Arithmetic Progressions Test

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Arithmetic Progressions Test - 37

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Weekly Quiz Competition

Question 1
1 / -0

If sum of $$n$$ terms of an A.P. is $$3n^2+5n$$ and $$T_m=164 $$, what is the value of m?

A
$$26$$

B
$$27$$

C
$$28$$

D
$$25$$

Solution

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
$${ T }_{ n }=6n+2 = 164$$
$$n = 27$$
Question 2
1 / -0

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.

A
$$72$$

B
$$68$$

C
$$54$$

D
$$89$$

Solution

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 $$row25 = 68$$.
Question 3
1 / -0

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

A
12

B
18

C
25

D
21

Solution

Here first term $$a=5$$ and common difference
$$d=12-5=7$$
Suppose nth term of the A.P. is 145
Now, $$T_n = a+(n-1)d$$
$$\therefore 145 =5 (n-1) (7)$$
$$\therefore 145-5 = 7(n-1)$$
$$\therefore 140 =7 (n-1)$$
$$\therefore 20 = n-1 \therefore n = 21$$
Question 4
1 / -0

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

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

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

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

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

Solution

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

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

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

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

$$ => d = \displaystyle \frac {2(S-na)}{n(n-1)} $$
Question 5
1 / -0

Find the sum of first $$11$$ positive numbers which are multiples of $$6$$.

A
$$314$$

B
$$396$$

C
$$452$$

D
$$245$$

Solution

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.
Question 6
1 / -0

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.

A
Rs. 49

B
Rs. 51

C
Rs. 53

D
Rs. 55

Solution

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 $$51$$.
Question 7
1 / -0

Find the sum of $$A.P.$$ whose first and last term is $$13$$ and $$216$$ respectively & common difference is $$7$$.

A
$$3434$$

B
$$3435$$

C
$$1545$$

D
$$3456$$

Solution

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 $$=3435$$.
Question 8
1 / -0

If $$t_{11}$$ and $$t_{16}$$ for an $$A.P.$$ are respectively $$38$$ and $$73$$, then $$t_{31}$$ is $$........$$

A
$$178$$

B
$$177$$

C
$$176$$

D
$$175$$

Solution

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.
Question 9
1 / -0

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

A
40

B
50

C
14

D
51

Solution

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.
$$\therefore n=40 $$
Question 10
1 / -0

The $$n^{th}$$ term of the sequence $$\displaystyle\frac{1}{p}$$, $$\displaystyle\frac{1 + 2p}{p}$$, $$\displaystyle\frac{1 + 4p}{p}$$,... is

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

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

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

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

Solution

Sequence, $$\displaystyle\frac{1}{p}$$, $$\displaystyle\frac{1 + 2p}{p}$$, $$\displaystyle\frac{1 + 4p}{p}$$........

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

$$\Rightarrow$$ $$d=\displaystyle\frac{1 + 2p - 1}{p}$$
$$\Rightarrow$$ $$d =2$$
$$\therefore t_n$$ $$=$$ $$\displaystyle\frac{1}{p} + (n - 1)2 $$ $$=$$ $$\displaystyle\frac{1}{p}$$ + $$\displaystyle\frac{2n - 2}{1}$$ $$\Rightarrow$$ $$t_n$$

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

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Arithmetic Progressions Test - 37

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Arithmetic Progressions Test - 37

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SHARING IS CARING

Question 1

$$26$$

$$27$$

$$28$$

$$25$$

Solution

Question 2

$$72$$

$$68$$

$$54$$

$$89$$

Solution

Question 3

12

18

25

21

Solution

Question 4

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

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

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

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

Solution

Question 5

$$314$$

$$396$$

$$452$$

$$245$$

Solution

Question 6

Rs. 49

Rs. 51

Rs. 53

Rs. 55

Solution

Question 7

$$3434$$

$$3435$$

$$1545$$

$$3456$$

Solution

Question 8

$$178$$

$$177$$

$$176$$

$$175$$

Solution

Question 9

40

50

14

51

Solution

Question 10

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

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

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

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

Solution

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