The arithmetic progression whose common difference is $$5$$ and $$6th$$ term is $$30,$$ is

$$5, 10, 15, 20, 25, 30, ......$$

$$4, 10, 14, 19, 23, 30, .......$$

$$3, 8, 13, 19, 23, 30, ......$$

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:

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

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

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

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

$46$ is a term of an

A P

$AP$

2, 5, 8, 11, . . . . . 65

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

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

State true/false:
The first negative term of an

A P

$AP$

35, 30, 25, 20, . . . . .

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

9 t h

$9th$ term.

0%
True
0%
False

Which of the following arithmetic progression has

0

$0$ 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

A P

$AP$

5, 8, 11, 14, 17, . . . . . .

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

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

The first three terms of an

A P

$AP$ are

57, 53

$57, 53$ and

49

$49$ . 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

A P

$AP$

- 7, - 3, 1, . . . . . ., 49

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

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

Check whether

6

$6$ is a term of an

A P

$AP$

34, 30, 26, 22, 18, . . .

$34, 30, 26, 22, 18, ...$ 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, . . . .

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

0

$0$ ?

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

The arithmetic progression whose common difference is

5

$5$ and

6 t h

$6th$ term is

30,

$30,$ 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

$132$ a term of an

A P

$AP$

7, 10, 13, 16, 19, . . . . . .

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

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

For an

A P,

$AP,$ if the sum of four consecutive terms is

42,

$42,$ and common difference is

1.

$1.$ 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

A P

$AP$ is

9

$9$ and sum of their squares is

45.

$45.$ 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

A P

$AP$ is

12

$12$ and

0

$0$ 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}, . . . .

$a_1, a_2, a_3, ....$ is an A.P. such that

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$ then

a_{1} + a_{2} + a_{3} + . . . + a_{23} + a_{24}

$a_1+a_2+a_3+ ... +a_{23}+a_{24}$ 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} (2 a + (n - 1) d)

$S_{n}=\frac {n}{2}(2a+(n-1)d)$

Since

a_{1} + a_{5} + a_{10} + a_{15} + a_{20} + a_{24} = 225

$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$

\Rightarrow 6 a + 69 d = 225

$\Rightarrow 6a+69d=225$

\Rightarrow 2 a + 23 d = 75

$\Rightarrow 2a+23d=75$
Now

S_{24} = a_{1} + a_{2} + a_{3} + . . . + a_{23} + a_{24}

$S_{24} = a_1+a_2+a_3+ ... +a_{23}+a_{24}$

S_{24} = 12 (2 a + 23 d)

$S_{24} =12(2a+23d)$

= 900

$=900$

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} = n \sum r = 0 a_{r} x^{r}

$(1+x)^n=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r } }$ . Then

(1 + \frac{a_{1}}{a_{0}}) (+ \frac{a_{2}}{a_{1}}) . . . . (1 + \frac{a_{n}}{a_{n - 1}})

$(1+\frac {a_1}{a_0})(+\frac {a_2}{a_1})....(1+\frac {a_n}{a_{n-1}})$ 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}

${ S }_{ 1 },{ S }_{ 2 },{ S }_{ 3 },.......,{ S }_{ r }$ are the sums of first n terms of r arithmetic progression whose first terms are

1, 2, 3......

$1,2,3......$ and whose common differences are

1, 3, 5, . . . . . . .

$1,3,5,.......$ respectively, then the value of

S_{1} + S_{2} + S_{3} + . . . . . . . + S_{r}

${ S }_{ 1 }+{ S }_{ 2 }+{ S }_{ 3 }+.......+{ S }_{ r }$ is

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

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

$\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

Step 1: By using the relation between sum of n terms and first term find the common difference d.

$\textbf{Step 1: By using the relation between sum of n terms and first term find the common difference d.}$

Given:

$\text{Given:}$

S_{1} = - 5; S_{2} = 4

${S_1=-5; S_2=4}$

The first term of A.P. is equal to

$\text{The first term of A.P. is equal to }$

S_{1}

${S_1}$

Hence, a_{1} = - 5

$\text{Hence, }a_1 = -5$

Now, a_{2} = S_{2} - S_{1}

$\text{Now, }a_2 = S_2-S_1$

\Rightarrow a_{2} = 4 - (- 5)

$\Rightarrow a_2 = 4-(-5)$

\Rightarrow a_{2} = 4 + 5

$\Rightarrow a_2 = 4+5$

\Rightarrow a_{1} = 9

$\Rightarrow a_1 = 9$

Therefore the A.P. is -5, 9,...

$\text{Therefore the A.P. is -5, 9,...}$

d = a_{2} - a_{1}

$d = a_2-a_1$

\Rightarrow d = 9 - (- 5)

$\Rightarrow d = 9-(-5)$

\Rightarrow d = 9 + 5

$\Rightarrow d = 9+5$

\Rightarrow d = 14

$\Rightarrow d = 14$

Hence, the value of d is 14 .

$\textbf{Hence, the value of d is 14}.$

The average of certain first consecutive even number is

101

$101$ . Find their sum?

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

{log}_{5} 2, {log}_{6} 2, {log}_{12} 2

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

0%
A.P
0%
G.P
0%
H.P
0%
none of these

In an A.P.,

S_{p} = q

$S_p=q$ and

S_{q} = p

$S_q=p$ , where

S_{r}

$S_r$ denotes the sum of the first

r

$r$ terms of an A.P. Then, the value of

S_{p + q}

$S_{p+q}$ is

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

Explanation

Let the first term of the sequence be

a

$a$ and its common difference be

d

$d$ .
Applying sum to

n

$n$ terms formula of A.P.,

S_{p} = \frac{p}{2} (2 a + (p - 1) d) = q

$S_p = \dfrac{p}2(2a+(p-1)d) = q$
On expanding and simplifying the above equation, we get

2 a p + p^{2} d - p d = 2 q

$2ap + p^2d - pd = 2q$ ...(1)

Similarly, for

S_{q}

$S_q$

2 a q + q^{2} d - q d = 2 p

$2aq + q^2d - qd = 2p$ ...(2)

Subtracting (2) from (1), we get

2 a (p - q) + d (p^{2} - q^{2}) - d (p - q) = - 2 (p - q)

$2a (p - q) + d(p^2 - q^2) - d(p - q) = -2(p - q)$
Dividing both sides by

(p - q)

$(p - q)$ , we get

2 a + d (p + q) - d = - 2

$2a + d(p + q) - d = -2$

\Rightarrow 2 a + (p + q - 1) d = - 2

$\Rightarrow {2a + (p + q - 1)d} = -2$ ...(3)

Sum to

p + q

$p+q$ terms formula of an A.P. is

S_{(p + q)} = \frac{p + q}{2} [2 a + (p + q - 1) d]

$S_ {(p+q)} = \dfrac{p+q}{2}[2a + (p + q - 1)d]$ ...(4)

Substituting (3) in (4), we get

S_{p + q} = \frac{p + q}{2} \times (- 2) = - (p + q)

$S_{p+q} = \dfrac{p+q}{2}\times(-2) = -(p + q)$

Hence, option B is the correct answer.

If the sum of the three consecutive terms of A.P is

48

$48$ and product of the first the last is

252

$252$ , 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} = ?

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

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

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

t_{n} = 2 n - 4

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

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^{t h}

$n^{th}$ 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

{l o g}_{c} a, {l o g}_{b} c, {l o g}_{a} b

${ log }_{ c }a,{ log }_{ b }c,{ log }_{ a }b$ 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

A P - 5, \frac{- 9}{2}, - 4,

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

0

$0$ ?

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

Explanation

Given,

a_{1} = - 5

${ a }_{ 1 }=-5$

d = \frac{- 9}{2} - (- 5)

$d=\frac { -9 }{ 2 } -\left( -5 \right)$

= \frac{- 9}{2} + 5

$=\frac { -9 }{ 2 } +5$

= \frac{1}{2}

$=\frac { 1 }{ 2 }$

Since

S_{n} = 0

${ { S }_{ n }= } 0$

Thus, sum of n terms of A.P. are calculated as,

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

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

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

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

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

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

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

$\therefore n=20+1$

$\therefore n=21$

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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