Arithmetic Progressions Test

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Arithmetic Progressions Test - 29

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Question 1
1 / -0

In an Arithmetic sequence, $$S_{n}$$ represents the sum to $$n$$ terms, what is $$S_{n} - S_{n - 1}$$?

A
$$t_{1} + t_{2} + .... t_{n - 1}$$

B
$$S_{n - 1}$$

C
$$\displaystyle \sum_{n = 1}^{n - 2} t_{n}$$

D
$$t_{n}$$

Solution

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

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

$$=\dfrac{1}{2}[2an+n^{2}d-nd-2an+2a-n^{2}d+3nd-2d]$$

$$=\dfrac{1}{2}[2a+2nd-2d]$$

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

$$=t_{n}$$
$$(Ans \to D)$$
Question 2
1 / -0

Find the sum of integers from $$1$$ to $$35$$.

A
$$1160$$

B
$$630$$

C
$$1360$$

D
$$1460$$

Solution

Given, $$n = 35$$
We know $$S_{n}=\dfrac{n}{2}[n+l]$$
$$S_{35}=\dfrac{35}{2}[35+1]$$
$$S_{35}=\dfrac {35[36]}{2}$$
$$S_{35}=\dfrac {1260}{2}=630$$
Question 3
1 / -0

The sum of first n terms of an A.P. is

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

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

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

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

Solution

The sum of first n terms of an A.P. is $$S_{n}=\frac{n}{2}[2a+(n-1)d]$$
Question 4
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If $$t_{n}$$ be the $$n^{th}$$ term of the A.P. $$-9, -14, -19, ....,$$ what is the value of $$t_{30} - t_{20}$$?

A
$$-35$$

B
$$-50$$

C
$$-55$$

D
$$-65$$

Solution

Given series: $$-9,\ -14,\ -19,\ .\ .\ .$$

Here $$a=-9$$

and $$d=(-14)-(-9)=-5$$

$$t_{30}=(-9)+(30-1)(-5)$$

$$=-154$$

$$t_{20}=(-9)+(20-1)(-5)$$

$$=-104$$

$$\therefore t_{30}-t_{20}=(-154)-(-104)$$

$$=-50$$
Question 5
1 / -0

What is the sum of 12 odd numbers $$1, 3, 5, 7, 9.....?$$

A
$$12$$

B
$$144$$

C
$$141$$

D
$$124$$

Solution

First term of the given arithmetic series $$= 1$$
Second term of the given arithmetic series $$= 3$$
Third term of the given arithmetic series $$= 5$$
Fourth term of the given arithmetic series $$= 7$$
Now, Second term - First term $$= 3 - 1 = 2$$
Third term - Second term $$= 5 - 3 = 2$$
Therefore, common difference of the given arithmetic series is $$2.$$
The number of terms of the given $$A. P.$$ series $$(n) = 12$$
We know that the sum of first n terms of the Arithmetic Progress, whose first term$$ = a$$ and common difference$$ = d$$ is
$$S_{n}=\dfrac{n}{2}[2a+(n-1)d]$$

$$S_{12}=\dfrac{12}{2}[2\times 1+(12-1)2]$$

$$S_{12}=6[2+22]$$

$$S_{12}=144$$
Question 6
1 / -0

What is the nth term of the arithmetic sequence $$1, 3, 5, 7, 9, 11...?$$

A
$$n + 1$$

B
$$n - 1$$

C
$$2n + 1$$

D
$$2n - 1$$

Solution

Clearly, the difference of successive terms of above sequence is constant which is 2
So given sequence is in AP with first term $$1$$ and common difference $$2$$
Hence general term is, $$a_n = a+(n-1)d=1+(n-1)2=2n-1$$
Hence option (D) is correct choice
Question 7
1 / -0

Find the nth term of an arithmetic sequence $$11, 22, 33, 44...$$.

A
$$n$$

B
$$n - 1$$

C
$$2n$$

D
$$11n$$

Solution

Clearly, the difference of successive terms of above sequence is constant which is 11
So given sequence is in AP with first term 11 and common difference $$11$$
Hence general term is, $$a_n = a+(n-1)d=11+(n-1)11=11n$$
Hence option 'D' is correct choice
Question 8
1 / -0

If the $$nth$$ term of AP is $$2n+5$$. Then find the $$AM$$ of first and last terms.

A
$$99$$

B
$$98$$

C
$$100$$

D
$$44$$

Solution

Given:
$$t_{n}=2n+5$$
First term $$=2\times 1+5=7$$
Last term $$=2\times 38+5=81$$
$$\therefore$$ Arithmetic mean$$(A.M).=\dfrac{7+81}{2}=44$$
Question 9
1 / -0

Obtain the sum of the first $$56$$ terms of an A.P. whose $$28^{th}$$ and $$29^{th}$$ terms are $$52$$ and $$148$$ respectively.

A
$$2100$$

B
$$5600$$

C
$$5200$$

D
$$2600$$

Solution

We know that, $${t}_{n}=a+(n-1)d$$

Here, $${t}_{28}=a+(28-1)d$$

$$\therefore$$ $$52=a+27d.......(i)$$

Also, $${t}_{29}=a+(29-1)d$$

$$\therefore$$ $$148=a+28d........(ii)$$

Adding equation $$(i)$$ and $$(ii)$$, we get:
$$a+27d=52$$
$$a+28d=148$$
----------------------
$$2a+55d=200.............(iii)$$

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

$${ S }_{ 56 }=\cfrac { 56 }{ 2 } \left[ 2\times a+(56-1)d \right] $$

$$\Rightarrow$$ $${S}_{56}=28(2a+55d)$$

$$\Rightarrow$$ $${S}_{56}=28\times 200$$

$$\Rightarrow$$ $${S}_{56}=5600$$

$$\therefore\ $$ The sum of the $$56$$ terms of the given A.P. is $$5600$$.
Question 10
1 / -0

If $$n^{th}$$ term of AP is $$4n+1$$, then AM of $$11^{th}$$ to $$ 20^{ th}$$ terms is

A
$$61.5$$

B
$$63$$

C
$$63.5$$

D
$$62$$

Solution

Given: $$t_{n}=4n+1$$
$$A.M.=\cfrac{t_{11}+t_{20}}{2}$$
$$=\cfrac{4\times 11+1+4\times 20+1}{2}$$
$$=63$$

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Arithmetic Progressions Test - 29

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Arithmetic Progressions Test - 29

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Question 1

$$t_{1} + t_{2} + .... t_{n - 1}$$

$$S_{n - 1}$$

$$\displaystyle \sum_{n = 1}^{n - 2} t_{n}$$

$$t_{n}$$

Solution

Question 2

$$1160$$

$$630$$

$$1360$$

$$1460$$

Solution

Question 3

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

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

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

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

Solution

Question 4

$$-35$$

$$-50$$

$$-55$$

$$-65$$

Solution

Question 5

$$12$$

$$144$$

$$141$$

$$124$$

Solution

Question 6

$$n + 1$$

$$n - 1$$

$$2n + 1$$

$$2n - 1$$

Solution

Question 7

$$n$$

$$n - 1$$

$$2n$$

$$11n$$

Solution

Question 8

$$99$$

$$98$$

$$100$$

$$44$$

Solution

Question 9

$$2100$$

$$5600$$

$$5200$$

$$2600$$

Solution

Question 10

$$61.5$$

$$63$$

$$63.5$$

$$62$$

Solution

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