MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 7

m

$m$ th term of an A.P. is

n

$n$ and

n

$n$ th term is

m

$m$ , then the

(m + n)

$(m + n)$ th term is

0%
$0$
0%
$m+n-1$
0%
$m+n$
0%
$\dfrac{mn}{m+n}$

Explanation

Using the formula for mth term, we get

a_{m} = a_{1} + (m - 1) d

$a_{m}=a_{1}+(m-1)d$

Hence

a_{1} + (m - 1) d = n

$a_{1}+(m-1)d=n$ ...(i)

Similarly

a_{1} + (n - 1) d = m

$a_{1}+(n-1)d=m$ ...(ii)

Subtracting (ii) from (i)

d (m - 1 - n + 1) = n - m

$d(m-1-n+1)=n-m$

d (m - n) = n - m

$d(m-n)=n-m$

d = - 1

$d=-1$ .

a_{1} + (m - 1) (- 1) = n

$a_{1}+(m-1)(-1)=n$

a_{1} - m + 1 = n

$a_{1}-m+1=n$

a_{1} = m + n - 1

$a_{1}=m+n-1$

Hence

(m + n)^{t h}

$(m+n)^{th}$ term is

a_{m + n} = a_{1} + (m + n - 1) d

$a_{m+n}=a_{1}+(m+n-1)d$

= (m + n - 1) - (m + n - 1)

$=(m+n-1)-(m+n-1)$

= 0

$=0$ .

If the sum to n terms of an A.P. is

3 n^{2} + 5 n

$3n^2+5n$ while

T_{m} = 164

$T_m=164$ , then value of m is

0%
$25$
0%
$26$
0%
$27$
0%
$28$

Explanation

S_{n} = 3 n^{2} + 5 n

$S_{n}=3n^{2}+5n$
Now

S_{1} = 3 + 5 = 8

$S_{1}=3+5=8$
Hence initial term is 8.
Now the second term is

T_{2} = S_{2} - S_{1}

$T_{2}=S_{2}-S_{1}$

= (3 (2)^{2} + 5 (2)) - 8

$=(3(2)^{2}+5(2))-8$

= 12 + 10 - 8

$=12+10-8$

= 14

$=14$ .
Hence common difference is

d = T_{2} - T_{1}

$d=T_{2}-T_{1}$

= 14 - 8 = 6

$=14-8=6$
Now

T_{n} = a + (n - 1) d

$T_{n}=a+(n-1)d$

= 8 + (n - 1) 6

$=8+(n-1)6$

= 6 n + 2

$=6n+2$ .
It is given that

T_{m} = 164

$T_{m}=164$
Or

164 = 6 m + 2

$164=6m+2$

162 = 6 m

$162=6m$

27 = m

$27=m$
Hence

m = 27

$m=27$ .

If the sum of all the terms of an A.P.:

25, 22, 19, . . . . . .,

$25, 22, 19, ......,$

t_{n}

$t_n$ is

116,

$116,$ then

n

$n$ is

0%
$8$
0%
$4$
0%
$2$
0%
$16$

Explanation

Let number of terms be

n

$n$

∴ \frac{n}{2} [50 + (n - 1) \times (- 3)] = 116

$\therefore \displaystyle \frac{n}{2} [50 + (n - 1) \times (-3)] = 116$

⟹ \frac{1}{2} [50 n - 3 n^{2} + 3 n] = 116

$\implies \displaystyle \frac 12[50n-3n^2+3n]=116$

⟹ 3 n^{2} - 53 n + 232 = 0 ⟹ n = \frac{29}{3} o r 8

$\implies 3n^2 - 53n + 232 = 0 \implies n = \dfrac{29}{3} \ or \ 8$

Hence,

n = 8

$n = 8$

m^{t h}

$m^{th}$ term of an A.P. is

n

$n$ and

n

$n$ th term is

m

$m$ , then the

r^{t h}

$r^{th}$ term is

0%
$m + n - r$
0%
$m + n + r$
0%
$m - n + r$
0%
$r - (m + n)$

Explanation

The

r^{t h}

$r^{th}$ term of an A.P with the first term as

^{'} a^{'}

$'a'$ and common difference

^{'} d^{'}

$'d'$ , is

a_{r} = a + (r - 1) d

$a_{r}=a+(r-1)d$ .

Hence

a + (m - 1) d = n

$a+(m-1)d=n$ ...........

(i)

$(i)$

a + (n - 1) d = m

$a+(n-1)d=m$ ..........

(i i)

$(ii)$

Subtracting

(i i)

$(ii)$ from

(i)

$(i)$ gives us

d (m - 1 - n + 1) = n - m

$d(m-1-n+1)=n-m$

d (m - n) = n - m

$d(m-n)=n-m$

d = - 1

$d=-1$

Hence

a = n - (m - 1) d

$a=n-(m-1)d$

a = n - (m - 1) (- 1)

$a=n-(m-1)(-1)$

a = n + m - 1

$a=n+m-1$ .

Hence

a_{r} = a + (r - 1) d

$a_{r}=a+(r-1)d$

= n + m - 1 + (r - 1) (- 1)

$=n+m-1+(r-1)(-1)$

= n + m - 1 - r + 1

$=n+m-1-r+1$

= n + m - r

$=n+m-r$

Hence

a_{r} = n + m - r

$a_{r}=n+m-r$ .

In an A.P., if the common difference is

2

$2$ and sum up to

n

$n$ terms is

49

$49$ and

7^{t h}

$7^{th}$ term is

13

$13$ , then value of

n

$n$ is equal to?

0%
$0$
0%
$5$
0%
$7$
0%
$13$

Explanation

Given:

a_{7} = a_{1} + 6 d

$a_{7}=a_{1}+6d$

It is given that the common difference is 2.

Therefore

a_{7} = a_{1} + 12

$a_{7}=a_{1}+12$

13 = a_{1} + 12

$13=a_{1}+12$

a_{1} = 1

$a_{1}=1$ .

Now sum of n terms is given by:

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

$S_{n}=\dfrac{n}{2}[2a_{1}+(n-1)d]$

49 = \frac{n}{2} [2 + (n - 1) 2]

$49=\dfrac{n}{2}[2+(n-1)2]$

49 = n (1 + (n - 1))

$49=n(1+(n-1))$

49 = n^{2}

$49=n^{2}$

n = 7

$n=7$

Hence the value of n is 7.

In an A.P., if

a = 3.5

$a = 3.5$ ,

d = 0

$d = 0$ ,

n = 101

$n = 101$ , then

a_{n}

$a_n$ will be

0%
$0$
0%
$3.5$
0%
$103.5$
0%
$104.5$

Explanation

a = 3.5, d = 0

$a = 3.5, d=0$ and

n = 101

$n= 101$

a_{n} = a + (n - 1) \times d

$a_n = a + (n-1)\times d$

a_{n} = 3.5 + 100 \times 0

$a_n = 3.5 + 100\times0$

a_{n} = 3.5

$a_n= 3.5$

If the first term of an AP is

5

$5$ and the common difference is

- 2,

$-2,$ then the sum of the first

6

$6$ terms is

0%
$0$
0%
$5$
0%
$6$
0%
$15$

Explanation

Sum of an AP is:

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

${\dfrac{n}{2}}\times[{2a+(n-1)d}]$

a = 5, d = - 2, n = 6

$a = 5, d= -2, n=6$

Sum

= \frac{6}{2} \times [(2 \times 5) + (6 - 1) \times (- 2)]

$={\dfrac{6}{2}}\times[{(2\times5)+(6-1)\times(-2)}]$

= 3 \times (10 - 10)

$= 3\times(10-10)$

= 0

$= 0$

The

11^{t h}

$11^{th}$ term of the series

- 5, \frac{- 5}{2}, 0, \frac{5}{2}, \dots

$\displaystyle -5,\frac{-5}{2},0,\frac{5}{2},\cdots$ is

0%
$20$
0%
$30$
0%
$40$
0%
$50$

Explanation

a = - 5, d = \frac{5}{2}

$a = -5, d=\dfrac{5}{2}$ and

n = 11

$n= 11$ .

a_{11} = - 5 + (11 - 1) \times d

$a_{11} = -5 + (11-1)\times d$

a_{11} = - 5 + 10 \times \frac{5}{2}

$a_{11} = -5 + 10\times\dfrac{5}{2}$

a_{11} = 20

$a_{11}= 20$

In an AP if

a = 1

$a = 1$ ,

a_{n} = 20

${ a }_{ n }=20$ and

S_{n} = 399

${ S }_{ n }=399$ , then

n

$n$ is

0%
$19$
0%
$21$
0%
$38$
0%
$42$

Explanation

a = 1

$a = 1$

a_{n} = 20

$a_n = 20$

S_{n} = 399

$S_n= 399$
Then,

\Rightarrow a_{n} = a + (n - 1) d

$\Rightarrow a_n=a+(n-1)d$

\Rightarrow 20 = 1 + (n - 1) d

$\Rightarrow 20=1 + (n-1)d$

\Rightarrow 19 = (n - 1) d

$\Rightarrow 19=(n-1)d$
Also given,

S_{n} = \frac{n}{2} \times

$S_n = \dfrac{n}{2} \times$

[2 a + (n - 1) d]

$[2a+(n-1)d]$
Substituting values,

399 \times 2 = n (2 + 19)

$399\times 2=n( 2+19)$

\Rightarrow 798

$\Rightarrow 798$ =

n \times 21

${n\times21}$

n = 38

$n = 38$

If the common difference of an AP is

5,

$5,$ then what is

a_{18} - a_{13}

$a_{18}-a_{13}$ ?

0%
$5$
0%
$20$
0%
$25$
0%
$30$

Explanation

Hint:

n^{t h}

$n^{th }$ term of an A.P. is

a_{n} = a + (n - 1) d

$a_n=a+(n-1)d$ where

a

$a$ is first term and

d

$d$ is common difference.

Given:
value of common difference

, d = 5

$,d=5$

Step 1 : Find the value of $18^{th}$ term i.e, $a_{18}$ by replacing $n$ with $18$ in the $n^{th}$ term of the A.P

\Rightarrow a_{n} = a + (n - 1) d

$\Rightarrow a_n=a+(n-1)d$

\Rightarrow a_{18} = a + (18 - 1) d

$\Rightarrow a_{18}=a+(18-1)d$

∴ a_{18} = a + 17 d

$\therefore a_{18}=a+17d$

Step 2 : Find the value of $13^{th}$ term i.e, $a_{13}$ by replacing $n$ with $13$ in the $n^{th}$ term of the A.P

\Rightarrow a_{n} = a + (n - 1) d

$\Rightarrow a_n=a+(n-1)d$

\Rightarrow a_{13} = a + (13 - 1) d

$\Rightarrow a_{13}=a+(13-1)d$

∴ a_{13} = a + 12 d

$\therefore a_{13}=a+12d$

Step 3 : Find the difference between $a_{18}$ and $a_{13}$

a_{18} - a_{13}

$a_{18}-a_{13}$

= [a + 17 d] - [a + 12 d]

$=[a+17d]-[a+12d]$

= a + 17 d - a - 12 d

$=a+17d-a-12d$

= 5 d = 5 \times 5 = 25

$=5d\\=5\times5\\=25$

Final step : Hence,

a_{18} - a_{13} = 25

$a_{18}-a_{13}=25$

Which term of the AP,

21, 42, 63, 84, . . .

$21, 42, 63, 84,...$ is

210 ?

$210?$

0%
$9^{th}$
0%
$10^{th}$
0%
$11^{th}$
0%
$12^{th}$

Explanation

First term of the AP,

a = 21

$a=21$
Common difference

d = 21

$d=21$

n^{t h}

${n}^{th}$ term is

210

$210$
Then,

210 = a + (n - 1) d

$210 = a + (n-1)d$

210 = 21 + (n - 1) \times 21

$210= 21 + (n-1)\times21$

210 = 21 n

$210= 21n$
Therefore,

n = 10

$n=10$
Hence, 210 is

10^{t h}

${ 10 }^{ th }$ term of the series.

If the

2^{n d}

${2}^{nd}$ term of an AP is

13

$13$ and the

5^{t h}

${5}^{th}$ term is

25,

$25,$ what is its

7^{t h}

${7}^{th}$ term?

0%
$30$
0%
$33$
0%
$37$
0%
$38$

Explanation

Let the first term of the AP be

a

$a$ , the common difference be

d

$d$ .
Then,

2^{n d}

${2}^{nd}$ term is

a + d = 13....... (1)

$a+d=13.......(1)$

5^{t h}

${5}^{th}$ term is

a + 4 d = 25........ (2)

$a+4d=25........(2)$

Solving for

d

$d$ ,

E q (2) - E q (1)

$Eq(2)-Eq(1)$

(a + 4 d) - (a + d) = 25 - 13

$(a+4d)-(a+d)=25-13$

\Rightarrow 3 d = 12

$\Rightarrow 3d=12$

\Rightarrow d = 4

$\Rightarrow d=4$

Hence, From

(1)

$(1)$

a = 13 - d = 13 - 4 = 9

$a=13-d=13-4=9$

Now,

7^{t h}

${7}^{th}$ term is

a + 6 d

$a+6d$

Substituting values of

a

$a$ and

d

$d$

7^{t h} t e r m = 9 + 6 \times 4 = 33

$7^{th} term=9+6\times4=33$

The

4^{t h}

$4^{th}$ term from the end of the AP:

11, 8, 5, . . ., - 49

$11, 8, 5, ..., -49$ is

0%
$-37$
0%
$-40$
0%
$-43$
0%
$-58$

Explanation

In the given A.P.

a = 11, d = - 3

$a=11,d=-3$ , last term

(l) = - 49

$(l)=-49$ ,
then

l = a + (n - 1) d

$l=a+(n-1)d$

\Rightarrow - 49 = 11 + (n - 1) (- 3)

$\Rightarrow -49=11+(n-1)(-3)$

\Rightarrow 3 n = 63 \Rightarrow n = 21

$\Rightarrow 3n=63\Rightarrow n=21$
Hence, there are

21

$21$ terms in AP.
The

4^{t h}

$4^{th}$ term from the last will be 18th term from beginning.
So,

a_{18} = a + (18 - 1) d

$a_{18}=a+(18-1)d$

\Rightarrow 11 + 17 (- 3)

$\Rightarrow 11+17(-3)$

\Rightarrow 11 - 51 = - 40

$\Rightarrow 11-51=-40$
SO

4^{t h}

$4^{th}$ term from the end of the AP:

11, 8, 5, . . ., - 49

$11, 8, 5, ..., -49$ is

- 40

$-40$ .

7

$7$ times the

7^{t h}

$7^{th}$ term of an AP is equal to

11

$11$ times its

11^{t h}

$11^{th}$ term, then its

18^{t h}

$18^{th}$ term will be

0%
$7$
0%
$11$
0%
$18$
0%
$0$

Explanation

${\textbf{Step - 1: Find 7th term of the AP}}$

${a_7} = a + (n - 1)d$

$= a + (7 - 1)d$

$= a + 6d$

${\text{The 11th term of the AP,}}$

${a_{11}} = a + (n - 1)d$

$= a + (11 - 1)d$

$= a + 10d$

${\textbf{Step - 2: According to the question 7 times of 7th term is equal to 11 times of 11th term}}$

$7(a + 6d) = 11(a + 10d)$

$\Rightarrow 7a + 42d = 11a + 110d$

$\Rightarrow 4a + 68d = 0$

${\textbf{Step - 3: Find value for a}}$

${\text{a = }}\dfrac{{ - 68}}{4}d$

${\textbf{Step - 4: Find value of 18th term }}$

$a_{18} = a + (18 - 1)d$

$= a + 17d$

${\textbf{Step - 5: Substitute the value of a in equation of }}{{\textbf{a}}_\textbf{18}}$

${a_{18}} = \dfrac{{ - 68}}{4}d + 17d$

$= \dfrac{{ - 68d + 68d}}{4} = 0$

${\textbf{Hence , the required value is (D) 0}}$

The

21^{s t}

$21^{st}$ term of the AP whose first term is

3

$3$ and second term is

4

$4$ , is

0%
$17$
0%
$19$
0%
$21$
0%
$23$

Explanation

Given,

a_{1} = 3

$a_1=3$ ,

a_{2} = 4

$a_2=4$ .

∴

$\therefore$ Common difference,

d = 1

$d=1$ .
Using the formula for

n^{t h}

$n^{th}$ term of an

A . P .

$A.P.$

a_{n} = a + (n - 1) d

$a_n=a+(n-1)d$

For,

n = 21

$n=21$ ,

a_{21} = 3 + (21 - 1) \times 1

$a_{21}=3+(21-1)\times 1$

a_{21} = 3 + 20 \times 1

$a_{21}=3+20\times 1$

a_{21} = 23

$a_{21}= 23$

The sum of the first

16

$16$ terms of the AP:

10, 6, 2, . . .

$10, 6, 2,...$ is

0%
$-320$
0%
$320$
0%
$-400$
0%
$400$

Explanation

Given,

10, 6, 2.........

$10,6,2.........$
Since, first term

= 10

$= 10$ and common difference

= 6 - 10 = - 4

$=6-10=-4$

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

$S_n=\dfrac n2[2a+(n-1)d]$

S_{16} = \frac{16}{2} [2 (10) + (16 - 1) (- 4)]

$S_{16}=\dfrac {16}2[2(10)+(16-1)(-4)]$

S_{16} = 8 [20 + (15) (- 4)]

$S_{16}=8[20+(15)(-4)]$

S_{16} = 8 [20 - 60]

$S_{16}=8[20-60]$

S_{16} = 8 \times (- 40) = - 320

$S_{16}=8\times (-40) = -320$
Option A is correct.

The sum of first five multiples of

3

$3$ is

0%
$45$
0%
$55$
0%
$65$
0%
$75$

Explanation

Sum of an AP is:

\frac{n}{2} \times (2 a + (n - 1) d)

${\dfrac{n}{2}}\times({2a+(n-1)d})$

a = 3, d = 3, n = 5

$a = 3, d= 3, n=5$
Then,
Sum =

\frac{5}{2} \times (2 \times 3 + (5 - 1) \times (3))

${\dfrac{5}{2}}\times{(2\times3+(5-1)\times(3)})$
=

\frac{5}{2} \times (6 + 12)

${\dfrac{5}{2}}\times{(6+12)}$
=

45

$45$

10^{t h}

$10^{th}$ term of AP:

2, 7, 12, . . . . .

$2, 7, 12, .....$ is

0%
$35$
0%
$47$
0%
$55$
0%
None of the above

Explanation

First term

(a_{1}) = 2

$(a_1)= 2$
difference

(d) = 5

$(d)= 5$

The

10

$10$ th term of A.P is

a_{10} = 2 + (10 - 1) \times (5)

$a_{10} = 2 +(10-1)\times(5)$

a_{10} = 2 + (9) \times (5)

$a_{10} = 2 +(9)\times(5)$

a_{10} = 2 + 45

$a_{10} = 2 + 45$

a_{10} = 47

$a_{10} = 47$

If the

n^{t h}

$n^{th}$ term of an A.P. is given by

a_{n} = 5 n - 3

$a_n=5n-3$ , then the sum of first

10

$10$ terms is

0%
$225$
0%
$245$
0%
$255$
0%
$270$

Explanation

Given,

a_{n} = 5 n - 3

$a_n = 5n-3$
Put

n = 1

$n=1$ ,

a_{1} = 5 - 3 = 2

$a_1= 5-3= 2$
Put

n = 10

$n=10$ ,

a_{10} = 5 \times 10 - 3 = 47

$a_{10}= 5\times10-3= 47$
Sum

=

$=$

\frac{n}{2} (a + l)

${\dfrac{n}{2}}{(a+l)}$

=

$=$

\frac{10}{2} (2 + 47)

${\dfrac{10}{2}}{(2+47)}$

=

$=$

49 \times 5

$49\times5$

= 245

$= 245$

Say true or false.

l

$l$ is the last term of the finite A.P., say with the

n

$n$ terms, and the first term is

a

$a$ then the sum of all terms of the A.P. is given by

S = \frac{n}{2} (a + l) .

$S=\dfrac {n}{2}(a+l).$

0%
True
0%
False

Explanation

General formula for sum of

n

$n$ terms of an AP

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

$={\dfrac{n}{2}}\times{(2a+(n-1)d)}$

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

$={\dfrac{n}{2}}\times{(a+ a + (n-1)d)}$

= \frac{n}{2} \times (a + l)

$={\dfrac{n}{2}}\times{(a+l)}$ , where

l

$l$ is the last term.

Therefore, given statement is true.

The sum to

200

$200$ terms of the series

1 + 4 + 6 + 5 + 11 + 6 + . . . . .

$1+4+6+5+11+6+.....$ is

0%
$29,600$
0%
$29,800$
0%
$30,200$
0%
None of these

Explanation

Given, series:

1 + 4 + 6 + 5 + 11 + 6....

$1+4+6+5+11+6....$ is a combination of two AP's i.e.,

(1 + 6 + 11 + . . . . .) + (4 + 5 + 6......)

$(1 + 6 + 11 + .....) + (4 + 5 + 6......)$
Now, each of the AP will have 100 terms.

⟹ {S = S}_{1} + S_{2} w h e r e S_{1} = 1 + 6 + 11... a n d S_{2} = 4 + 5 + 6... N o w, S_{1} = \frac{n}{2} [2 a + (n - 1) d] = \frac{100}{2} [2 \times 1 + (100 - 1) (5)] ⟹ S_{1} = 50 (497) = 24850 A g a i n, S_{2} = \frac{n}{2} [2 a + (n - 1) d] = \frac{100}{2} [2 \times 4 + (100 - 1) (1)] ⟹ S_{1} = 50 (107) = 5350 ∴ {S = S}_{1} + S_{2} = 24850 + 5350 = 30, 200

$\Longrightarrow \quad { S=S }_{ 1 }+{ S }_{ 2 }\quad where{ \quad S }_{ 1 }=1+6+11...\quad and\quad { S }_{ 2 }=4+5+6...\quad \\ Now,{ S }_{ 1 }=\dfrac { n }{ 2 } [2a+(n-1)d]=\dfrac { 100 }{ 2 } [2\times 1+(100-1)(5)]\\ \Longrightarrow { S }_{ 1 }=50(497)=24850\\ Again,{ S }_{ 2 }=\dfrac { n }{ 2 } [2a+(n-1)d]=\dfrac { 100 }{ 2 } [2\times 4+(100-1)(1)]\\ \Longrightarrow { S }_{ 1 }=50(107)=5350\quad \\ \therefore { \quad S=S }_{ 1 }+{ S }_{ 2 }=24850+5350=30,200$
Hence, the sum of 200 terms to the given series is 30,200.

The sum of

11

$11$ terms of an A.P. whose middle term is

30,

$30,$ is

0%
$320$
0%
$330$
0%
$340$
0%
$350$

Explanation

Step -1: Formulating for middle term.

$\textbf{Step -1: Formulating for middle term.}$

Given that, an A.P. of 11 terms, whose middle term is 30.

$\text{Given that, an A.P. of 11 terms, whose middle term is 30.}$

Middle term = 6^{t h} term.

$\text{Middle term = }6^{th}\text{ term.}$

∴ a_{n} = a + (n - 1) d

$\therefore a_n = a+ (n-1)d$

Where, a_{n} is the n^{t h} term.

$\text{Where, }a_n\text{ is the }n^{th}\text{ term.}$

where a is the first term and d is the common difference.

$\text{where }a \text{ is the first term and }d\text{ is the common difference.}$

a_{6} = a + (6 - 1) d = 30

$a_6 = a + (6-1)d = 30$

\Rightarrow a + 5 d = 30

$\Rightarrow a + 5d = 30$

. . . (i)

$...\text{(i)}$

Step -2: Finding the sum of 11 terms.

$\textbf{Step -2: Finding the sum of 11 terms.}$

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

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

∴ S_{11} = \frac{11}{2} (2 a + (11 - 1) d)

$\therefore S_{11} = \dfrac{11}{2}(2a + (11-1)d)$

\Rightarrow S_{11} = \frac{11}{2} (2 a + 10 d)

$\Rightarrow S_{11} = \dfrac{11}{2}(2a + 10d)$

\Rightarrow S_{11} = \frac{11}{2} \times 2 (a + 5 d)

$\Rightarrow S_{11} = \dfrac{11}{2}\times 2(a + 5d)$

From (i), a + 5 d = 30

$\text{From (i), }a + 5d = 30$

∴ S_{11} = 11 \times 30

$\therefore S_{11} = 11\times 30$

\Rightarrow S_{11} = 330

$\Rightarrow S_{11} = 330$

Hence option B is correct.

$\textbf{Hence option B is correct.}$

If the sum of the series

54 + 51 + 48 + . . . . .

$54+51+48+.....$ is

513

$513$ , then the number of terms are

0%
$18$
0%
$20$
0%
$17$
0%
None of these

Explanation

Sum of an AP =

\frac{n}{2} (2 a + (n - 1) \times d)

${\dfrac{n}{2}}{(2a+(n-1)\times d)}$ = 513
where

a = 54

$a= 54$ and

d = - 3

$d=-3$

\frac{n}{2} (2 \times 54 + (n - 1) \times (- 3))

${\dfrac{n}{2}}{(2\times54+(n-1)\times(-3))}$ = 513

\frac{n}{2} (111 - 3 n)

${\dfrac{n}{2}}{(111-3n)}$ = 513

111 n - 3 n^{2} = 1026

$111n-3n^2 = 1026$
Solving for n,

n = 18

$n=18$

There are

60

$60$ terms in an A.P. of which the first term is

8

$8$ and the last term is

185.

$185.$ The

31^{s t}

$31^{st}$ term is

0%
$56$
0%
$94$
0%
$85$
0%
$98$

Explanation

First term of given AP

a = 8

$a= 8$

60^{t h}

$60^{th}$ term of AP

t_{60} = 185

$t_{60}= 185$
Formula for

n^{t h}

$n^{th}$ term

t_{n} = a + (n - 1) d

$t_{n}=a+(n-1)d$
Hence,

185 = 8 + (60 - 1) d

$185= 8 + (60-1)d$

\Rightarrow 177 = 59 d

$\Rightarrow 177= 59d$

\Rightarrow d = 3

$\Rightarrow d= 3$
From

n^{t h}

$n^{th}$ term formula,

31^{s t}

$31^{st}$ term

= 8 + (31 - 1) \times 3

$= 8 + (31-1)\times3$
Thus ,

t_{31} = 98

$t_{31} = 98$

t_{n} = 6 n + 5

$t_n=6n+5$ , then

t_{n + 1} =

$t_{n+1}=$

0%
$6(n+1)+17$
0%
$6(n-1)+17$
0%
$6n+11$
0%
$6n-11$

Explanation

Given,

t_{n} = 6 n + 5

$t_n=6n+5$

t_{n + 1} = 6 (n + 1) + 5 = 6 n + 11

$t_{n+1} = 6(n+1)+5= 6n+11$
Now, check other options as well.

We see that

6 (n - 1) + 17

$6(n-1)+17$ is also equal to

6 n + 11

$6n+11$ .

Hence, options B and C are correct.

If the sum of the series

2 + 5 + 8 + 11.....

$2+5+8+11.....$ is

60100

$60100$ , then the number of terms are

0%
$100$
0%
$200$
0%
$250$
0%
$300$

Explanation

Given AP:

2 + 5 + 8 + 11 + . . .

$2+5+8+11+...$

Here, first term

, a = 2

$,\ a=2$

Common difference

, d = 3

$,\ d=3$

Given: Sum of terms of the AP

= 60100

$=60100$

We know that sum of terms of an AP

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

$={\dfrac{n}{2}}{(2a+(n-1)\times d)}$

\Rightarrow \frac{n}{2} (2 \times 2 + (n - 1) \times 3) = 60100

$\Rightarrow {\dfrac{n}{2}}{(2\times2+(n-1)\times3)}=60100$

\Rightarrow \frac{n}{2} (4 + 3 n - 3) = 60100

$\Rightarrow {\dfrac{n}{2}}{(4+3n-3)}=60100$

\Rightarrow \frac{n}{2} (3 n + 1) = 60100

$\Rightarrow {\dfrac{n}{2}}{(3n+1)}=60100$

\Rightarrow 3 n^{2} + n = 120200

$\Rightarrow 3n^2 + n = 120200$

\Rightarrow 3 n^{2} + n - 120200 = 0

$\Rightarrow 3n^2+n-120200=0$

\Rightarrow 3 n^{2} + 601 n - 600 n - 120200 = 0

$\Rightarrow 3n^2+601n-600n-120200=0$

\Rightarrow n (3 n + 601) - 200 (3 n + 601) = 0

$\Rightarrow n(3n+601)-200(3n+601)=0$

\Rightarrow (n - 200) (3 n + 601) = 0

$\Rightarrow (n-200)(3n+601)=0$

\Rightarrow n = 200

$\Rightarrow n= 200$ or

n = \frac{- 601}{3}

$n=\dfrac{-601}{3}$

Since

n

$n$ cannot be negative, hence

n = 200

$n=200$

The sum of 24 terms of the series

1, - 3, 4, - 6, 7, - 9, 10, . . .

$1, -3, 4, -6, 7, -9, 10, ...$ is

0%
24
0%
-24
0%
16
0%
-16

Explanation

The given series S can be divided into two sub-series

S_{1}

$S_{1}$ and

S_{2}

$S_{2}$
where

S_{1} = 1, 4, 7, 10..

$S_1 = 1,4,7,10..$ and

S_{2} = - 3, - 6, - 9, . . .

$S_2 = -3,-6,-9,...$
24 terms of S would consist of first 12 terms of

S_{1}

$S_1$ and first 12 terms of

S_{2}

$S_2$

∴

$\therefore$ Sum of

S

${S}$ = Sum of

S_{1}

${S_1}$ + Sum of

S_{2}

${S_2}$

So using sum of A.P formula

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

$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$

= \frac{12}{2} (2 \times 1 + 11 \times 3) + \frac{12}{2} (2 \times - 3 + 11 \times - 3)

$= \frac{12}{2}(2\times1+11\times3)+ \frac{12}{2}(2\times-3+11\times-3)$

= 210 - 234 = - 24

$= 210 -234 = -24$

Following two given series are in A.P.

2, 4, 6, 8 . . .

$2, 4, 6, 8 ...$

3, 6, 9, 12 . . .

$3, 6, 9, 12 ...$
First series contains

30

$30$ terms, while the second series contains

20

$20$ terms. Both of the above given series contains some terms, which are common to both of them.The sum of both the above given A.P. are respectively:

0%
$(930, 630)$
0%
$(630, 930)$
0%
$(870, 580)$
0%
$(580, 870)$

Explanation

The given series are

2, 4, 6, 8, . . . .

$2,4,6,8,....$ and

3, 6, 9, 12, . . . .

$3,6,9,12,....$

For first AP:

a = 2, d = 2, n = 30

$a= 2, d= 2, n=30$
Sum

=

$=$

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

${\dfrac{n}{2}}{(2a +(n-1)d)}$

=

$=$

\frac{30}{2} (4 + 29 \times 2)

${\cfrac{30}{2}}{(4+29\times2)}$

=

$=$

\frac{30}{2} (4 + 29 \times 2)

${\cfrac{30}{2}}{(4+29\times2)}$

=

$=$

31 \times 30

$31\times30$

= 930

$= 930$

For second AP:

a = 3, d = 3, n = 20

$a= 3, d= 3, n=20$
Sum

=

$=$

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

${\cfrac{n}{2}}{(2a +(n-1)d)}$

=

$=$

\frac{20}{2} (6 + 19 \times 3)

${\cfrac{20}{2}}{(6+19\times3)}$

=

$=$

\frac{20}{2} (6 + 19 \times 3)

${\dfrac{20}{2}}{(6+19\times3)}$

=

$=$

10 \times 63

$10\times63$

= 630

$= 630$

A ladder has rungs

25

$25$ cm apart. The rungs decrease uniformly in the length from

45

$45$ cm at the bottom to

25

$25$ cm at the top. If the top and the bottom rungs are

2 \frac{1}{2}

$\displaystyle 2 \frac{1}{2}$ m apart, what is the length of the wood required for the rungs

0%
$280$ cm
0%
$320$ cm
0%
$250$ cm
0%
$385$ cm

Explanation

The distance between the top and the bottom rungs

= 2 \frac{1}{2}

$=2\dfrac {1}{2}$ m

= 250

$=250$ cm and the distance between two consecutive rungs

= 25

$=25$ cm.

Therefore, the number of gaps between the rungs

= \frac{250}{25} = 10

$=\dfrac {250}{25}=10$ .

So, the total number of rungs in the ladder

= 11

$=11$

The length of the bottom rung is

45

$45$ cm and the lengths of rungs decrease uniformly from bottom to top.

Therefore, the numbers involved in the lengths (in cm) of the rungs from bottom to top form an AP with

a = 45, l = 25

$a=45, l=25$ and

n =

$n=$ total number of rungs

= 11

$=11$ .

Using the formula for sum of an A.P. to find the total length of the woods required

S_{n}

$S_n$ :

S_{n} = \frac{n}{2} (a + l)

$S_n=\dfrac {n}{2}(a+l)$

= \frac{11}{2} (45 + 25)

$=\dfrac {11}{2}(45+25)$

= \frac{11}{2} \times 70

$=\dfrac {11}{2}\times 70$

= 11 \times 35 = 385

$=11\times 35=385$

Hence, the length of the wood required for rungs

= 385

$=385$ cm.

The sum of

12

$12$ terms of an A.P., whose first term is

4,

$4,$ is

256.

$256.$ What is the last term?

0%
$35$
0%
$36$
0%
$37$
0%
None

Explanation

Given that, for an A.P.,

n = 12, a = 4

$n=12,\ a=4$ and

S_{12} = 256

$S_{12}=256$

To find out: The last term of the A.P.

We know that, sum of

n

$n$ terms of an A.P. is

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

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

∴ 256 = \frac{12}{2} [2 \times 4 + (12 - 1) d]

$\therefore \ 256=\dfrac {12}{2}[2\times 4+(12-1)d]\\$

\Rightarrow 256 = 6 [8 + 11 d]

$\Rightarrow 256=6[8+11d]\\$

\Rightarrow 256 = 48 + 66 d

$\Rightarrow 256=48+66d\\$

\Rightarrow 208 = 66 d

$\Rightarrow 208=66d\\$

\Rightarrow d = \frac{104}{33}

$\Rightarrow d=\dfrac {104}{33}\\$

We also know that, the

n^{t h}

$n^{th}$ term of an A.P. is

t_{n} = a + (n - 1) d

$t_n=a+(n-1)d$

∴ t_{12} = 4 + (12 - 1) \frac{104}{33}

$\therefore \ t_{12}=4+(12-1)\dfrac {104}{33}\\$

\Rightarrow 4 + 11 \times \frac{104}{33}

$\Rightarrow 4+11\times \dfrac {104}{33}\\$

\Rightarrow \frac{12 + 104}{3}

$\Rightarrow \dfrac {12+104}{3}\\$

∴ t_{12} = \frac{116}{3}

$\therefore \ t_{12}= \dfrac {116}{3}\\$

Hence, the last term of the given A.P. is

\frac{116}{3}

$\dfrac {116}{3}$ and therefore, option D is correct.

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 7 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 7 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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