MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 7

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

$n$ and

$n$ th term is

$m$ , then the

$(m + n)$ 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

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

Hence

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

Similarly

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

Subtracting (ii) from (i)

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

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

$d=-1$ .

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

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

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

Hence

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

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

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

$=0$ .

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

$3n^2+5n$ while

$T_m=164$ , then value of m is

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

Explanation

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

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

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

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

$=12+10-8$

$=14$ .
Hence common difference is

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

$=14-8=6$
Now

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

$=8+(n-1)6$

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

$T_{m}=164$
Or

$164=6m+2$

$162=6m$

$27=m$
Hence

$m=27$ .

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

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

$t_n$ is

$116,$ then

$n$ is

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

Explanation

Let number of terms be

$n$

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

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

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

Hence,

$n = 8$

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

$n$ and

$n$ th term is

$m$ , then the

$r^{th}$ term is

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

Explanation

The

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

$'a'$ and common difference

$'d'$ , is

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

Hence

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

$(i)$

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

$(ii)$

Subtracting

$(ii)$ from

$(i)$ gives us

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

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

$d=-1$

Hence

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

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

$a=n+m-1$ .

Hence

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

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

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

$=n+m-r$

Hence

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

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

$2$ and sum up to

$n$ terms is

$49$ and

$7^{th}$ term is

$13$ , then value of

$n$ is equal to?

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

Explanation

Given:

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

It is given that the common difference is 2.

Therefore

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

$13=a_{1}+12$

$a_{1}=1$ .

Now sum of n terms is given by:

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

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

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

$49=n^{2}$

$n=7$

Hence the value of n is 7.

In an A.P., if

$a = 3.5$ ,

$d = 0$ ,

$n = 101$ , then

$a_n$ will be

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

Explanation

$a = 3.5, d=0$ and

$n= 101$

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

$a_n = 3.5 + 100\times0$

$a_n= 3.5$

If the first term of an AP is

$5$ and the common difference is

$-2,$ then the sum of the first

$6$ terms is

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

Explanation

Sum of an AP is:

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

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

Sum

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

$= 3\times(10-10)$

$= 0$

The

$11^{th}$ term of the series

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

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

Explanation

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

$n= 11$ .

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

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

$a_{11}= 20$

In an AP if

$a = 1$ ,

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

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

$n$ is

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

Explanation

$a = 1$

$a_n = 20$

$S_n= 399$
Then,

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

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

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

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

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

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

$\Rightarrow 798$ =

${n\times21}$

$n = 38$

If the common difference of an AP is

$5,$ then what is

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

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

Explanation

Hint:

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

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

$a$ is first term and

$d$ is common difference.

Given:
value of common difference

$,d=5$

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

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

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

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

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

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

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

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

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

$a_{18}-a_{13}$

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

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

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

Final step : Hence,

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

Which term of the AP,

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

$210?$

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

Explanation

First term of the AP,

$a=21$
Common difference

$d=21$

${n}^{th}$ term is

$210$
Then,

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

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

$210= 21n$
Therefore,

$n=10$
Hence, 210 is

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

If the

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

$13$ and the

${5}^{th}$ term is

$25,$ what is its

${7}^{th}$ term?

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

Explanation

Let the first term of the AP be

$a$ , the common difference be

$d$ .
Then,

${2}^{nd}$ term is

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

${5}^{th}$ term is

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

Solving for

$d$ ,

$Eq(2)-Eq(1)$

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

$\Rightarrow 3d=12$

$\Rightarrow d=4$

Hence, From

$(1)$

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

Now,

${7}^{th}$ term is

$a+6d$

Substituting values of

$a$ and

$d$

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

The

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

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

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

Explanation

In the given A.P.

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

$(l)=-49$ ,
then

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

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

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

$21$ terms in AP.
The

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

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

$\Rightarrow 11+17(-3)$

$\Rightarrow 11-51=-40$
SO

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

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

$-40$ .

$7$ times the

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

$11$ times its

$11^{th}$ term, then its

$18^{th}$ 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

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

$3$ and second term is

$4$ , is

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

Explanation

Given,

$a_1=3$ ,

$a_2=4$ .

$\therefore$ Common difference,

$d=1$ .
Using the formula for

$n^{th}$ term of an

$A.P.$

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

For,

$n=21$ ,

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

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

$a_{21}= 23$

The sum of the first

$16$ terms of the AP:

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

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

Explanation

Given,

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

$= 10$ and common difference

$=6-10=-4$

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

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

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

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

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

The sum of first five multiples of

$3$ is

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

Explanation

Sum of an AP is:

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

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

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

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

$45$

$10^{th}$ term of AP:

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

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

Explanation

First term

$(a_1)= 2$
difference

$(d)= 5$

The

$10$ th term of A.P is

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

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

$a_{10} = 2 + 45$

$a_{10} = 47$

If the

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

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

$10$ terms is

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

Explanation

Given,

$a_n = 5n-3$
Put

$n=1$ ,

$a_1= 5-3= 2$
Put

$n=10$ ,

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

$=$

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

$=$

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

$=$

$49\times5$

$= 245$

Say true or false.

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

$n$ terms, and the first term is

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

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

0%
True
0%
False

Explanation

General formula for sum of

$n$ terms of an AP

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

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

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

$l$ is the last term.

Therefore, given statement is true.

The sum to

$200$ terms of the series

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

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

Explanation

Given, series:

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

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

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

The sum of

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

$30,$ is

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

Explanation

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

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

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

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

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

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

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

$\Rightarrow a + 5d = 30$

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

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

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

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

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

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

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

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

$\Rightarrow S_{11} = 330$

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

If the sum of the series

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

$513$ , then the number of terms are

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

Explanation

Sum of an AP =

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

$a= 54$ and

$d=-3$

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

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

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

$n=18$

There are

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

$8$ and the last term is

$185.$ The

$31^{st}$ term is

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

Explanation

First term of given AP

$a= 8$

$60^{th}$ term of AP

$t_{60}= 185$
Formula for

$n^{th}$ term

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

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

$\Rightarrow 177= 59d$

$\Rightarrow d= 3$
From

$n^{th}$ term formula,

$31^{st}$ term

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

$t_{31} = 98$

$t_n=6n+5$ , then

$t_{n+1}=$

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

Explanation

Given,

$t_n=6n+5$

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

We see that

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

$6n+11$ .

Hence, options B and C are correct.

If the sum of the series

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

$60100$ , then the number of terms are

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

Explanation

Given AP:

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

Here, first term

$,\ a=2$

Common difference

$,\ d=3$

Given: Sum of terms of the AP

$=60100$

We know that sum of terms of an AP

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

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

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

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

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

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

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

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

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

$\Rightarrow n= 200$ or

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

Since

$n$ cannot be negative, hence

$n=200$

The sum of 24 terms of the series

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

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

Explanation

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

$S_{1}$ and

$S_{2}$
where

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

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

$S_1$ and first 12 terms of

$S_2$

$\therefore$ Sum of

${S}$ = Sum of

${S_1}$ + Sum of

${S_2}$

So using sum of A.P formula

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

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

$= 210 -234 = -24$

Following two given series are in A.P.

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

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

$30$ terms, while the second series contains

$20$ 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

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

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

For first AP:

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

$=$

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

$=$

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

$=$

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

$=$

$31\times30$

$= 930$

For second AP:

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

$=$

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

$=$

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

$=$

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

$=$

$10\times63$

$= 630$

A ladder has rungs

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

$45$ cm at the bottom to

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

$\displaystyle 2 \frac{1}{2}$ 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

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

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

$=25$ cm.

Therefore, the number of gaps between the rungs

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

So, the total number of rungs in the ladder

$=11$

The length of the bottom rung is

$45$ 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

$a=45, l=25$ and

$n=$ total number of rungs

$=11$ .

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

$S_n$ :

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

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

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

$=11\times 35=385$

Hence, the length of the wood required for rungs

$=385$ cm.

The sum of

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

$4,$ is

$256.$ What is the last term?

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

Explanation

Given that, for an A.P.,

$n=12,\ a=4$ and

$S_{12}=256$

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

We know that, sum of

$n$ terms of an A.P. is

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

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

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

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

$\Rightarrow 208=66d\\$

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

We also know that, the

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

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

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

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

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

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

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

$\dfrac {116}{3}$ 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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