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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 5 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 5
Choose the correct statement(s):
$$A$$: Every sequence is a progression.
$$B$$: Every progression is a sequence.
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Only $$A$$
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Only $$B$$
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Both $$A$$ and $$B$$
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None
Explanation
The difference between a progression and a sequence is that a progression has a specific rule to calculate its next term from its previous term, whereas a sequence can be based on a logical rule like 'a group of prime numbers'.
Thus, every progression is a sequence but every sequence is not a sequence.
Find the sum of first $$32$$ multiples of $$4$$.
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$$2,112$$
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$$2,712$$
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$$2,110$$
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$$2,111$$
Explanation
The multiples of $$4$$ are, $$4, 8, 12, 16,$$ ............
Therefore, $$a= 4, d = 4, n=32$$
We have $$S_{32} = \dfrac{n}{2} [2a + (n - 1)d]$$
$$= \dfrac{32}{2} [2 \times 4 + (32 - 1)4]$$
$$= 16 [8 + 31 \times 4]$$
$$= 16 [8 + 124]$$
Therefore, $$ S_{20} = 2,112$$
Constant is subtracted from each term of an A.P. the resulting sequence is also an ______
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0%
Geometric Sequence
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Arithmetic Progression
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Both A and B
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None of the above
Explanation
If a constant is subtracted from each term of an A.P. the resulting sequence is also an arithmetic progression.
For example : Let the sequence be $$\{ a_1, a_2, a_3 ...... a_4 \}$$ be an arithmetic progression with common difference $$d$$.
Again subtract $$k$$ to the above sequence.
Then the resulting sequence is $$a_1 -k, a_2 - k, a_3 - k ...... $$ be an A.P.
Find the difference of an A.P. $$a_n$$, if $$a_4 = 16$$ and $$a_2 = 8$$
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$$3$$
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$$4$$
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$$2$$
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$$1$$
Explanation
Given, $$a_4=16, a_2=8$$
$$a_4 = a + 3d=16$$ ...... (1)
$$a_2 = a + d=8$$ ...... (2)
Multiplying equation (2) by $$3$$ and then subtract the equation, we get
$$16 = a + 3d$$
$$24 = 3a + 3d$$
(-) (-) (-)
----------------
$$-8 = - 2a$$
$$a = 4$$
Put $$a= 4$$ in equation (1), we get
$$16 = 4 + 3d$$
$$\Rightarrow 16 - 4 = 3d$$
$$\Rightarrow d = 4$$
The last term of the AP $$21, 18, 15$$ ....... is $$-351$$. Find the nth term.
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$$213$$
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$$123$$
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$$312$$
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$$-231$$
A train can travel $$200$$ m in the first hour, $$400$$ m the next hour, $$600$$ m the third hour and so on in an arithmetic sequence. What is the total distance the train travels in $$5$$ hours?
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$$2,000$$
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$$3,000$$
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$$4,000$$
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$$5,000$$
Explanation
The sequence is $$200, 400, 600 .....$$
$$a = 200, d = 200$$
First find the common difference:
$$a_n = a_1 + (n - 1) d$$
$$a_5 = 200 + (5- 1) 200$$
$$= 200 + 4 \times 200$$
$$a_5 = 1,000$$
We know that, $$S_n = \dfrac{n}{2} $$ [First term $$+$$ Last term]
$$= \dfrac{5}{2} [200 + 1,000]$$
$$= 2.5 [1,200]$$
$$S_5 = 3,000$$
Hence, the train will travel $$3,000$$ m in $$5$$ hours.
Find the sum of first $$20$$ multiples of $$13$$.
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$$2,720$$
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$$2,730$$
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$$2,740$$
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$$2,750$$
Explanation
The multiples of $$13$$ are, $$13, 26, 39, .....$$
Therefore, $$a = 13, d = 13,n=20$$
We have $$S_{20} = \dfrac{n}{2} [2a + (n - 1)d]$$
$$= \dfrac{20}{2} [2 \times 13 + (20 - 1)13]$$
$$= 10 [26 + 19 \times 13]$$
$$ = 10 [26 + 247]$$
Therefore, $$S_{20} = 2,730$$
Find the sum of the first 20 terms of the arithmetic series if $$a_{1} = 10$$ and $$a_{20}=100$$.
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0%
1000
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1100
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1200
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1300
Explanation
To find the sum of the first n terms of an arithmetic series use the formula,
$$S_{n}=\dfrac{n(a_{1}+a_{n})}{2}$$
$$S_{20}=\dfrac{20(10+100)}{2}$$
$$S_{20}=\dfrac{2200}{2}$$
$$S_{20}=1100$$
What is the sum of $$t_n= (2n-5)$$ from n =10 to 150?
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$$22650$$
0%
$$21855$$
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$$23250$$
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$$21250$$
Explanation
$$T_n = (2n-5)$$
$$T_{10} = 2\times10-5 = 15$$
$$T_{150} = 2\times15- 5 = 295$$
Sum of AP, if first terms and last term is given:
$$S = \dfrac n2(\text{first term } +\text{ last term})$$
Here, $$n = 150-10+1 = 141$$
$$S = \dfrac{141}2(15+295)$$
$$S = 141\times155 = 21855$$
Calculate the $$15^{\text{th}}$$ term of the A.P. $$ -3, -4, -5, -6, -7....$$
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0%
$$-13$$
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$$-15$$
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$$-17$$
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$$-19$$
Explanation
Let the general form of A.P., $$a_n = a + (n-1)d$$
Common difference, $$d = -1$$
$$a_{15}=?$$
$$a_n = a + (n-1)d$$
$$a_{15} = -3 + (15-1)\times -1$$
$$a_{15} = -3 -14$$
$$a_{15} = -17$$
Find the $$8^{th}$$ term of the A.P. : $$11, 14, 17, 20.....$$
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$$11$$
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$$-17$$
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$$17$$
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$$32$$
Explanation
Given series is $$11,14,17,20,....$$
Let the general form of A.P., $$a_n = a + (n-1)d$$
Common difference, $$d = 3$$ and first term $$a=11$$
$$a_{8}=?$$
$$a_n = a + (n-1)d$$
$$a_{8} = 11 + (8-1)\times 3$$
$$a_{8} = 11 +21$$
$$a_{8} = 32$$
3, 5, 7, 9, 11, 13, 15.... is an
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0%
Geometric progression
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Harmonic progression
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Arithmetic progression
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None of above
Explanation
3, 5, 7, 9, 11, 13, 15.... is an arithmetic progression.
Here the common difference between two consecutive terms is 2.
A sequence in which the difference between any two consecutive terms is a constant is called as arithmetic progression.
A sequence of numbers in which each term is related to its predecessor by same law is called
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0%
arithmetic series
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progression
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geometric series
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none of these
Explanation
A sequence of numbers in which each term is related to its predecessor by same law is called progression
Example: 1, 2, 3, 4.... is an example of sequence or progression.
Since the given sequence follows a same rule or law through out the sequence and there is a relation between each term and it's previous one.
_______ is a series of successive events.
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Series
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Preceding sequence
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Progression
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Geometric progression
Explanation
Progression is a series of successive events.
Which of the following is not in the form of A.P.?
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$$1, -1, -3, -5, -7...$$
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$$0, 3, 6, 9, 12...$$
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$$4, 5, 7, 10, 14...$$
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$$-1, 2, 5, 8, 11..$$
Explanation
For any series to be in Arithmetic Progression, the common difference (i.e., the difference between any two consecutive terms should be the same).
$$4, 5, 7, 10, 14...$$ is not in the form of A.P.
Here the common difference is not constant.
________ can be defined as arrangement of terms in which sequence of terms follow some conditions.
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Series
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Preceding sequence
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Progression
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Geometric progression
Explanation
Progression can be defined as arrangement of terms in which sequence of terms follow some conditions.
For given A.P. $$-\dfrac{1}{2}, -\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{3}{2}, ..$$ find the common difference.
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$$-1$$
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$$-\dfrac{1}{2}$$
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$$\dfrac{3}{2}$$
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$$1$$
Explanation
The general form of A.P. is $$a, a + d, a + 2d, a + 3d.....$$
$$-\dfrac{1}{2}, -\dfrac{3}{2}, \dfrac{1}{2}, -\dfrac{3}{2}, ..$$
Here the common difference is $$-1$$.
Which of the following is in the form of $$A.P$$?
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$$1, -1, -3, -5, -7,\dots$$
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$$0, 3, 2, 1, -2,\dots$$
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$$4, 5, 7, 10, 14,\dots$$
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$$-2, 2, -2, 2, -2,\dots$$
Explanation
$$1, -1, -3, -5, -7$$ is in the form of $$A.P$$ because the common difference between consecutive terms of this sequence is constant and that is equal to $$-2$$.
Find the sum of all the odd positive integers less than $$100$$.
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$$2400$$
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$$2500$$
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$$2525$$
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$$2600$$
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$$2650$$
Explanation
The possible odd positive integers: $$1,3,5,7,9,11..........99$$
The first term, $$a$$ $$=1$$
The difference between two consecutive terms $$=3-1=2$$
The total no. of terms $$=50$$
Applying sum of arithmetic progression formula,
$$S_n=\dfrac{n}{2}[2a+(n-1)d]$$
$$=\dfrac{50}{2}[2\times1+(50-1)2]$$
$$=25(2+98)=2500$$
Hence, choice B is correct.
In an Arithmetic sequence, $$S_{n}$$ represents the sum to $$n$$ terms, what is $$S_{n} - S_{n - 1}$$?
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$$t_{1} + t_{2} + .... t_{n - 1}$$
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$$S_{n - 1}$$
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$$\displaystyle \sum_{n = 1}^{n - 2} t_{n}$$
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$$t_{n}$$
Explanation
We know,
$$S_{n}=\dfrac{n}{2}[2a+(n-1)d]$$
$$\therefore S_{n-1}=\dfrac{n-1}{2}[2a+(n-2)d]$$
$$\therefore S_{n}-S_{n-1}=\dfrac{n}{2}[2a+(n-1)d]-\dfrac{n-1}{2}[2a+(n-2)d]$$
$$=\dfrac{1}{2}[2an+n^{2}d-nd-2an+2a-n^{2}d+3nd-2d]$$
$$=\dfrac{1}{2}[2a+2nd-2d]$$
$$=a+(n-1)d]$$
$$=t_{n}$$
$$(Ans \to D)$$
The sum of first n terms of an A.P. is
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$$S_{n+1}=\frac{n}{2}[2a+(n-1)d]$$
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$$S_{n}=\frac{n}{2}[2a+(n-1)d]$$
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$$S_{n}=\frac{n}{2}[a+(n-1)d]$$
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$$S_{n}=\frac{n}{2}[2a+(n+1)d]$$
Explanation
The sum of first n terms of an A.P. is $$S_{n}=\frac{n}{2}[2a+(n-1)d]$$
Find the sum of integers from $$1$$ to $$35$$.
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$$1160$$
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$$630$$
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$$1360$$
0%
$$1460$$
Explanation
Given, $$n = 35$$
We know $$S_{n}=\dfrac{n}{2}[n+l]$$
$$S_{35}=\dfrac{35}{2}[35+1]$$
$$S_{35}=\dfrac {35[36]}{2}$$
$$S_{35}=\dfrac {1260}{2}=630$$
If $$t_{n}$$ be the $$n^{th}$$ term of the A.P. $$-9, -14, -19, ....,$$ what is the value of $$t_{30} - t_{20}$$?
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0%
$$-35$$
0%
$$-50$$
0%
$$-55$$
0%
$$-65$$
Explanation
Given series: $$-9,\ -14,\ -19,\ .\ .\ .$$
Here $$a=-9$$
and $$d=(-14)-(-9)=-5$$
$$t_{30}=(-9)+(30-1)(-5)$$
$$=-154$$
$$t_{20}=(-9)+(20-1)(-5)$$
$$=-104$$
$$\therefore t_{30}-t_{20}=(-154)-(-104)$$
$$=-50$$
What is the sum of 12 odd numbers $$1, 3, 5, 7, 9.....?$$
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$$12$$
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$$144$$
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$$141$$
0%
$$124$$
Explanation
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
$$S_{n}=\dfrac{n}{2}[2a+(n-1)d]$$
$$S_{12}=\dfrac{12}{2}[2\times 1+(12-1)2]$$
$$S_{12}=6[2+22]$$
$$S_{12}=144$$
What is the nth term of the arithmetic sequence $$1, 3, 5, 7, 9, 11...?$$
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$$n + 1$$
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$$n - 1$$
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$$2n + 1$$
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$$2n - 1$$
Explanation
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
Find the nth term of an arithmetic sequence $$11, 22, 33, 44...$$.
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$$n$$
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$$n - 1$$
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$$2n$$
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$$11n$$
Explanation
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
If the $$nth$$ term of AP is $$2n+5$$. Then find the $$AM$$ of first and last terms.
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0%
$$99$$
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$$98$$
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$$100$$
0%
$$44$$
Explanation
Given:
$$t_{n}=2n+5$$
First term $$=2\times 1+5=7$$
Last term $$=2\times 38+5=81$$
$$\therefore$$ Arithmetic mean$$(A.M).=\dfrac{7+81}{2}=44$$
Obtain the sum of the first $$56$$ terms of an A.P. whose $$28^{th}$$ and $$29^{th}$$ terms are $$52$$ and $$148$$ respectively.
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$$2100$$
0%
$$5600$$
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$$5200$$
0%
$$2600$$
Explanation
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 $$5600$$.
If $$n^{th}$$ term of AP is $$4n+1$$, then AM of $$11^{th}$$ to $$ 20^{ th}$$ terms is
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0%
$$61.5$$
0%
$$63$$
0%
$$63.5$$
0%
$$62$$
Explanation
Given:
$$t_{n}=4n+1$$
$$A.M.=\cfrac{t_{11}+t_{20}}{2}$$
$$=\cfrac{4\times 11+1+4\times 20+1}{2}$$
$$=63$$
The common difference of A.P.: $$1, -1, -3, ....$$ is _______
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$$-1$$
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$$+2$$
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$$-2$$
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$$+1$$
Explanation
The common difference d of A.P. is given by the difference between second term and first term.
$$\therefore d=-1-1=-2$$
Option $$C$$ is correct.
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