Arithmetic Progressions Test

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Arithmetic Progressions Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

$${\log _5}2,\,\,{\log _6}\,2,\,\,{\log _{12}}\,\,2\,\,$$ are in

A
A.P

B
G.P

C
H.P

D
none of these
Question 2
1 / -0

$$
\ln \text { an } A . P \mathrm{t}_{7}=4, \mathrm{d}=-4 \text { and } n=101, \text { find the value of } a
$$

A
404

B
912

C
95

D
393
Question 3
1 / -0

The average of certain first consecutive even number is $$101$$. Find their sum?

A
$$25,000$$

B
$$33,600$$

C
$$10100$$

D
$$24,960$$

Solution
Question 4
1 / -0

In an A.P., $$S_p=q$$ and $$S_q=p$$, where $$ S_r $$ denotes the sum of the first $$r$$ terms of an A.P. Then, the value of $$S_{p+q}$$ is

A
$$0$$

B
$$-(p+q)$$

C
$$p+q$$

D
$$pq$$

Solution

Let the first term of the sequence be $$a$$ and its common difference be $$d$$.
Applying sum to $$n$$ terms formula of A.P.,
$$S_p = \dfrac{p}2(2a+(p-1)d) = q$$
On expanding and simplifying the above equation, we get
$$2ap + p^2d - pd = 2q$$ ...(1)

Similarly, for $$S_q$$
$$2aq + q^2d - qd = 2p$$ ...(2)

Subtracting (2) from (1), we get
$$2a (p - q) + d(p^2 - q^2) - d(p - q) = -2(p - q)$$
Dividing both sides by $$(p - q)$$, we get
$$2a + d(p + q) - d = -2$$
$$\Rightarrow {2a + (p + q - 1)d} = -2$$ ...(3)

Sum to $$p+q$$ terms formula of an A.P. is
$$S_ {(p+q)} = \dfrac{p+q}{2}[2a + (p + q - 1)d] $$ ...(4)

Substituting (3) in (4), we get
$$S_{p+q} = \dfrac{p+q}{2}\times(-2) = -(p + q)$$

Hence, option B is the correct answer.
Question 5
1 / -0

Which of the following are APs ? If they an AP. find the common difference of and write three more terms.

A
2.4,8,16...

B
$$2,\frac { 5 }{ 2 } ,3,\frac { 7 }{ 2 } ,...$$

C
-1.2, -3.2, -5.2, -7.2,...

D
-10, -6, -2, 2, ...

E
3, 3 + $$\sqrt { 2 } $$, 3 + $$2\sqrt { 2 } $$, 3 + $$3\sqrt { 2 } $$
Question 6
1 / -0

If $$a_1, a_2, a_3, ....$$ is an A.P. such that $$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$$ then $$a_1+a_2+a_3+ ... +a_{23}+a_{24}$$ is equal to-

A
$$909$$

B
$$75$$

C
$$750$$

D
$$900$$

Solution

Here we use the following formula to calculate the sum of first n terms of an AP:
$$S_{n}=\frac {n}{2}(2a+(n-1)d)$$
Since $$a_1+a_5+a_{10}+a_{15}+a_{20}+a_{24}=225$$
$$\Rightarrow 6a+69d=225$$
$$\Rightarrow 2a+23d=75$$
Now $$ S_{24} = a_1+a_2+a_3+ ... +a_{23}+a_{24}$$
$$S_{24} =12(2a+23d)$$
$$=900$$
Question 7
1 / -0

The sum of digits of all numbers from 1 to 300 is equal to

A
3000

B
3003

C
3033

D
none of these

Solution
Question 8
1 / -0

Let $$(1+x)^n=\sum _{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r } } $$. Then $$(1+\frac {a_1}{a_0})(+\frac {a_2}{a_1})....(1+\frac {a_n}{a_{n-1}})$$ is equal to:

A
$$\frac {(n+1)^{n+1}}{n!}$$

B
$$\frac {(n+1)^{n}}{n!}$$

C
$$\frac {(n)^{n-1}}{(n-1)!}$$

D
$$\frac {(n+1)^{n+1}}{(n-1)!}$$
Question 9
1 / -0

Choose the correct alternative for questions:
For an A.P., if d=11, then $${ t }_{ 17 }-{ t }_{ 15 }=?$$

A
2

B
22

C
33

D
13

Solution
Question 10
1 / -0

What is the common difference of an A.P. whose nth term is $${ t }_{ n }=2n-4$$

A
3

B
2

C
-2

D
-3

Solution