MCQ Questions for 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

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?

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

0%
$a_{n - 5} = a_{n - 4} + d$
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.

0%
True
0%
False

In an A.P

$a_{n + 5} = 50$ and

$a_{n + 1} = 38$ then common difference

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?

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

0%
$3n$
0%
$4n$
0%
$2$
0%
$3$

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 17 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 17 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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