If $$S=\cfrac { n }{ 2 } [2a+(n-1)d]$$ ; find $$d$$ , when $$a=8, S=380$$ and $$n=10$$.

$$\displaystyle 6 \dfrac{2}{3}$$

$$\displaystyle 7 \frac{1}{6}$$

$$\displaystyle 4 \frac{5}{2}$$

$$\displaystyle 2\dfrac{1}{6}$$

MCQ Questions for Class 10 Maths Arithmetic Progressions Quiz 8

Find the sum of all natural numbers not exceeding

$1000,$ which are divisible by

$4$ but not by

$8.$

0%
$62500$
0%
$62800$
0%
$64000$
0%
$65600$

Explanation

The natural numbers that are divisible by 4 but not 8 below 1000 are:

$4,\ 12,\ 20,\ 28,\ ... ,\ 996$

These numbers form an AP

Here

$a=4, d=8,$

And,

$a_n = 996$

As we know for an AP,

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

$\Rightarrow 996 = 4+(n-1) \times 8$

$\Rightarrow 8n - 8 + 4 = 996$

$\Rightarrow n =125$

For the terms in an AP:

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

$\therefore\ \ S_{125} = \dfrac{125}{2} [2 \times 4 + (125 - 1) \times 8]$

$=\dfrac{125}{2}\times [8+992]$

$=\dfrac{125}{2}\times 1000$

$= 62500$

$30$ trees are planted in a straight line at intervals of

$5\ m.$ To water them, the gardener needs to bring water for each tree, separately from a well, which is

$10\ m$ from the first tree in line with the trees. How far will he have to walk in order to water all the trees beginning with the first tree? Assume that he starts from the first well, and he can carry enough water to water only one tree at a time.

0%
$4785\ m$
0%
$4795\ m$
0%
$4800\ m$
0%
None of these

Explanation

To water the first tree he has to walk

$= 10$ metres

To water the second tree he has to walk

$= 25$ metres

To water the third tree he has to walk

$= 35$ metres

To water the fourth tree he has to walk

$= 45$ metres

So, if we observe the above distances except for the first they form an arithmetic progression,

$25,35,45,55,\dots\ \ \text{upto 29 terms}$

Let the total distance travelled by the gardener be

$d.$ So,

$d= 10 + 25 + 35 +\dots$ to

$30$ terms

$= 10 + (25+35+\dots29 \ \text{terms})$

$=10 + \dfrac{29}{2}\left[2\times25 + (29-1)\times10\right]$

$=10 + \dfrac{29}{2}\left(50 + 28\times10\right)$

$=10 + 4785$

$=4795\ m$

Hence, option

$B$ is correct.

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is incorrect but Reason is correct

Explanation

When

$n =1$ then

$t_1 = 4$
when

$n= 15$ then

$t_{15} = -66$

Sum

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

$\therefore S_{15} =\dfrac{15}{2}{(4-66)}$
=

$-465$

The assertion and reason are both correct and reason is the correct explanation of the assertion.

Identify the progression:

$A : 4, 7, 10, 13, 16, 19, 22, 25, .....$

$B : 4, 7, 9, 10, 13, 14, .........$

0%
Only $A$
0%
Only $B$
0%
Both $A$ and $B$
0%
None of these

Explanation

Only sequence

$A$ is progression as its terms follow same pattern

If sum of

$n$ terms of an A.P. is

$3n^2+5n$ and

$T_m=164$ , what is the value of m?

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

Explanation

Since

${ S }_{ n }={ 3n }^{ 2 }+5n$
Replace

$n$ by

$(n-1)$ ; we get

${ S }_{ n-1 }={ 3(n-1) }^{ 2 }+5(n-1)$

$= (n-1) [3(n-1)+5]$

$= (n-1) [3n-3+5]$

$= (n-1) (3n+2)$
Now,

${ T }_{ n }={ S }_{ n }-{ S }_{ n-1 }= 6n+2$
Substituting

${ S }_{ n }$ and

${ S }_{ n-1 }$ ; we get

${ T }_{ n }=6n+2 = 164$

$n = 27$

Check if the series is an

$AP.$ Find the common difference

$d$ . Also, find the next three terms.

$-10, -6, -2 , 2.....$ .

0%
It is an $AP$ and $d=4$ , other terms $6,10,14$
0%
It is an $AP$ and $d=\dfrac{3}{5}$ , other terms $5,10,15$
0%
It is not an $AP$ , other terms $3,4,5$
0%
None of these

Explanation

Given series is

$-10, -6,-2,2,.......$

Clearly, the given sequence is an AP with first term

$a=-10$ and common difference

$d=-6-(-10)=4$

$n^{th}$ term

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

$\Rightarrow a_5=a+4d=-10+4\times4=6$

$\Rightarrow a_6=a+5d=-10+5\times4=10$

$\Rightarrow a_7=a+6d=-10+6\times4=14$

If the

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

$(2n-1),$ then the sum of its first

$n$ terms will be

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

Explanation

Given,

$\Rightarrow n^{th}$ term of AP

$= 2n-1$

$\Rightarrow 1^{st}$ term of AP

$= 2(1)-1= 1$

$\Rightarrow 2^{nd}$ term of AP

$= 2(2)-1 = 4-1= 3$

$\Rightarrow$ Common difference

$(d)=$

$2^{nd}$ term -

$1^{st}$ term

$= 3-1 = 2$

$\Rightarrow$ Sum of first

$n$ terms of AP

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

$\Rightarrow$ Sum

$=$

${\dfrac{n}{2}}{[2\times1 + (n-1)2]}$

$=$

${\dfrac{n}{2}}\times {(2n)}$

$=$

$n^2$

$=$

$n^2 +(2n+1)-(2n+1)$ (adding and subtracting

$2n+1$ )

$=$

$(n^2-2n+1) +2n-1$

$=$

$(n-1)^2 + (2n-1)$

Correct options

$: B,D$

Let

$S_n$ denote the sum of the first

$'n'$ terms of an A.P.

$S_{2n} = 3S_n$ . Then, the ratio

$\dfrac{S_{3n}}{S_n}$ is equal to :

0%
$4$
0%
$6$
0%
$8$
0%
$10$

Explanation

We know,

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

$a$ is the first term and

$d$ is the common difference]

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

$S_{3n}=\dfrac{3n}{2}[2a+(3n-1)d]$
Given,

$S_{2n}=3S_n$

$\Rightarrow n[2a+2nd-d]$

$=3 \left [ \dfrac{n}{2} (2a+nd-d) \right ]$
Therefore,

$\displaystyle d(n+1)= 2a$

The value of ratio,

$\frac{S_{3n}}{S_n}= \frac{{\frac{3n}{2}}{(2a + (3n-1)d)}}{{\frac{n}{2}}{(2a + (n-1)d)}}$
=

$\frac{{\frac{3n}{2}}{(nd+d + (3n-1)d)}}{{\frac{n}{2}}{(nd+d + (n-1)d)}}$
=

$\frac{{\frac{3n}{2}}{(4nd)}}{{\frac{n}{2}}{(2nd)}}$
= 6

$\displaystyle \frac{3+5+7+......+ n(terms)}{5+8+11+.....+ 10(terms)}=7$ , then the value of

$n$ is

0%
$35$
0%
$36$
0%
$37$
0%
$40$

Explanation

$S_n =$ Sum of n terms of an A.P.

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

where

$a=$ first term

$d=$ common difference

$\therefore \displaystyle \frac{3+5+7+ ...... + n terms}{5+8+11+...... + 10 terms}=7$

$\Rightarrow \displaystyle \dfrac{\dfrac{n}{2}[2\times 3 + (n-1) \times 2]}{\dfrac{10}{2} [2 \times 5 + (10-1) \times 3]}=7$

$\Rightarrow \displaystyle \frac{n(2n+4)}{370}=7$

$\Rightarrow 2n^2 + 4n - 2590 = 0$

$\Rightarrow n^2+2n-1295=0$

$\Rightarrow n^2+37n-35n-1295=0$

$\Rightarrow n(n+37)-35(n+37)=0$

$\Rightarrow (n-35)(n+37)=0$

$\Rightarrow n=35$

The sum up to

$9$ terms of the series

$\displaystyle \frac{1}{2}+\frac{1}{3}+\frac{1}{6}+ ...$ is

0%
$\displaystyle -\frac{5}{6}$
0%
$\displaystyle -\frac{1}{2}$
0%
$1$
0%
$\displaystyle -\frac{3}{2}$

Explanation

Given that series is an A.P with first term,

$a=\dfrac{1}{2}$ and
Common difference,

$d=\dfrac{1}{3} -\dfrac{1}{2} =\dfrac{1}{6}-\dfrac{1}{3} =\dfrac{-1}{6}$
So using sum of A.P formula

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

$n =9 \Rightarrow S_9=\dfrac{9}{2}\left ( 2. \dfrac{1}{2} + ( 9 - 1 ).\dfrac{-1}{6}\right )=\dfrac{-3}{2}$

$S=\cfrac { n }{ 2 } [2a+(n-1)d]$ ; find

$d$ , when

$a=8, S=380$ and

$n=10$ .

0%
$\displaystyle 7 \frac{1}{6}$
0%
$\displaystyle 4 \frac{5}{2}$
0%
$\displaystyle 2\dfrac{1}{6}$
0%
$\displaystyle 6 \dfrac{2}{3}$

Explanation

We know,

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

Given,

$a=8, S=380$ and

$n=10$

$\therefore 380= \dfrac { 10 }{ 2 } (2(8)+9d)$

$380= 5(16 +9d)$

$300= 45d$

$\therefore d =\dfrac { 300 }{ 45 }$

$= 6\dfrac { 30 }{ 45 }$

$= 6\dfrac { 2 }{ 3 }$

A sprinter runs 6 meters in the first second of a certain race and increase her speed by 25 cm/sec. in each succeeding second. (This means that she goes 6m 25 cm. the second second, 6m 50 cm. the third second, and so on.) How far does she go during the eight second?

0%
8.75 m
0%
7.75 m
0%
8.25 m
0%
9.25 m

Explanation

Since the sprinter increase her speed by

$25cm/sec$ , so we have an AP corresponding to givem problem as below

$6, 6.25, 6.50, 6.75, .......$

$\therefore$

$a = 6, d= 0.25, n = 8$

Required distance

$= 6 + (8-1) \times 0.25$

$=6 + 1.75 = 7.75 m$

The

$p^{th}$ and

$q^{th}$ terms an

$A.P$ are respectively

$a$ and

$b.$ Then sum of

$(p+q)$ terms is

0%
$\dfrac {p+q}{2}\left(a+b+\dfrac {a-b}{p-q}\right)$
0%
$\dfrac {p+q}{2}\left(a-b-\dfrac {a+b}{p+q}\right)$
0%
$(p+q)\left(a+b+\dfrac {p-q}{a-b}\right)$
0%
$\dfrac {p+q}{2}\left(a+b+\dfrac {p-q}{a-b}\right)$

Explanation

Let the first term of the A.P be '

$c$ ' and its common difference be '

$d$ '.

So its

$p^{th}$ term is

$c+(p-1)d=a$

$...(1)$

and its

$q^{th}$ term is

$c+(q-1)d=b$

$...(2)$

Now, add equation

$(1)$ and

$(2)$ ; we get

$\Rightarrow 2c+(p+q-2)d=a+b$

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

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

$...(3)$

Now, subtracting

$(1)$ from

$(2)$ ; we get

$\Rightarrow (p-q)d=a-b$

$\Rightarrow d=\dfrac { (a-b) }{ (p-q) }$

$...(4)$

Now, the sum of

$(p+q)$ terms is

$S_{(p+q)}=\dfrac { (p+q) }{ 2 } \left[ 2c+(p+q-1)d \right]$

$...(5)$

Substitute equation

$(3)$ and

$(4)$ in

$(5)$ ; we get

$S_{(p+q)}=\dfrac { (p+q) }{ 2 } \left[ a+b+d \right]$

$S_{(p+q)}=\dfrac { (p+q) }{ 2 } \left[ a+b+\dfrac { (a-b) }{ (p-q) } \right]$

STATEMENT -

$1$ : The sum of first

$11$ terms of the A.P:

$2, 6, 10, 14,\dots$ is

$242.$
STATEMENT -

$2$ : The sum of first

$n$ terms of the A.P. is given by

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

0%
Statement - $1$ is True, Statement - $2$ is True, Statement - $2$ is a correct explanation for Statement - $1$
0%
Statement - $1$ is True, Statement - $2$ is True : Statement $2$ is NOT a correct explanation for Statement - $1$
0%
Statement - $1$ is True, Statement - $2$ is False
0%
Statement - $1$ is False, Statement - $2$ is True

Explanation

Checking Statement

$1$ :

Given series:

$2, 6, 10, 14,\dots$

$n=11$

$a=2$

$d=6-2=4$

$\therefore S_{11}=\dfrac{11}{2}[2.2+(11-1)4]=242$

$\therefore$ Statement

$1$ is True

$\therefore$ Statement

$2$ is also True (general formula)

Statement

$2$ is used to prove Statement

$1$

Hence, statement

$2$ is correct explanation for statement

$1.$

There is an auditorium with

$35$ rows of seats. There are

$20$ seats in the first row,

$22$ seats in the second row,

$24$ seats in the third row, and so on. Find the number of seats in the twenty fifth row.

0%
$72$
0%
$68$
0%
$54$
0%
$89$

Explanation

Given the number of seats in

$row1, row2$ and

$row3$ is

$20, 22$ and

$24$ .

We note that the number of seats in each subsequent row forms an AP.

$20, 22, 24, .....$ and so on.

We need to find the number of seats in

$row25$ .

So, number of seats in

$row25$

$= { T }_{ 25 }\quad =\quad a\quad +\quad (n-1)d\\ \\$

where

$'a'$ is the number of seats in first row, n is the number of rows and

$'d'$ is the difference in seats in each row.

$\quad \qquad \qquad { T }_{ 25 }\quad =\quad 20\quad +\quad (25-1)2\\ \\ \Longrightarrow \qquad { T }_{ 25 }\quad =\quad 20\quad +\quad (24)2\qquad =\quad 68.\\ \\$

Number of seats in

$row25 = 68$ .

Which term of the A.P. 5, 12, 19, 26, ............ is 145

0%
12
0%
18
0%
25
0%
21

Explanation

Here first term

$a=5$ and common difference

$d=12-5=7$
Suppose nth term of the A.P. is 145
Now,

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

$\therefore 145 =5 (n-1) (7)$

$\therefore 145-5 = 7(n-1)$

$\therefore 140 =7 (n-1)$

$\therefore 20 = n-1 \therefore n = 21$

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

$d$ the subject of formula.

0%
$\displaystyle d=\frac{\left ( 2S-a \right )}{n\left ( n-1 \right )}$
0%
$\displaystyle d=\frac{\left ( 2S-na \right )}{n\left ( n+1 \right )}$
0%
$\displaystyle d=\frac{2\left ( S-na \right )}{n\left ( n-1 \right )}$
0%
$\displaystyle d=\frac{2\left ( S-a \right )}{n\left ( n+1 \right )}$

Explanation

Given,

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

$=> \displaystyle \frac {2S}{n} = 2a + (n-1)d$

$=> \displaystyle \frac {2S}{n} - 2a = (n-1)d$

$=> \displaystyle \frac {2(S-na)}{n} = (n-1)d$

$=> d = \displaystyle \frac {2(S-na)}{n(n-1)}$

Find the sum of first

$11$ positive numbers which are multiples of

$6$ .

0%
$314$
0%
$396$
0%
$452$
0%
$245$

Explanation

Clearly, the numbers are

$6,12,18,...$

This is an AP with first term

$a=6$ , common difference

$d=6$ and

$n=11$ .

We have to find the sum of

$11$ terms of the given AP

Putting

$a=6$ ,

$d=6, n=11$ in

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

$S_{11}=\dfrac{11}2[2\times 6+(11-1)\times6]$

$\therefore S_{11}=11\times36$

Hence,

$S_{11}=396$

Option B is correct.

A man arranges to pay off a debt of Rs. 3600 in 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid he dies leaving one-third of the debt unpaid. Find the value of the first instalment.

0%
Rs. 49
0%
Rs. 51
0%
Rs. 53
0%
Rs. 55

Explanation

Given that, debt of

$\text{Rs. }3600$ is to be paid in

$40$ annual installments.

$\therefore \ S_{40}=3600........(i)$

Also, given that, after

$30$ installments one-third of the debt is unpaid

$\therefore \ S_{30}=3600-\dfrac{1}{3} \text{ of } 3600$

$=3600-1200$

$\left[\because \ \dfrac{1}{3}\times 3600=1200\right]$

$\therefore \ S_{30}=2400........(ii)$

From

$(i)$ , we get:

$S_{40}=\dfrac{40}{2}[2a+39d]=3600$

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

$\Rightarrow 2a+39d=180.....(iii)$

Similarly, from

$(ii)$ , we get:

$S_{30}=2400$

$\Rightarrow \dfrac{30}{2}[2a+29d]=2400$

$\Rightarrow 2a+29d=160......(iv)$

Subtracting

$(iv)$ from

$(iii)$ , we get:

$(2a+39d)-(2a+29d)=180-160$

$\Rightarrow 10d=20$

$\Rightarrow d=2$

Rewriting

$(iii)$ , we get:

$2a=180-39d$

$=180-39\times 2$

$=180-78$

$=2a=102$

$\Rightarrow a=\dfrac{102}{2}=51$

Hence, the value of the first instalment is

$51$ .

Find the sum of

$A.P.$ whose first and last term is

$13$ and

$216$ respectively & common difference is

$7$ .

0%
$3434$
0%
$3435$
0%
$1545$
0%
$3456$

Explanation

Let a and d be the first term and the common difference of the given AP respectively. Let there are n terms in the given AP.
Given

$a=13, d=7$ and

$a_n=216$

$\Rightarrow a+(n-1)d=216\Rightarrow 13+(n-1)\times7=216\Rightarrow n=30$
Therefore, there are 30 terms in the given AP.
Now, Required sum=

$S_{30}=\frac{30}2[2\times13+(30-1)\times7]=15\times229=3435$
Hence, Required sum

$=3435$ .

$t_{11}$ and

$t_{16}$ for an

$A.P.$ are respectively

$38$ and

$73$ , then

$t_{31}$ is

$........$

0%
$178$
0%
$177$
0%
$176$
0%
$175$

Explanation

Given:

${t}_{11}=38,{t}_{16}=73$

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

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

According to the given condition,

$\Rightarrow a+10d=38$

$...(1)$

$\Rightarrow a+15d=73$

$...(2)$

where

$a$ is the first term and

$d$ is the common difference of given AP.
On subtracting

$(1)$ from

$(2)$ , we get,

$5d=35$

$\Rightarrow d=7$

substituting the value of

$d$ in

$(1)$ , we get

$a+10\times7=38\Rightarrow a+70=38$

$\Rightarrow a=-32$

Therefore,

${t}_{31}=a+30d\\=-32+30\times7\\=-32+210\\=178$
Hence, option

$(A)$ is correct.

In the A.P. 7, 14, 21, ... How many terms are to be considered for getting sum 5740.

0%
40
0%
50
0%
14
0%
51

Explanation

Given AP is 7,14,21.....
Since, first term

$a=7$ and common difference

$d=7$
Suppose no. of terms in given AP=

$n$
Sum of

$n$ terms

$S_{n}=5740$
Since

$S_{n}=\frac n2 [2a+(n-1)d]$

$\Rightarrow 5740=\frac n2[2\times 7+(n-1)7]$

$\Rightarrow 5740=\frac n2[14+7n-7]$

$\Rightarrow 5740=\frac n2[7+7n]$

$\Rightarrow 5740=\frac n2\times 7[1+n]$

$\Rightarrow 5740\times 2= 7n[1+n]$

$\Rightarrow 11480=7n[1+n]$

$\Rightarrow \frac{11480}{7}=n[n+1]$

$\Rightarrow 1640=n^2+n$

$\Rightarrow n^2+n-1640=0$

$\Rightarrow (n+41)(n-40)=0$

$\Rightarrow n=-41, n=40$
no. of terms

$n$ can't be negative.

$\therefore n=40$

The

$n^{th}$ term of the sequence

$\displaystyle\frac{1}{p}$ ,

$\displaystyle\frac{1 + 2p}{p}$ ,

$\displaystyle\frac{1 + 4p}{p}$ ,... is

0%
$\displaystyle\frac{1 + 2np + 2p}{p}$
0%
$\displaystyle\frac{1 - 2np - 2p}{p}$
0%
$\displaystyle\frac{1 + 2np - 2p}{p}$
0%
$\displaystyle\frac{1 + 2np}{p}$

Explanation

Sequence,

$\displaystyle\frac{1}{p}$ ,

$\displaystyle\frac{1 + 2p}{p}$ ,

$\displaystyle\frac{1 + 4p}{p}$ ........

Here

$a =\displaystyle\frac{1}{p}$ ,

$d$

$=$

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

$-\displaystyle\frac{1}{p}$

$\Rightarrow$

$d=\displaystyle\frac{1 + 2p - 1}{p}$

$\Rightarrow$

$d =2$

$\therefore t_n$

$=$

$\displaystyle\frac{1}{p} + (n - 1)2$

$=$

$\displaystyle\frac{1}{p}$ +

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

$\Rightarrow$

$t_n$

$=$

$\displaystyle\frac{1 + 2np - 2p}{p}$

Find the

$n^{th}$ term of the sequence

$m -1, m - 3, m - 5,.....$

0%
$m - n + 1$
0%
$m + 2n + 1$
0%
$m - 2n + 1$
0%
$m - 2n$

Explanation

Clearly, the given sequence is an AP with first term

$a=m-1$ and common difference

$d=(m-3)-(m-1)=-2$ .

$\therefore$ nth term

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

$\Rightarrow a_n=(m-1)+(n-1)\times-2$

$=m-1-2n+2$

$=m-2n+1$

$(p + q)^{th}$ and

$(p - q)^{th}$ terms of an A.P. are respectively

$m$ and

$n.$ The

$p^{th}$ term is

0%
$\displaystyle\frac{1}{2}(m + n)$
0%
$\displaystyle\sqrt{mn}$
0%
$m + n$
0%
$mn$

Explanation

Let

$a$ be the first term and

$d$ be the common difference of the given

$AP$ then, by using

$n^{th}$ term formula

nth term

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

$(p+q)^{th}\space=m\Rightarrow a+(p+q-1)d=m$ ...(1)

$(p-q)^{th}\space=n\Rightarrow a+(p-q-1)d=n$ ...(2)

Adding (1) and (2), we get

$2a+(2p-2)d=m+n$

$\Rightarrow a+(p-1)d=\dfrac{m+n}2$

$\Rightarrow p^{th}\space term=\dfrac{m+n}2$

So, Option A is correct.

Sum of first

$5$ terms of an A.P. is one fourth of the sum of next five terms. If the first term is

$2,$ then the common difference of the A.P. is

0%
$6$
0%
$-6$
0%
$3$
0%
None of these

Explanation

let the A.P. be

$a,a+d,a+2d.......$
We have

$T_{1}+T_{2}+T_{3}+T_{4}+T_{5}=\dfrac{1}{4}\left [ T_{6}+T_{7}+T_{8}+T_{9}+T_{10} \right ]$

$\because$ Sum

$=\dfrac{n}{2} ($ first term

$+$ last term

$)$
Then

$\dfrac{5}{2}\left ( T_{1}+T_{5} \right )=\dfrac{1}{4}\times\dfrac{5}{2}\left ( T_{6}+T_{10} \right )$

$\Rightarrow \dfrac{5}{2}\left ( a+\left ( a+4d \right ) \right )=\dfrac{5}{8}\left [ \left (a+5d \right )\left (a+9d \right ) \right ]$

$\Rightarrow 2a+4d=\dfrac{1}{4}\left [ 2a+14d \right ]$

$\Rightarrow 4a+8d=a+7d$

$\Rightarrow d=-3a$ ....

$(a=2)$

$\Rightarrow d=-3\times 2=-6$

In an A.P. S

$_3$

$= 6$ , S

$_6$

$= 3$ , then it's common difference is equal to ?

0%
$3$
0%
$-1$
0%
$1$
0%
None of these

Explanation

We know that for an A.P., sum of n terms is

${ S }_{ n }=\dfrac { n }{ 2 } [2a+(n-1)d]$ , where

$a$ is the first term and

$d$ is the common difference.

Now, here

${ S }_{ 3 }=6$ and

${ S }_{ 6 }=3$

${ S }_{ 3 }=\dfrac { 3 }{ 2 } [2a+(3-1)d]=6$

$=>\dfrac { 3 }{ 2 } [2a+2d]=6$

$=>\dfrac { 6 }{ 2 } [a+d]=6$

$=>3(a+d)=6$

$=>a+d=2$ -------------(i)

Also,

${ S }_{ 6 }=\dfrac { 6 }{ 2 } [2a+(6-1)d]=3$

$=>3[2a+5d]=3$

$=>2a+5d=1$

$=>2(2-d)+5d=1$ (using (i))

$=>4-2d+5d=1$

$=>3d=1-4$

$=>d=-1$

Thus, the common difference is

$-1$

The sum of all

$2$ -digit odd number is

0%
$2475$
0%
$2530$
0%
$4905$
0%
$5049$

Explanation

We have to find the sum

$11+13+15+...+99.$

Clearly, the terms of this series form an AP with

first term

$a=11$ ,

common difference

$d=2$ and

last term

$l=a_n=99$

$\Rightarrow a+(n-1)d=99\Rightarrow 11+(n-1)2=99 \Rightarrow n=45$

$\therefore$ Required sum

$=S_n=\dfrac n2[a+l]$

$=\dfrac{45}2[11+99]=45\times55=2475$

Therefore, option A is correct.

The first term of an A.P. of consecutive integers is

$p$

$^2$

$+$ The sum

$2p + 1$ terms of this series can be expressed as

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

Explanation

$\textbf{Hint: Sum of n terms of an AP is}$

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

Correct Option:

$\textbf{B}$

Solution:

$\textbf{Step 1: Find the common difference of the given A.P}$

It is given that the first term of an A.P of cosecutive integers is

$\ a=p^{2}+1$

$\because$ given A.P is of consecutive integers

$\therefore$ common difference

$\ d=1$

$\textbf{Step 2: Use the formula of sum of n terms of an AP.}$

$\because$ Sum of n terms of an A.P is

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

$\therefore$ Sum of

$\ 2p+1$ terms of an A.P is

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

$\Rightarrow$

$S_{2p+1} = \dfrac {2p+1}{2}[2 p^{2}+2+2p]$

$\Rightarrow$

$S_{2p+1} = \left ( 2p+1\right)\left (p^{2}+1+p\right)$

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

The sum of first 24 terms of the sequence whose n

$^{th}$ term is given by

${a}_{n} = 3 + \displaystyle\frac{2n}{3}$ , is

0%
$278$
0%
$272$
0%
$270$
0%
$268$

Explanation

Given,

${a}_{n} = 3 + \displaystyle\frac{2n}{3}$

Hence,

${a}_{n+1} = 3 + \displaystyle\frac{2n+2}{3}$

$\therefore\$ Common difference,

$d={a}_{n+1}-{a}_{n}$

$=3 + \displaystyle\frac{2n+2}{3}-3 - \displaystyle\frac{2n}{3}$

$=\dfrac{2}{3}$

First term,

$a = {a}_{1} = 3 + \displaystyle\frac{2}{3}$

$= \displaystyle\frac{11}{3}$

We know that, sum of

$n$ terms of an arithmetic progression is

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

$\Rightarrow {S}_{24} = \displaystyle\frac{24}{2}\left[\displaystyle\frac{22}{3} + 23\left(\displaystyle\frac{2}{3}\right)\right]$

$= 12\left[\displaystyle\frac{22 + 46}{3}\right]$

$= \displaystyle\frac{12\times 68}{3}$

$=4\times 68$

$=272$

Hence, the sum of first

$24$ terms of the given series is

$272$ .

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 8 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 8 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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