In $$n^{th}$$ term of an AP is 7n-1 then

Sum of n terms $$\dfrac{n(7n+5)}{2}$$

MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 16

Check whether

$46$ is a term of an

$AP$

$2, 5, 8, 11, ..... 65$ or not.

0%
Yes
0%
No
0%
Can't say anything
0%
None of these

State true/false:
The first negative term of an

$AP$

$35, 30, 25, 20, .....$ is

$9th$ term.

0%
True
0%
False

Which of the following arithmetic progression has

$0$ as its term?

0%
$9, 7, 5, 3, ......$
0%
$15, 11, 7, ....$
0%
$12, 7, 2, .....$
0%
$6, 4, 2, .....$

Which of the following is the term of given

$AP$

$5, 8, 11, 14, 17, ......$ ?

0%
$19$
0%
$20$
0%
$21$
0%
$22$

The first three terms of an

$AP$ are

$57, 53$ and

$49$ . Which of the following term will be its first negative term?

0%
$14th$
0%
$15th$
0%
$16th$
0%
None of these

The middle most term of an

$AP$

$-7, -3, 1, ......, 49$ is

0%
$a_7 = 17$
0%
$a_8 = 17$
0%
$a_7 = 21$
0%
$a_8 = 21$

Check whether

$6$ is a term of an

$AP$

$34, 30, 26, 22, 18, ...$ or not?

0%
Yes
0%
No
0%
Can't say anything
0%
Nine of these

Which of the following term of the arithmetic progression

$36, 33, 30, 27, ....$ is

$0$ ?

0%
$12th$ term
0%
$13th$ term
0%
$14th$ term
0%
$15th$ term

The arithmetic progression whose common difference is

$5$ and

$6th$ term is

$30,$ is

0%
$5, 10, 15, 20, 25, 30, ......$
0%
$4, 10, 14, 19, 23, 30, .......$
0%
$3, 8, 13, 19, 23, 30, ......$
0%
None of these

$132$ a term of an

$AP$

$7, 10, 13, 16, 19, ......$ ?

0%
Yes
0%
No
0%
Can't say anything
0%
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

0%
$9$
0%
$10$
0%
$11$
0%
$12$

5,5,5,... forms a ______

0%
Series
0%
Sequence
0%
Progression
0%
None of the above

The sum of three terms in an

$AP$ is

$9$ and sum of their squares is

$45.$ Find the terms.

0%
$0, 3, 6$
0%
$0, 4, 5$
0%
$0, 2, 7$
0%
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?

0%
$2, 3, 7$
0%
$1, 5, 7$
0%
$0,4,8$
0%
None of these

Whole numbers from a sequence/progression, as they are formed by the fixed rule of adding 1 to previous whole number.

0%
True
0%
False

$a_1, a_2, a_3, ....$ 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-

0%
$909$
0%
$75$
0%
$750$
0%
$900$

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

0%
3000
0%
3003
0%
3033
0%
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:

0%
$\frac {(n+1)^{n+1}}{n!}$
0%
$\frac {(n+1)^{n}}{n!}$
0%
$\frac {(n)^{n-1}}{(n-1)!}$
0%
$\frac {(n+1)^{n+1}}{(n-1)!}$

${ S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 },.......,{ S }_{ r }$ 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

0%
$\frac { (nr-1)(nr+1) }{ 2 }$
0%
$\frac { (nr+1)nr }{ 2 }$
0%
$\frac { (nr-1)nr }{ 2 }$
0%
$\frac { n(nr+1) }{ 2 }$

$\text{In an A.P, if $S_{2}=4$ and $S_{1}=-5$ then d is equal to }$

0%
9
0%
14
0%
-9
0%
-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?

0%
$25,000$
0%
$33,600$
0%
$10100$
0%
$24,960$

${\log _5}2,\,\,{\log _6}\,2,\,\,{\log _{12}}\,\,2\,\,$ are in

0%
A.P
0%
G.P
0%
H.P
0%
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

0%
$0$
0%
$-(p+q)$
0%
$p+q$
0%
$pq$

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=........

0%
2
0%
3
0%
4
0%
16

Choose the correct alternative for questions:
For an A.P., if d=11, then

${ t }_{ 17 }-{ t }_{ 15 }=?$

0%
2
0%
22
0%
33
0%
13

What is the common difference of an A.P. whose nth term is

${ t }_{ n }=2n-4$

0%
3
0%
2
0%
-2
0%
-3

The sum of five consecutive odd numbers isThe smallest number is

0%
51
0%
53
0%
55
0%
57

$n^{th}$ term of an AP is 7n-1 then

0%
First term=6
0%
Common difference=7
0%
Sum of n terms $\dfrac{n(7n+5)}{2}$
0%
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

0%
$\cfrac { 1 }{ 2 }$
0%
$\cfrac { 3 }{ 2 }$
0%
$\cfrac { 5 }{ 2 }$
0%
$\cfrac { 7 }{ 2 }$

How many terms of the

$AP -5, \frac{-9}{2}, -4,$ .... will give the sum

$0$ ?

0%
$21$
0%
$18$
0%
$23$
0%
$16$

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,

${ { S }_{ n }= } \frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right]$

$\therefore 0=\frac { n }{ 2 } \left[ \left( 2\times (-5) \right) +\left( n-1 \right) \frac { 1 }{ 2 } \right]$

$\therefore \left( 2\times (-5) \right) +\left( n-1 \right) \frac { 1 }{ 2 } =0$

$-10+\frac { \left( n-1 \right) }{ 2 } =0$

$\frac { \left( n-1 \right) }{ 2 } =10$

$\left( n-1 \right) =20$

$\therefore n=20+1$

$\therefore n=21$

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 16 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 16 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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