Arithmetic Progressions Test

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Arithmetic Progressions Test - 32

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Question 1
1 / -0

The common difference of the A.P. $$3, 5, 7,....$$ is _________.

A
$$4$$

B
$$2$$

C
$$3$$

D
$$-2$$

Solution

$$d = a_2-a_1\\ = 5-3 = 2 $$,
where $$ a_1 $$ & $$a_2 $$ are the first and second terms of the given sequence respectively.
Question 2
1 / -0

Which term of the sequence 4, 9, 14, 19, ..., is 124 ?

A
25

B
30

C
15

D
35

Solution

The given sequence is an A.P. with first term $$a=4$$ and common difference $$d=5$$. Let 124 be the nth term of the given sequence. Then,
$$a_n=124$$
$$a+(n-1)d=124$$
$$4+(n-1) \times 5 =124$$
$$n=25$$
Hence, 25th term of the given sequence is 124.
Question 3
1 / -0

Find the common difference of $$4,\dfrac{15}{2}, 11$$

A
$$\frac { 7 }{ 2 } $$

B
$$\frac { 2 }{ 7 } $$

C
$$\frac { 11 }{ 41 } $$

D
$$\frac { 41 }{ 11 } $$

Solution

The series is $$4,\dfrac{15}{2},11$$
The common difference is
$$\dfrac{15}{2}-4\\\dfrac{15-8}{2}=\dfrac 72$$
Question 4
1 / -0

The common difference of the A.P.: 3, 5, 7, ..... is __.

A
$$4$$

B
$$2$$

C
$$3$$

D
$$-2$$

Solution

The given AP is $$3,\ 5,\ 7,\ .....$$
To find out: The common difference of the A.P.

We know that, the common difference of an A.P. is the difference between any two consecutive terms of the A.P.
$$\therefore \ $$ The common difference $$=5-3=2$$.

Hence, the common difference of the given A.P. is $$2$$.
Question 5
1 / -0

A conference hall has nine rows of chairs which are in AP. If the sum of chairs of first five consecutive rows is $$230$$ and $$7th$$ row has $$52$$ chairs, then the number of chairs in first row is

A
$$40$$

B
$$42$$

C
$$44$$

D
$$46$$

Solution

Since, the number of chairs in the rows of a conference hall form an AP and the sum of chairs of five consecutive rows in a conference hall is 230
$$\therefore (a-2d)+(a-d)+(a)+(a+d)+(a+2d) = 230$$
$$\implies 5a= 230 \implies a = 46$$
Also, by given we have $$a_7 = 52 \implies a+6d = 52 \implies 46 +6d = 52 \implies 6d = 6 \implies d = 1$$
Thus, the number of chairs in consecutive five rows are $$44, 45, 46, 47, 48, 48, 50, 51, 52$$
So, the number of chairs in first row $$=44$$.
Question 6
1 / -0

Riya arranged her magazines collection in three columns so that the number of magazines in columns form an $$AP$$. If the sum of number of magazines in all three columns is $$18$$ and the number of magazines in last column is $$9,$$ then first column has ____ number of magazines.

A
$$2$$

B
$$3$$

C
$$4$$

D
None of these

Solution

Since, the number of magazines in columns form an AP
So by given, we have $$(a-d)+(a)+(a+d) = 18$$
$$\implies 3a = 18 \implies a =6$$
Also, the number of magazines in last column $$=9$$
$$\implies (a+d) = 9 \implies 6+d = 9 \implies d = 3$$
Thus, number of magazines in all three columns are $$3, 6, 9$$
Question 7
1 / -0

If the sum of first $$2n$$ terms of the A.P. 2,5,8,... is equal to the sum of first $$n$$ terms of the A.P. 57, 59, 61,... ,then $$n$$ equals

A
10

B
12

C
11

D
13

Solution

Sum to n terms of the an A.P is

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

Now equating sum of 2n terms of first AP to sum of n terms of second a.p we get;

$$ \frac{2n}{2}{(2a_1\;+\;(2n\;-\;1)d_1)}\;=\;\frac{n}{2}{(2a_2\;+\;(n\;-\;1)d_2)}$$

$$ \frac{2n}{2}{(4\;+\;(2n\;-\;1)3)}\;=\;\frac{n}{2}{(114\;+\;(n\;-\;1)2)} $$

On solving, we get $$n =11$$
Question 8
1 / -0

If $$m^{th}$$ term of an A.P. is $$n$$ and $$n$$th term is $$m$$, then the $$r^{th}$$ term is

A
$$m + n - r $$

B
$$m + n + r$$

C
$$m - n + r$$

D
$$r - (m + n)$$

Solution

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
$$a_{r}=n+m-r$$.
Question 9
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

A
$$25$$

B
$$26$$

C
$$27$$

D
$$28$$

Solution

$$S_{n}=3n^{2}+5n$$
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
$$m=27$$.
Question 10
1 / -0

If $$m$$th term of an A.P. is $$n$$ and $$n$$th term is $$m$$, then the $$(m + n)$$th term is

A
$$0$$

B
$$m+n-1$$

C
$$m+n$$

D
$$\dfrac{mn}{m+n}$$

Solution

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
$$a_{m+n}=a_{1}+(m+n-1)d$$
$$=(m+n-1)-(m+n-1)$$
$$=0$$.

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Re-Attempt Weekly Quiz Competition

Arithmetic Progressions Test - 32

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Arithmetic Progressions Test - 32

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SHARING IS CARING

Question 1

$$4$$

$$2$$

$$3$$

$$-2$$

Solution

Question 2

25

30

15

35

Solution

Question 3

$$\frac { 7 }{ 2 } $$

$$\frac { 2 }{ 7 } $$

$$\frac { 11 }{ 41 } $$

$$\frac { 41 }{ 11 } $$

Solution

Question 4

$$4$$

$$2$$

$$3$$

$$-2$$

Solution

Question 5

$$40$$

$$42$$

$$44$$

$$46$$

Solution

Question 6

$$2$$

$$3$$

$$4$$

None of these

Solution

Question 7

10

12

11

13

Solution

Question 8

$$m + n - r $$

$$m + n + r$$

$$m - n + r$$

$$r - (m + n)$$

Solution

Question 9

$$25$$

$$26$$

$$27$$

$$28$$

Solution

Question 10

$$0$$

$$m+n-1$$

$$m+n$$

$$\dfrac{mn}{m+n}$$

Solution

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