Arithmetic Progressions Test

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Arithmetic Progressions Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

Find the sum of odd numbers between $$0$$ and $$50$$.

A
$$800$$

B
$$745$$

C
$$625$$

D
$$520$$
Question 2
1 / -0

Find the A.P. whose sum to $$n$$ terms is $$2n$$$$^2$$ $$+ n$$

A
The required A.P. is $$2, 6, 10, 14,...$$

B
The required A.P. is $$3, 7, 11, 15,...$$

C
The required A.P. is $$4, 8, 12, 16,...$$

D
The required A.P. is $$5, 10, 15, 20,...$$

Solution

Given, $$S_n=2n^2+n$$
Now, $$a_1=S_1=2(1)^2+1=3$$
$$a_2=S_2-S_1=2(2)^2+2-3=7$$
$$a_3=S_3-S_2=[2(3)^2+3]-[2(2)^2+2]=21-10=11$$
$$a_4=S_4-S_3=[2(4)^2+4]-[2(3)^2+3]=36-21=15$$ and so on.
Therefore the required A.P. is $$3,7,11,15...$$
Question 3
1 / -0

Find the $$S_{15} $$ for an AP whose $$nth$$ term is given by $$a_n$$ $$= 3 + 4n$$.

A
$$420$$

B
$$525$$

C
$$630$$

D
$$675$$

Solution

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$$S_{15}=\frac{15}2[2a+(15-1)d]$$
$$\Rightarrow S_{15}=\displaystyle \frac{15}2[2\times7+14\times4]$$
$$=15\times35$$
$$=525$$
Question 4
1 / -0

In an AP, given $$a_n = 4, d = 2, S_n = -14,$$ find $$n$$ and $$a$$.

A
$$n=11$$ and $$a=-3$$

B
$$n=10$$ and $$a=-3$$

C
$$n=7$$ and $$a=-8$$

D
$$n=10$$ and $$a=3$$

Solution

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
$$\therefore a=6-2\times7=-8$$
Question 5
1 / -0

How many terms of the A.P.: $$9, 17, 25,...$$ must be taken to give a sum of $$636$$?

A
$$6$$

B
$$8$$

C
$$12$$

D
$$16$$

Solution

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 $$n=12$$
Question 6
1 / -0

In an AP,
Given $$l = 28, S = 144$$ and there are total $$9$$ terms. Find $$a$$.

A
$$a=3$$

B
$$a=4$$

C
$$a=5$$

D
$$a=7$$

Solution

Given, $$l=28, S= 144, n=9$$
$$\therefore S=144\Rightarrow S_9$$

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

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

$$a+28=32$$
$$\therefore a=4$$
Question 7
1 / -0

Find the sum of first $$51$$ terms of an AP, whose second and third terms are $$14$$ and $$18$$ respectively.

A
$$5610$$

B
$$5800$$

C
$$6120$$

D
$$6680$$

Solution
Question 8
1 / -0

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.

A
$$15^{th}$$ term of AP is $$-14$$

B
$$15^{th}$$ term of AP is $$-86$$

C
$$15^{th}$$ term of AP is $$-46$$

D
$$15^{th}$$ term of AP is $$-35$$

Solution

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 $$-46$$.
Question 9
1 / -0

Write first four terms of the AP, when the first term $$a$$ and the common difference $$d$$ are given as follows $$a = -1.25, d = -0.25.$$

A
First four terms of the given AP are $$-1.25, -1.70, -1.95, -2.00$$

B
First four terms of the given AP are $$-1.25, -1.50, -1.75, -2.00$$

C
Not an AP

D
None of these

Solution

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 $$-1.25,-1.50,-1.75,-2.00$$
Question 10
1 / -0

Which term of the AP : $$3,8, 13, 18,.. $$is $$78?$$

A
$$t_{16}$$

B
$$t_{18}$$

C
$$t_{14}$$

D
None of these

Solution

Given :- A.P. is $$3, 8 , 13, 18, ....$$

To get :- which term is $$78$$.

solution :-

First term $$= 3 = a_1$$

common diff. $$= d = (8 - 3) = (13 - 8) = (18 - 13) = 5$$

$$\therefore $$ Let the required term be n

$$\therefore a_n = 78$$

$$\therefore a_n = a_1 + (n - 1) d$$

$$\Rightarrow 78 = 3 + (n - 1) d$$

$$\Rightarrow 78 = 3 + (n - 1) 5$$

$$\Rightarrow 78 = 3 + 5n - 5$$

$$\Rightarrow 78 = -2 + 5n$$

$$\Rightarrow 5n = 78 + 2 = 80$$

$$\Rightarrow 5n = 80$$

$$\Rightarrow n = 16$$

$$\therefore$$ so the required term is $$16^{th}$$ term.

i.e. $$t_{16}$$

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Arithmetic Progressions Test - 41

Result

Arithmetic Progressions Test - 41

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Question 1

$$800$$

$$745$$

$$625$$

$$520$$

Question 2

The required A.P. is $$2, 6, 10, 14,...$$

The required A.P. is $$3, 7, 11, 15,...$$

The required A.P. is $$4, 8, 12, 16,...$$

The required A.P. is $$5, 10, 15, 20,...$$

Solution

Question 3

$$420$$

$$525$$

$$630$$

$$675$$

Solution

Question 4

$$n=11$$ and $$a=-3$$

$$n=10$$ and $$a=-3$$

$$n=7$$ and $$a=-8$$

$$n=10$$ and $$a=3$$

Solution

Question 5

$$6$$

$$8$$

$$12$$

$$16$$

Solution

Question 6

$$a=3$$

$$a=4$$

$$a=5$$

$$a=7$$

Solution

Question 7

$$5610$$

$$5800$$

$$6120$$

$$6680$$

Solution

Question 8

$$15^{th}$$ term of AP is $$-14$$

$$15^{th}$$ term of AP is $$-86$$

$$15^{th}$$ term of AP is $$-46$$

$$15^{th}$$ term of AP is $$-35$$

Solution

Question 9

First four terms of the given AP are $$-1.25, -1.70, -1.95, -2.00$$

First four terms of the given AP are $$-1.25, -1.50, -1.75, -2.00$$

Not an AP

None of these

Solution

Question 10

$$t_{16}$$

$$t_{18}$$

$$t_{14}$$

None of these

Solution

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