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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 17 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 17
The number of common terms in two AP's 2, 5, 8, 11,..., 179 and 3, 5, 7, 9, .. 101 are
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0%
15
0%
16
0%
17
0%
18
The sum of n terms of an AP is $$(3n^2 + 5n).$$ Which of its terms is 164?
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0%
28th
0%
27th
0%
26th
0%
29th
Explanation
Given sum of n terms is,
$${ S }_{ n }=3{ n }^{ 2 }+5n$$
Put n=1, we get,
$${ S }_{ 1 }=a_{ 1 }=3{ \left( 1 \right) }^{ 2 }+5\left( 1 \right) $$
$$\therefore a_{ 1 }=3+5$$
$$\therefore a_{ 1 }=8$$
Put n=2 in above equation, we get,
$${ S }_{ 2 }={ a }_{ 1 }+{ a }_{ 2 }=3{ \left( 2 \right) }^{ 2 }+5\left( 2 \right) $$
$$\therefore 8+{ a }_{ 2 }=\left( 3\times 4 \right) +10$$
$$\therefore 8+{ a }_{ 2 }=22$$
$$\therefore { a }_{ 2 }=14$$
$$d=14-8$$
$$\therefore d=6$$
Now, Given $${ a }_{ n }=164$$
$${ a }_{ n }={ a }_{ 1 }+\left( n-1 \right) d$$
$$\therefore 164=8+\left( n-1 \right) 6$$
$$\therefore 156=6\left( n-1 \right) $$
$$\therefore n-1=26$$
$$\therefore n=27$$
In an AP, the correct relation is
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$$a_{n - 5} = a_{n - 4} + d$$
0%
$$a_{n - 5} = a_{n - 6} + d$$
0%
$$a_{n - 5} = a_{n} + d$$
0%
$$a_{n + 5} = a_{n - 5} - d$$
Explanation
We know that, in an A.P., the common difference $$(d)$$ is the difference between any two consecutive terms.
If we take one of the terms as $$a_{n-6}$$, we will need either $$a_{n-7}$$ or $$a_{n-5}$$ for finding the common difference.
If we take $$a_{n-6}$$ and $$a_{n-5}$$, then,
$$d={ a }_{ n-5 }-{ a }_{ n-6 }$$
$$\therefore \ { a }_{ n-5 }={ a }_{ n-6 }+d$$
Hence, option B is correct.
Set of even numbers form a progression.
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0%
True
0%
False
In an A.P $$a_{n + 5} = 50$$ and $$a_{n + 1} = 38$$ then common difference
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0%
$$3$$
0%
$$2$$
0%
$$3n$$
0%
$$2n$$
Explanation
nth term of A.P. is given by,
$${ a }_{ n }=a+\left( n-1 \right) d$$ (1)
1) Put $$n=n+5$$ in (1), we get,
$${ a }_{ n+5 }=a+\left( n+5-1 \right) d$$
$$\therefore 50=a+\left( n+4 \right) d$$
$$\therefore \dfrac { 50-a }{ d } =n+4$$
$$\therefore n=\dfrac { 50-a }{ d } -4$$ (2)
2) Put $$n=n+1$$ in (1), we get,
$${ a }_{ n+1 }=a+\left( n+1-1 \right) d$$
$$\therefore 38=a+nd$$
$$\therefore n=\dfrac { 38-a }{ d } $$ (3)
From (2) and (3), we get,
$$\dfrac { 50-a }{ d } -4=\dfrac { 38-a }{ d }$$
$$\dfrac { 50-a }{ d } -\dfrac { 38-a }{ d } =4$$
$$\therefore \dfrac { 50-a-\left( 38-a \right) }{ d } =4$$
$$\therefore \dfrac { 50-a-38+a }{ d } =4$$
$$\therefore \dfrac { 50-38 }{ d } =4$$
$$\therefore \dfrac { 12 }{ d } =4$$
$$\therefore d=3$$
How many terms of the AP 6, 12, 18, 24,.... must be take to make the sum 816?
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0%
16
0%
18
0%
14
0%
22
Explanation
Given,
$${ a }_{ 1 }=6$$
$$d=12-6=6$$
Since $$ S_n =816$$
Sum of n terms of A.P. is given as,
$$ S_n =\dfrac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] $$
$$\therefore 816=\dfrac { n }{ 2 } \left[ \left( 2\times 6 \right) +\left( n-1 \right) 6 \right] $$
$$\therefore 1632=n\left[ 12+6n-6 \right] $$
$$\therefore 1632=n\left[ 6n+6 \right] $$
$$\therefore 1632=6{ n }^{ 2 }+6n$$
$$\therefore 6{ n }^{ 2 }+6n-1632=0$$
Divide both sides by 6, we get,
$${ n }^{ 2 }+n-272=0$$
$$\therefore n=\dfrac { -1\pm \sqrt { { \left( 1 \right) }^{ 2 }-4\times 1\times \left( -272 \right) } }{ 2\times 1 } $$
$$\therefore n=\dfrac { -1\pm \sqrt { 1+1088 } }{ 2 } $$
$$\therefore n=\dfrac { -1\pm \sqrt { 1089 } }{ 2 } $$
$$\therefore n=\dfrac { -1\pm 33 }{ 2 } $$
Consider only positive root.
$$\therefore n=\dfrac { -1+33 }{ 2 } $$
$$\therefore n=\dfrac { 32 }{ 2 } $$
$$\therefore n=16$$
In an arithmetic progression $$a_{n + 5} = 35 \,\&\, a_{n + 1} = 23$$ then common difference is
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0%
$$3n$$
0%
$$4n$$
0%
$$2$$
0%
$$3$$
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