MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 9

$7^{th}$ and

$13^{th}$ terms of an A.P. are 34 and 64, respectively, then its

$18^{th}$ term is

0%
$87$
0%
$88$
0%
$89$
0%
$90$

Explanation

$7^{th}$ term of the A.P. is

$= 34 = a_7$

$13^{th}$ term of the A.P. is

$= 64 = a_{13}$

Let

$a$ be the 1st term

$d$ be the common diff.

$\therefore a_7 = 34$

$\Rightarrow a+ 6d = 34$ ____(i)

again

$a_{13} = 64$

$\Rightarrow a+ 12d = 64$ ___(ii)

(ii) - (i)

$\Rightarrow 6d = 30$

$\Rightarrow d = 5$

$\therefore a = 34 - 6d$

$= 34 - 30$

$= 4$

$\therefore a = 4$

Now,

$18^{th}$ term

$= a + (18 - 1) d$

$= a + 17 d$

$= a + 17 \times 5$

$= 4 + 17 \times 5$

$= 4 + 85$

$= 89$

$\therefore a_{18} = 89$

Find the

$12^{th}$ term from the end of the arithmetic progression

$3, 5, 7, 9,...201?$

0%
$179$
0%
$175$
0%
$172$
0%
$174$

Explanation

Clearly, the given sequence is an AP with first term

$a=3$ and common difference

$d=2$ .
Let there are

$n$ terms in this AP. Therefore,

$a_n=201\Rightarrow a+(n-1)d=201$

$\Rightarrow 3+(n-1)2=201$

$\Rightarrow (n-1)2=198$

$\Rightarrow n-1=99$

$\Rightarrow n=100$
Now

$12^{th}$ term from end=

$(100-12+1)^{th}$ term from beginning.

$\Rightarrow 12^{th}$ term from end=

$89^{th}$ term from beginning=

$3+(89-1)\times2=3+176=179.$

In an A.P.,

$S_1 = 6, S_7 = 105,$ then

$S_n$ :

$S_{n-3}$ is same as

0%
$(n + 3) : (n - 3)$
0%
$(n + 3) : n$
0%
$n : (n - 3)$
0%
None of these

Explanation

Let a and d be the first term and common difference of the given AP respectively.
Given,

$S_1=a=6$ and

$S_7=105\Rightarrow \frac72[2a+(7-1)d]=105$

$\Rightarrow \frac72[2\times6+6d]=105$

$\Rightarrow 6+3d=15$

$\Rightarrow d=3$
Now,

$S_n=\dfrac n2[2a+(n-1)d]=\dfrac n2[12+(n-1)3]=\dfrac {3n}2[n+3]$ ...(1)

$S_{n-3}=\dfrac {n-3}2[2a+((n-3)-1)d]=\dfrac {n-3}2[12+(n-4)3]=\dfrac {n-3}2[3n]$ ...(2)
From (1) and (2), we get

$\dfrac{S_n}{S_{n-3}}=\dfrac{\dfrac {3n}2[n+3]}{\dfrac {n-3}2[3n]}=\dfrac{n+3}{n-3}$
Therefore,

$S_n:S_{n-3}::(n+3):(n-3)$

The sum of all two digit numbers which leave remainder

$1$ when divided by

$3$ is

0%
$1616$
0%
$1602$
0%
$1605$
0%
None of these

Explanation

$\textbf{Step 1:Find first term, common difference and last term of an A.P}$

$\text{Clearly, the two digits numbers which leave remainder $$1$$ when divided by $$3$$ are $$10,13,16,...,97$$. }$

$\text{This is an AP with first term $$a=10$$, common difference $$d=3$$ and last term $$l=97$$.}$

$\textbf{Step 2:Find total number of terms of anA.P}$

$\text{Let there be $$n$$ terms in this AP, then}$

$a_n=97 = a+(n-1)d$

$\therefore 10+(n-1)\times3=97$

$10+3n-3=97$

$3n=97+3-10$

$3n=90$

$\therefore n=30$

$\textbf{Step 3:Sum of n terms of an A.P}$

$\therefore$

$\text{Required sum }$

$=\displaystyle \frac n2[a+l]=\frac{30}2[10+97]=15\times107=1605$

$\textbf{Hence, Option C is correct.}$

If the

$nth$ term of the AP

$9, 7, 5..$ is same as the

$nth$ term of the AP

$15, 12, 9,...$ , find

$n.$

0%
$7$
0%
$8$
0%
$9$
0%
$10$

Explanation

$n th$ term of A.P

$9,7,5....=a+(n-1)d$

$\Rightarrow 9+(n-1)-2$

$\Rightarrow 9-2n+2\Rightarrow 11-2n$

$n th$ term of A.P

$15,12,9,....=a+(n-1)d$

$\Rightarrow 15+(n-1)-3$

$\Rightarrow 15-3n+3\Rightarrow 18-3n$

$\therefore 11-2n=18-3n$

$\Rightarrow 3n-2n=18-11$

$\Rightarrow n=7$

Find

$n$ if the given value of

$x$ is the

$n^{th}$ term of the AP:

$5\dfrac{1}{2}, 11, 16\dfrac{1}{2},22,...$ ;

$x=550$ .

0%
$100$
0%
$110$
0%
$200$
0%
$220$

Explanation

Here, first term $a=5\displaystyle \frac12=\frac{11}2$ and the common difference $d=11-\displaystyle \frac{11}2=\frac{11}2.$

Given, $a_n=x=550$
$\Rightarrow a+(n-1)d=550$

$\Rightarrow \dfrac{11}2+(n-1)\times\dfrac{11}2=550$

$\Rightarrow 1+(n-1)=550\times\dfrac2{11}$

$\Rightarrow n=100$

Which term of the AP:

$3, 10, 17,..$ will be

$84$ more than its

$13^{th}$ term?

0%
$25^{th}$
0%
$24^{th}$
0%
$35^{th}$
0%
$34^{th}$

Explanation

Clearly, the given sequence is an AP with first term

$a=3$ and common difference

$d=10-3=7$ .
Given

$n^{th}$ term is

$84$ more than its

${13}^{th}$ term.

$\Rightarrow t_n=t_{13}+84$

$\Rightarrow a+(n-1)d=(a+12d)+84$

$\Rightarrow (n-1)d=12d+84$

$\Rightarrow (n-1)7=12\times7+84$

$\Rightarrow (n-1)7=7(12+12)$

$\Rightarrow n-1=24$

$\Rightarrow n=25$
Therefore

${25}^{th}$ term is the required term.

How many terms are there in the A.P.,

$7, 13, 19, ...205$ ?

0%
$32$
0%
$33$
0%
$34$
0%
$35$

Explanation

Given sequence is

$7,13,19,...,205$

The first term

$a=7$ and

the common difference

$d=13-7=6$

the last term is

$205$

Let the last term be the

$n^{th}$ term

We know that the

$n^{th}$ term of the arithmetic progression is given by

$a+(n-1)d$

Therefore,

$a+(n-1)d=205$

$\implies 7+(n-1)\times(6)=205$

$\implies 7-6+6n=205$

$\implies 1+6n=205$

$\implies 6n=205-1$

$\implies n=\dfrac{204}{6}$

$\implies n=34$

Therefore, the number of terms in the given sequence is

$34$

If the sum of

$n$ terms of an AP is

$2n$

$^2$

$+ 5n$ , then its

$n$ th term is

0%
$4n - 3$
0%
$3n - 4$
0%
$4n + 3$
0%
$3n + 4$

Explanation

Given

${ S }_{ n }=2{ n }^{ 2 }+5n$

${ S }_{ n-1 }=2{ (n-1) }^{ 2 }+5(n-1)$

$=2({ n }^{ 2 }-2n+1)+5n-5$

$=2{ n }^{ 2 }-4n+2+5n-5$

$=2{ n }^{ 2 }+n-3$

Now,

${ t }_{ n }={ S }_{ n }-{ S }_{ n-1 }$

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

$=2{ n }^{ 2 }+5n-2{ n }^{ 2 }-n+3$

$=4n+3$

Find the sum of all even integers between

$101$ and

$999.$

0%
$246950$
0%
$249650$
0%
$256950$
0%
$276950$

Explanation

$\textbf{Step -1: Find the number of terms.}$

$\text{Even integers between }101\text{ and }999 \text{ are:}$

$102,104,106,\ldots998.$

$\text{It is an A.P. sequence whose first term}(a)=102,$

$\text{common difference}(d)=2$

$\text{and last term}(l)=998.$

$\text{Let this sequence has }n \text{ terms.}$

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

$\Rightarrow 998=102+(n-1)2$

$\Rightarrow 2n-2=896$

$\Rightarrow 2n=898$

$\Rightarrow n=449$

$\textbf{Step -2: Find the required sum.}$

$\because \text{Sum of an A.P.}(S_n)=\dfrac{n}{2}(a+l).$

$\therefore\text{The sum of all even integers between }101\text{ and }998,$

$=\dfrac{449}{2}(102+998)$

$=\dfrac{449\times1100}{2}$

$=449\times550$

$=246950$

$\textbf{Hence,The correct option is A.}$

Which term of the A.P.,

$84, 80, 76,..$ is

$0?$

0%
$18^{th}$ term
0%
$20^{th}$ term
0%
$22^{nd}$ term
0%
$24^{th}$ term

Explanation

The given sequence is an AP in which first term $a=84$ and common difference $d=-4$ .

We have to find value of n for which $a_n$ is $0$ .

Then,
$a+(n-1)d=0$
$\Rightarrow 84+(n-1)\times-4=0$
$\Rightarrow 88-4n=0$

$\Rightarrow n=22$
Hence,

${22}^{nd}$ term is

$0$ .

Find the sum of first

$n$ odd natural numbers.

0%
$n^3$
0%
$n(n+1)$
0%
$n(n+1)^2$
0%
$n^2$

Explanation

We have to find sum of the series :

$1+3+5+...$ to

$n$ terms
Clearly, terms of the given series form an AP with first term

$a=1$ and common difference

$d=2$ .

Therefore,

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

$\Rightarrow S_n=\dfrac n2[2\times1+(n-1)\times2]$

$\Rightarrow S_n=\dfrac n2[2+2n-2]$

$\Rightarrow S_n=n^2$

The

$6^{th}$ and

$17^{th}$ terms of an AP are

$19$ and

$41$ respectively, find the

$40^{th}$ term.

0%
$87$
0%
$77$
0%
$85$
0%
$89$

Explanation

Let

$a$ and

$d$ be the first term and the common difference of the given AP respectively. Then,

$a_6=19\Rightarrow a+5d=19$

$...(1)$

$a_{17}=41\Rightarrow a+16d=41$

$...(2)$
On subtracting

$(2)$ from

$(1),$ we get

$11d=22\Rightarrow d=2$
Putting

$d=2$ in

$(1),$ we get

$a=19-5\times2=9$
Now,

$a_{40}=a+39d=9+39\times2=87$

Find the second term and

$nth$ term of an AP whose

$6th$ term is

$12$ and

$8th$ term is

$22.$

0%
$a_2$ $= -9,$ $a_n$ $= 3n -43$
0%
$a_2$ $= 7,$ $a_n$ $= 3n -52$
0%
$a_2$ $= -8,$ $a_n$ $= 5n -18$
0%
$a_2$ $= -9,$ $a_n$ $= 5n -38$

Explanation

Let

$a$ and

$d$ be the first term and the common difference of the given AP respectively. Then,

$a_6=12\Rightarrow a+5d=12$ ...(1)

$a_8=22\Rightarrow a+7d=22$ ...(2)
On subtracting (2) from (1), we get

$2d=10\Rightarrow d=5$
Putting

$d=5$ in (1), we get

$a=12-5\times5=-13$
Now,

$a_2=a+d=-13+5=-8$ and

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

$=-13+(n-1)\times5$

$=5n-18$

Find the sum of all multiples of

$9$ lying between

$300$ and

$700$ .

0%
$21978$
0%
$22978$
0%
$21970$
0%
$21960$

Explanation

${\textbf{Step 1: Write multiple of 9 as A}}{\text{.P}}{\textbf{. having common difference as 9.}}$

${\text{Multiple of 9 between 300 and 700 are as follows:}}$

$306,315,324, \ldots ,693$

${\text{So, the formed A}}{\text{.P}}{\text{. is}}$ $306,315,324, \ldots ,693$

${\text{Where,}}$ $a = 306:$ ${\text{First term of A}}{\text{.P}}{\text{.}}$

$d = 9:$ ${\text{common difference of A}}{\text{.P}}{\text{.}}$

$l = 639:$ ${\text{Last term of A}}{\text{.P}}{\text{.}}$

${\text{Now we know that, formula for }}{{\text{n}}^{{\text{th}}}}{\text{ term of an A}}{\text{.P}}{\text{. is given by,}}$

${a_n} = a + \left( {n - 1} \right)d \ldots \left( 1 \right)$

${\text{Where,}}$ $a =$ ${\text{First term of A}}{\text{.P}}{\text{.,}}$ $d =$ ${\text{common difference of A}}{\text{.P}}{\text{.,}}$ ${a_n} =$ ${{\text{n}}^{{\text{th}}}}{\text{term of an A}}{\text{.P}}{\text{.}}$

$n =$ ${\text{Total number of terms in A}}{\text{.P}}{\text{.}}$

${\text{Substitute the known values in equation }}\left( 1 \right)$

$\Rightarrow 693 = 306 + \left( {n - 1} \right)9$

$\Rightarrow 693 - 306 = \left( {n - 1} \right)9$

$\Rightarrow 387 = \left( {n - 1} \right)9$

$\Rightarrow \left( {n - 1} \right) = \dfrac{{387}}{9}$

$\Rightarrow n - 1 = 43$

$\Rightarrow n = 44$

${\text{There are total 44 numbers between 300 to 700 which are multiple of 9}}{\text{.}}$

${\textbf{Step 2: Find the sum of multiple of 9 lying between 300 to 700.}}$

${\text{Sum of the n terms of an A}}{\text{.P}}{\text{. having first term 'a' and last term 'l' is given by, }}$

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

$\Rightarrow {S_{44}} = \dfrac{{44}}{2}\left( {306 + 693} \right)$

$\Rightarrow {S_{44}} = 22\left( {999} \right)$

$\Rightarrow {S_{44}} = 21978$

${\textbf{Final Answer: Hence, sum of all multiple of 9 lying between 300 to 700 is 21978.}}$

${\textbf{Therefore, option (A) 21978 is correct answer.}}$

Which of the following are

$AP's$ ? If they form an

$AP$ , find the common difference

$d$ and the first

$3$ terms.

$3, 3 +$

$\sqrt{2}$ ,

$3 +$ 2

$\sqrt{2}$ ,

$3 +$ 3

$\sqrt{2}$ ,...

0%
It is not an $A.P.$
0%
It is an $A.P$ and $d=\sqrt2$ , $t_5=3 + 4\sqrt {2}$ , $t_6 = 3 + 5\sqrt{2}$ , $t_7 =3 + 6\sqrt{2}$
0%
It is an $A.P$ and $d=\sqrt3$ , $t_5 = 3 + 4\sqrt {2}$ , $t_6 = 4 + 5\sqrt {2}$ , $t_7$= 5 + 6\sqrt {2}$
0%
None of these

Explanation

Given series is

$3,3+\sqrt2, 3+2 \sqrt 2, 3+3 \sqrt2,.......$

Clearly, the given sequence is an AP with first term

$a=3$ and common difference

$d=3+\sqrt2-3=\sqrt2$

$n^{th}\space term=a_n=a+(n-1)d$

$\therefore a_5=a+4d=3+4\times\sqrt2=3+4\sqrt2$

$a_6=a+5d=3+5\times\sqrt2=3+5\sqrt2$

$a_7=a+6d=3+6\times\sqrt2=3+6\sqrt2$

In a garden bed, there are

$23$ rose plants in the first row, twenty one in the second row, nineteen in the third row and so on. There are five plants in the last row. How many rows are there of rose plants?

0%
$12$ rows
0%
$10$ rows
0%
$14$ rows
0%
$16$ rows

Explanation

Clearly, rose plants in the rows from an AP

$23,21,19,...,5$ with first term

$a=23$ and common difference

$d=-2$ .
Let there be

$n$ rows. Then,

$a_n=5\Rightarrow a+(n-1)d$

$5= 23+(n-1)\times-2$

$\therefore n=10$
Therefore, there are

$10$ rows of rose plants.

Find

$a_{30} -a_{20}$ for the

$AP$

$: a, a + d, a + 2d, a + 3d,...$

0%
$60d$
0%
$50d$
0%
$10d$
0%
$\dfrac{3}{2}d$

Explanation

Here the first term is

$a$ and the common difference is

$d$ .

$n^{th}$ term of the AP is given by,

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

$\therefore a_{20}=a+19d$ ...

$(1)$

$a_{30}=a+29d$ ...

$(2)$
On subtracting

$(1)$ from

$(2)$ , we get

$a_{30}-a_{20}=10d$

The sum of the first

$9$ terms of an

$A.P$ is

$81$ and the sum of it's first

$20$ terms is

$400.$ Find the first term, the common difference and the sum upto

$15th$ term.

0%
$a = 1, d =2,$ $S_{15} = 235$
0%
$a = 3, d =4, S_{15} = 215$
0%
$a = 5, d =3, S_{15} = 205$
0%
None of these

Explanation

Sum of the term

$S_{n} = \dfrac{n}{2}\left[2\times a + \left(n-1 \right)d \right]$
Given sum of

$9$ terms

$= 81$

$81 = \dfrac{9}{2} \left[2a + \left(9-1 \right)d \right] ...(1)$
Sum of

$20$ terms

$= 400$

$400 = \dfrac{20}{2}\left[2a+\left(20-1 \right)d \right]...(2)$

$\Rightarrow 81 = \dfrac{9}{2} \left[2a+8d\right]$

$\Rightarrow 400 = \dfrac{9}{2} \left[2a+19d\right]$

$\Rightarrow 162 = 18a+72d...(3)$

$\Rightarrow 400=20a+190d....(4)$
Multiply eq

$(3)$ by

$20$ and and eq

$(4)$ by

$18$ and subtract both the equation

$\Rightarrow 360a+3420d = 7200$

$\Rightarrow 360a+1440d = 3240$

$1980d = 3690$

$\Rightarrow d = 2$
Now substitute the

$d = 2$ in eq

$(4)$

$\Rightarrow 400= 20a+190\times 2$

$\Rightarrow 400 = 20a+380$

$\Rightarrow 20a = 20$

$\Rightarrow a =1$
Sum of the 15th Term

$S_{15} = \dfrac{15}{2} \left[2\times 1+\left(15-1 \right)2 \right]$

$S_{15} = \dfrac{15}{2} \left[2+\left(14\right)2 \right]$

$S_{15} = \dfrac{15}{2}\left[2+28 \right]$

$S_{15} = \dfrac{15}{2} \left[30\right]$

$S_{15} = 15 \times 15 = 225$
Hence

$a =1 , d = 2 , S_{15} = 225$

A man saved Rs.

$16500$ in ten years. In each year after the first he saved Rs.

$100$ more than he did in the preceding year. How much did he save in the first year?

0%
Rs. $1400$
0%
Rs. $1600$
0%
Rs. $1500$
0%
Rs. $1200$

Explanation

$S_{n}=16500, n=10, d=100$

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

$\therefore 16500=\dfrac{10}{2}[2a+(10-1)100]$

$16500=5[2a+900]$

$16500=10a+4500$

$10a=12000$

$\therefore a=1200$
Hence he saves Rs.

$1200$ in first year.

How many terms are there in the AP whose first and fifth terms are

$-14$ and

$2$ respectively and the sum of the terms is

$40?$

0%
$5$
0%
$10$
0%
$30$
0%
$55$

Explanation

Given,

$a=-14$ and

$a_5=2$ .

$\Rightarrow a+4d=2\Rightarrow 4d=2+14\Rightarrow d=4$
Let the given sequence has

$n$ terms. Therefore,

$S_n=\dfrac n2[2a+(n-1)d]=40$ (given)

$\Rightarrow \dfrac n2[2\times-14+(n-1)4]=40$

$\Rightarrow n[-14+(n-1)2]=40$

$\Rightarrow n(2n-16)=40$

$\Rightarrow n^2-8n-20=0$

$\Rightarrow (n-10)(n+2)=0$

$\Rightarrow n=10,-2$
But,

$n=-2$ is not possible.
Therefore

$n=10$ .

Find the sum of odd numbers between

$0$ and

$50$ .

0%
$800$
0%
$745$
0%
$625$
0%
$520$

In an AP,
Given

$a =5, d = 3,$

$a_n$

$= 50$ , find

$n$ and

$S_n$ .

0%
$330$
0%
$390$
0%
$440$
0%
$560$

Explanation

As we know nth term,

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

& Sum of first

$n$ terms,

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

$a$ &

$d$ are the first term amd common difference of an AP.

Given,

$a =5, d =3, a_n = 50$

$\Rightarrow a + (n -1)d = 50$

$\Rightarrow 5 + (n -1)3 = 50$

$\Rightarrow 5 + 3n -3 = 50\Rightarrow 3n = 48\Rightarrow n =16$

$\therefore S_{16} = \dfrac{16}{2} [2a + (16-1)d]=8[2\times5+15\times3]=440$
Hence,

$n=16,S_{16}=440$

Find the

$S_{15}$ for an AP whose

$nth$ term is given by

$a_n$

$= 3 + 4n$ .

0%
$420$
0%
$525$
0%
$630$
0%
$675$

Explanation

We know that nth term,

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

& Sum of first

$n$ terms,

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

$a$ &

$d$ are the first term amd common difference of an AP.

Since,

$a_n=3+4n$ ...(1)

On putting

$n=1,2,3,..$ in successively, we get

$a_1=3+4\times1=7$

$a_2=3+4\times2=11$

$a_3=3+4\times3=15$ and so on..
Thus the sequence is

$7,11,15,.....$
Clearly it is an AP with first term

$a=7$ and common difference

$d=11-7=4$
Now, Required sum

$S_{15}=\frac{15}2[2a+(15-1)d]$

$\Rightarrow S_{15}=\displaystyle \frac{15}2[2\times7+14\times4]$

$=15\times35$

$=525$

In an AP, given

$a_n = 4, d = 2, S_n = -14,$ find

$n$ and

$a$ .

0%
$n=11$ and $a=-3$
0%
$n=10$ and $a=-3$
0%
$n=7$ and $a=-8$
0%
$n=10$ and $a=3$

Explanation

As we know nth term,

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

& Sum of first

$n$ terms,

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

$a$ &

$d$ are the first term amd common difference of an AP.

Since,

$a_n=4\Rightarrow a+(n-1)d=4$

$\Rightarrow a+(n-1)2=4$

$\Rightarrow a=6-2n$ ...(1)
Also,

$S_n=-14$

$\Rightarrow \dfrac n2[2a+(n-1)d]=-14$

$\Rightarrow \dfrac n2[2(6-2n)+(n-1)2]=-14$ (

$\because from\ (1)$ )

$\Rightarrow n[12-4n+2n-2]=-28$

$\Rightarrow n[n-5]=14$

$\Rightarrow n^2-5n-14=0$

$\Rightarrow (n-7)(n+2)=0$

$\Rightarrow n=7,-2$
But,

$n=-2$ is not possible.

$\therefore n=7$
Putting

$n=7$ in (1), we get

$\therefore a=6-2\times7=-8$

How many terms of the A.P.:

$9, 17, 25,...$ must be taken to give a sum of

$636$ ?

0%
$6$
0%
$8$
0%
$12$
0%
$16$

Explanation

First term,

$a = 9$
Common difference,

$d=17-9=25-17=8$

$d = 8$
Sum

$S_{n} = 636$

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

$636=\dfrac{n}{2}[2 \times 9+(n-1)8]$

$1272=n(18+8n-8)$

$1272=10n+8n^2$

$8n^2+10n-1272=0$

$4n^2+5n-636=0$

upon solving the quadratic equation, we get,

$n=12,-\dfrac{53}{4}$

n cannot be negative,

Therefore

$n=12$

In an AP,
Given

$l = 28, S = 144$ and there are total

$9$ terms. Find

$a$ .

0%
$a=3$
0%
$a=4$
0%
$a=5$
0%
$a=7$

Explanation

Given,

$l=28, S= 144, n=9$

$\therefore S=144\Rightarrow S_9$

$\displaystyle \frac92[a+l]=144$

$\displaystyle \frac92[a+28]=144$

$a+28=32$

$\therefore a=4$

Find the sum of first

$51$ terms of an AP, whose second and third terms are

$14$ and

$18$ respectively.

0%
$5610$
0%
$5800$
0%
$6120$
0%
$6680$

The

$6^{th}$ term of an arithmetic progression is

$-10$ and the

$10^{th}$ term is

$-26.$ Determine the

$15^{th}$ term of the AP.

0%
$15^{th}$ term of AP is $-14$
0%
$15^{th}$ term of AP is $-86$
0%
$15^{th}$ term of AP is $-46$
0%
$15^{th}$ term of AP is $-35$

Explanation

Let

$a$ and

$d$ be the first term and common difference of the given AP respectively.

$\therefore a_6=-10\Rightarrow a+5d=-10$ ...(1)
&

$a_{10}=-26\Rightarrow a+9d=-26$ ...(2)
On subtracting (1) from (2), we get

$4d=-16\Rightarrow d=-4$
Putting

$d=-4$ in (1), we get

$a=-10-(5\times-4)=10$
Now,

$a_{15}=a+14d=10+14\times-4=-46$
Hence,

${15}^{th}$ term of the given AP is

$-46$ .

Write first four terms of the AP, when the first term

$a$ and the common difference

$d$ are given as follows

$a = -1.25, d = -0.25.$

0%
First four terms of the given AP are $-1.25, -1.70, -1.95, -2.00$
0%
First four terms of the given AP are $-1.25, -1.50, -1.75, -2.00$
0%
Not an AP
0%
None of these

Explanation

Given,

$a=-1.25, d=-0.25$

$nth$ term=

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

$\therefore a_1=a=-1.25$

$a_2=a+d=-1.25+(-0.25)=-1.50$

$a_3=a+2d=-1.25+2\times(-0.25)=-1.75$

$a_4=a+3d=-1.25+3\times(-0.25)=-2.00$
Hence, four terms are

$-1.25,-1.50,-1.75,-2.00$

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 9 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 9 - MCQExams.com

0:0:1

Practice Class 10 Maths Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.