $$
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$$\frac { (nr-1)nr }{ 2 } $$
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$$\frac { n(nr+1) }{ 2 } $$
$$\text{In an A.P, if $S_{2}=4$ and $S_{1}=-5$ then d is equal to }$$
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9
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14
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-9
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-4
Explanation
$$\textbf{Step 1: By using the relation between sum of n terms and first term find the common difference d.}$$
$$\text{Given:}$$ $${S_1=-5; S_2=4}$$
$$\text{The first term of A.P. is equal to }$$ $${S_1}$$
$$\text{Hence, }a_1 = -5$$
$$\text{Now, }a_2 = S_2-S_1$$
$$\Rightarrow a_2 = 4-(-5)$$
$$\Rightarrow a_2 = 4+5$$
$$\Rightarrow a_1 = 9$$
$$\text{Therefore the A.P. is -5, 9,...}$$
$$d = a_2-a_1$$
$$\Rightarrow d = 9-(-5)$$
$$\Rightarrow d = 9+5$$
$$\Rightarrow d = 14$$
$$\textbf{Hence, the value of d is 14}.$$
The average of certain first consecutive even number is $$101$$. Find their sum?
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$$25,000$$
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$$33,600$$
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$$10100$$
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$$24,960$$
$${\log _5}2,\,\,{\log _6}\,2,\,\,{\log _{12}}\,\,2\,\,$$ are in
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A.P
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G.P
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H.P
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none of these
In an A.P., $$S_p=q$$ and $$S_q=p$$, where $$ S_r $$ denotes the sum of the first $$r$$ terms of an A.P. Then, the value of $$S_{p+q}$$ is
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$$0$$
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$$-(p+q)$$
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$$p+q$$
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$$pq$$
Explanation
Let the first term of the sequence be $$a$$ and its common difference be $$d$$.
Applying sum to $$n$$ terms formula of A.P.,
$$S_p = \dfrac{p}2(2a+(p-1)d) = q$$
On expanding and simplifying the above equation, we get
$$2ap + p^2d - pd = 2q$$ ...(1)
Similarly, for $$S_q$$
$$2aq + q^2d - qd = 2p$$ ...(2)
Subtracting (2) from (1), we get
$$2a (p - q) + d(p^2 - q^2) - d(p - q) = -2(p - q)$$
Dividing both sides by $$(p - q)$$, we get
$$2a + d(p + q) - d = -2$$
$$\Rightarrow {2a + (p + q - 1)d} = -2$$ ...(3)
Sum to $$p+q$$ terms formula of an A.P. is
$$S_ {(p+q)} = \dfrac{p+q}{2}[2a + (p + q - 1)d] $$ ...(4)
Substituting (3) in (4), we get
$$S_{p+q} = \dfrac{p+q}{2}\times(-2) = -(p + q)$$
Hence, option B is the correct answer.
If the sum of the three consecutive terms of A.P is $$48$$ and product of the first the last is $$252$$, and d=........
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2
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3
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4
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16
Choose the correct alternative for questions:
For an A.P., if d=11, then $${ t }_{ 17 }-{ t }_{ 15 }=?$$
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2
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22
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33
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13
What is the common difference of an A.P. whose nth term is $${ t }_{ n }=2n-4$$
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3
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2
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-2
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-3
The sum of five consecutive odd numbers isThe smallest number is
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51
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53
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55
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57
In $$n^{th}$$ term of an AP is 7n-1 then
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First term=6
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Common difference=7
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Sum of n terms $$\dfrac{n(7n+5)}{2}$$
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Second term=15
Let a, b, c be three distinct positive numbers which are in G.P, if $${ log }_{ c }a,{ log }_{ b }c,{ log }_{ a }b$$ are in A.P., then the common difference of the A.P is
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$$\cfrac { 1 }{ 2 } $$
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$$\cfrac { 3 }{ 2 } $$
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$$\cfrac { 5 }{ 2 } $$
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$$\cfrac { 7 }{ 2 } $$
How many terms of the $$ AP -5, \frac{-9}{2}, -4, $$ .... will give the sum $$0$$?
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$$21$$
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$$18$$
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$$23$$
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$$16$$
Explanation
Given,
$${ a }_{ 1 }=-5$$
$$d=\frac { -9 }{ 2 } -\left( -5 \right) $$
$$=\frac { -9 }{ 2 } +5$$
$$=\frac { 1 }{ 2 } $$
Since $$ { { S }_{ n }= } 0$$
Thus, sum of n terms of A.P. are calculated as,
$$ { { S }_{ n }= } \frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] $$
$$\therefore 0=\frac { n }{ 2 } \left[ \left( 2\times (-5) \right) +\left( n-1 \right) \frac { 1 }{ 2 } \right] $$
$$\therefore \left( 2\times (-5) \right) +\left( n-1 \right) \frac { 1 }{ 2 } =0$$
$$-10+\frac { \left( n-1 \right) }{ 2 } =0$$
$$\frac { \left( n-1 \right) }{ 2 } =10$$
$$\left( n-1 \right) =20$$
$$\therefore n=20+1$$
$$\therefore n=21$$
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