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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 2 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 2
Strikers at a plant were ordered to return to work and were told they would be fined Rs. $$50$$ the first day they failed to do so, Rs. $$75$$ the second day, Rs. $$100$$ the third day, and so on. If the strikers stayed out for $$6$$ days, what was the fine for the sixth day?
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$$200$$
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$$150$$
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$$125$$
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$$175$$
Explanation
Strikers on plant fined Rs.$$50$$ on first day,& $$75$$ on second ,Rs$$100$$ on third and so on,
difference in fine from day$$1$$ to day$$2=Rs.25$$
For $${ 6 }^{ th }$$ day,$$n=6,a=50$$ and $$d=25$$,
$$=a+(n-1)d\\ =50+(6-1)25\\ =50+(5\times 25)\\ =50+125\\ =175$$
Answer is$$(D)$$
Write the sum of first five terms of the following Arithmetic Progressions where, the common difference $$d$$ and the first term $$a$$ are given:
$$a = 6, d = 6$$
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$$88$$
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$$89$$
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$$90$$
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$$91$$
Explanation
Here the first term $$a=6$$ and common difference $$d=6$$.
Now $$a_n=a+(n-1)d$$
$$\therefore a_1=a=6$$
$$a_2=a+d=6+6=12$$
$$a_3=a+2d=6+2\times6=18$$
$$a_4=a+3d=6+3\times6=24$$
$$a_5=a+4d=6+4\times6=30$$
$$\therefore$$ series in A.P is $$6,12,18,24,30,....$$
Thus, the sum of first five terms $$=6+12+18+24+30=90$$
Hence, the answer is $$90$$.
Or
We can use the formula for sum of A.P $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$\therefore S_5$$ =$$\dfrac{5}{2} \times [2(6)+(5-1)(6)]=90$$
Find the twenty fifth term of the A.P. : 12, 16, 20, 24, ..........
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108
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112
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116
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120
Explanation
Let $$a$$ be the first term and $$d$$ be the common difference of given AP.
Hence $$a=12$$ and $$d=16-12=4$$.
$$t_n=a+(n-1)d$$
$$\Rightarrow t_{25}=12+24\times4$$
$$\Rightarrow t_{25}=12+96$$
$$\Rightarrow t_{25}=108$$
Find the sum of first 20 natural numbers.
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210
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220
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230
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240
Explanation
We have to find the sum n terms of the AP 1,2,3,...
Here
$$a=1,d=1$$.
Therefore, Required sum $$S_n=\frac n2[2a+(n-1)d]=\frac n2[2\times1+(n-1)\times1]$$
$$\Rightarrow S_n=\displaystyle \frac n2[2+n-1]$$
$$\Rightarrow S_n=\displaystyle \frac{n(n+1)}2$$ ...(1)
Put $$n=20$$ in (1), we get
$$S_{20}=\displaystyle \frac{20(20+1)}2=210$$
Find $$t_{11}$$ from the given A.P. 4, 9, 14, ...
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54
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55
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51
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21
Explanation
Given series is $$4,9,14,....$$
Clearly, the given sequence is an AP with first term $$a=4$$ and the common difference $$d=5$$.
We know, $$a_n=a+(n-1)d$$
Now, $$t_{11}=a+10d$$
$$\therefore t_{11}=4+10\times5=54$$
The sum of the first $$100$$ natural numbers is
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$$-5050$$
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$$5005$$
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$$5050$$
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$$-5005$$
Explanation
We know, sum of a arithmetic progression, with first term as $$a$$ and common difference $$d$$ is
$${S}_{n}=n[a+\frac{1}{2}(n-1)d]$$
Here, given, $$a=1$$, $$d=1$$ and $$n=100$$
$${S}_{100}=100[1+\displaystyle\dfrac{1}{2}(100-1)1]$$
$$ =100[1+\dfrac{99}{2}]$$
$$=100\times \displaystyle\frac{101}{2}$$
$$=5050$$
Write the sum of the first five terms of the Arithmetic Progression for which the common difference $$d$$ and the first term $$a$$ are given below:
$$a = 5, d = 2$$
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$$40$$
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$$45$$
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$$50$$
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$$55$$
Explanation
Given: the first term $$a=5$$ and common difference $$d=2$$.
Now $$a_n=a+(n-1)d$$
$$\therefore a_1=a=5$$
$$a_2=a+d=5+2=7$$
$$a_3=a+2d=5+2\times2=9$$
$$a_4=a+3d=5+3\times2=11$$
$$a_5=a+4d=5+4\times2=13$$
$$\therefore $$ The series in A.P. is $$5,7,9,11,13,.....$$
Sum of the first five terms is $$=5+7+9+11+13=45$$
Hence, the answer is $$45$$.
Or
We can also use the formula for sum of A.P => $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$\therefore S_5$$ =$$\dfrac{5}{2} \times [2(5)+(5-1)(2)]=45$$
If the following sequence is an arithmetic progression, find its general term.
-5, 2, 9, 16, 23, 30, .......... is
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$$7n-12$$
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$$6n-12$$
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$$5n-12$$
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$$4n-12$$
Explanation
We can observe the difference between two consecutives terms is equal:
$$2-(-5)=9-2= 16-9=23-16=7$$
Hence, the sequence is an arithmetic progression with first term $$a=-5$$ and common difference $$d= 7$$.
General term of an AP is $$t_n=a+(n-1)d$$
$$t_n=-5+(n-1)\times 7$$
$$t_n=7n-12$$
Find $$S_{10}$$, if a = 6 and d = 3.
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0%
180
0%
185
0%
190
0%
195
Explanation
Given First term, $$a=6$$,
Common difference, $$d=3$$
Number of terms, $$n=10$$
Sum of $$n$$ terms is given as
$$S_n$$= $$\displaystyle \frac n2[2a+(n-1)d]$$
Substituting the values, we get
$$S_{10} =\displaystyle \frac {10}2[2\times 6+(10-1)\times3]$$
$$=5(39)$$
$$=195$$
How many terms of the sequence $$18, 16, 14,....$$ should be taken so that their sum is zero?
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$$19$$
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$$17$$
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$$18$$
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$$16$$
Explanation
Here $$a=18,d=-2$$.
Let there are $$n$$ terms so that the sum is zero.
Now, $$S_n=\dfrac n2[2a+(n-1)d]=0$$
$$\Rightarrow \dfrac n2[2\times18+(n-1)\times-2]=0$$
$$\Rightarrow 36+(n-1)\times-2=0$$
$$\Rightarrow n-1=18$$
$$\Rightarrow n=19$$
Therefore $$19$$ terms of the sequence has to be taken so that their sum is zero.
How many terms are there in the A.P.: $$-1,\displaystyle\frac{-5}{6},\displaystyle-\frac{2}{3},\displaystyle\frac{-1}{2}, ....., \displaystyle\frac{10}{3}$$ ?
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$$27$$
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$$18$$
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$$16$$
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$$9$$
Explanation
Given A.P. is $$-1,\ \dfrac{-5}{6},\ \dfrac{-2}{3},\ \dfrac{-1}{2},\ .....,\ \dfrac{10}{3}$$
Here, first term $$(a)=-1$$ and the common difference $$(d)=\displaystyle \dfrac{-5}6-(-1)=\dfrac{1}6.$$
Let $$n$$ be the number of terms of the given A.P.
We know that the $$n_{th}$$ term of an A.P. is:
$$a_n=a+(n-1)d\\$$
Here, $$a_n=\dfrac{10}3$$
$$\therefore \ a+(n-1)d=\dfrac{10}3$$
$$\Rightarrow -1+(n-1)\times\dfrac{1}6=\dfrac{10}3$$
$$\Rightarrow (n-1)\times\dfrac{1}6=\dfrac{13}3$$
$$\Rightarrow n-1=26$$
$$\Rightarrow n=27$$
Therefore, there are $$27$$ terms in the give A.P.
Find the sum of $$7 + 10\dfrac{1}{2} + 14+...+84$$.
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$$\displaystyle\frac{2093}{2}$$
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$$\displaystyle\frac{1093}{2}$$
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$$\displaystyle\frac{2193}{2}$$
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$$\displaystyle\frac{3093}{2}$$
Explanation
Clearly, terms of the given series form an AP with first term and last term $$a=7$$ & $$ l= a_n = 84$$ respectively and common difference $$d=\displaystyle \frac{21}2-7=\frac72$$.
Let the given series has $$n$$ terms.
Therefore,
$$a_n=84=a+(n-1)d$$
$$84= 7+(n-1)\times\frac72$$
$$ (n-1)\times\frac72=77$$
$$\therefore n=23$$
Here we use the following formula to calculate the sum of terms:
$$S_n =\dfrac {n}{2} (a+l) $$
$$\therefore$$
Required sum $$=S_{23}=\displaystyle \dfrac{23}2(7+84)$$
$$\Rightarrow S_{23}=\displaystyle \frac{23\times91}2=\dfrac{2093}2$$
The fourth term of an A.P. is $$4$$. Then the sum of the first $$7$$ terms is ?
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$$4$$
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$$28$$
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$$16$$
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$$40$$
Explanation
Let a and d be the first term and common difference of A.P. Then,
$$a_4=4\Rightarrow a+3d=4$$ ...(1)
Now,
$$S_n=\frac{n}{2}[2a+(n-1)d]$$
$$ \therefore S_7=\frac72[2a+(7-1)d]$$
$$\Rightarrow S_7=\frac72(2a+6d)$$
$$\Rightarrow S_7=7(a+3d)=7\times4=28$$ ( $$\because of (1)$$)
Therefore option B is correct
Write the arithmetic progression when first term $$a = -1.5$$ and common difference $$d = -0.5$$.
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$$-1.5, -2, -2.5, -3,.....$$
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$$-1.5, -1, -0.5, 0,.....$$
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$$-1.5, -2, -2.5, -5,.....$$
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$$-1.5, 2, 2.5, 3,.....$$
Explanation
Here $$a=-1.5$$ and $$d=-0.5$$.
The $$n^{th}$$ term of the AP is $$a_n=a+(n-1)d$$
$$\therefore a_1=a=-1.5$$
$$a_2=a+d\\=-1.5+(-0.5)\\=-2$$
$$a_3=a+2d\\=-1.5+2\times(-0.5)\\=-2.5$$
$$a_4=a+3d\\=-1.5+3\times(-0.5)\\=-3$$ and so on.
Therefore, the required AP is
$$-1.5,-2,-2.5,-3,...$$
Find the nth term and 9th term of the sequence 3, 7, 11, 15.........
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$$4n-1$$ and $$T_9=35$$
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$$4n+1$$ and $$T_9=37$$
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$$2n-1$$ and $$T_9=35$$
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$$4n-1$$ and $$T_9=39$$
Explanation
We know,
$$n^{th}$$ term $$a_{n}= a+(n-1)d $$
$$a =$$ First term
$$d =$$ Common difference
$$n =$$ number of terms
$$a_n = n^{th}$$ term
Here,
$$a=3\ and\ d=7-3\ or\ 11-7=4$$
$$a_n=a+(n-1)d=3+(n-1)4$$
$$a_n=4n-1$$
For $$9$$th term put $$n=9$$
$$T_9=a_9=4\times9-1=35$$
Find the $$18^{th}$$ term of the A.P., $$\sqrt{2}, 3\sqrt{2}, 5\sqrt{2}$$.....
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$$36\sqrt{2}$$
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$$35\sqrt{2}$$
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$$34\sqrt{2}$$
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$$33\sqrt{2}$$
Explanation
Here $$a=\sqrt2$$ and $$d=3\sqrt2-\sqrt2=2\sqrt2$$.
The nth term of the AP is $$a_n=a+(n-1)d$$
$$\therefore a_{18}=a+17d$$
$$=\sqrt2+17\times2\sqrt2$$
$$=35\sqrt2$$
Is $$-150$$ a term of the $$A.P$$ : $$11, 8, 5, 2$$ ?
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No
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Yes
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Data Insufficient
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Can't say
Explanation
Clearly the given sequence is an AP with first term $$a=11$$ and the common difference $$d=8-11=-3$$.
Let $$-150$$ be the $$n^{th}$$ term of given AP. Then,
$$a_n=-150\Rightarrow t_n= a+(n-1)d=-150$$
$$\Rightarrow 11+(n-1)\times-3=-150$$
$$\Rightarrow (n-1)\times-3=-161$$
$$\Rightarrow n-1=\dfrac{-161}{-3}$$
$$\Rightarrow n=\dfrac{161}{3}+1=\dfrac{164}3$$
Clearly $$n$$ is not an natural number, Hence, $$-150$$ is not a term of given AP.
Write the first term $$a$$ and the common difference $$d$$ of the AP : $$-5, -1, 3, 7......$$
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$$a = -5, d = 4$$
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$$a =-5, d = -6$$
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$$a = -5, d = -4$$
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$$a = 5, d = -4$$
Explanation
Given series is $$-5, -1, 3,7,......$$
Clearly first term is $$a=-5$$
And common difference $$d=-1-(-5)=4$$
Find out whether the sequence $$1^2, 3^2, 5^2, 7^2$$,... is an AP. If it is, find out the common difference.
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No
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Yes, $$d=8$$
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Yes, $$d=-8$$
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Yes, $$d=9$$
Explanation
Given series is $$1^2, 3^2, 5^2, 7^2,......$$
We have, $$3^2-1^2=9-1=8$$
$$5^2-3^2=25-9=16$$
$$7^2-5^2=49-25=24$$
This shows that the difference of a term and the preceding term is not always same.
Hence, the given sequence is not an AP.
Check if the series is an A.P. Find the common difference $$d$$ and write three more terms of the the following series.
$$2,\displaystyle\frac{5}{2}$$, $$3,\displaystyle\frac{7}{2}, ...$$
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$$a=2$$; $$d=\dfrac{1}{2}$$; other terms $$4,$$ $$\dfrac{9}{2},5$$
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$$a=3$$; $$d=\dfrac{1}{2}$$;
other terms $$6, 8, 10$$
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$$a=2$$; $$d=\dfrac{3}{2}$$;
other terms $$7, 8, 15$$
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$$a=3$$; $$d=\dfrac{3}{2}$$;
other terms $$4, 5, 15$$
Explanation
It is an A.P. because $$a = 2$$, $$d = \dfrac12$$
$$\because [(t_2 - t_1) = (t_3 - t_2) = (t_4 - t_3) = \dfrac12]$$
$$t_5 = \dfrac{7}{2} + \dfrac{1}{2} = 4$$
$$t_6 = 4 + \dfrac{1}{2} = \dfrac{9}{2}$$
$$t_7 = \dfrac{9}{2} + \dfrac{1}{2}= 5$$
Check if the series is an AP. Find the common difference $$d$$ and write three more terms of the the following series.
$$2, 4, 8, 16,....$$
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Common difference: $$2$$ and Other three terms are: $$32, 64, 128$$
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Common difference: $$4$$ and Other three terms are: $$24, 32, 40$$
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Common difference: $$8$$ and Other three terms are: $$20, 24, 28$$
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None of these
Explanation
Given series is $$2,4,8,16,.....$$
In an AP series, the difference between the consecutive terms remain same:
$$a_2-a_1=4-2=2$$
$$a_3-a_2=8-4=4$$
$$a_3-a_2\neq a_2-a_1$$
This shows that the difference of a term and the preceding terms is not always same. Hence, the given sequence is not an AP.
For the following AP, write the first term and the common difference
$$3, 1, -1, -3$$
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First term: $$-3$$ and Common difference: $$2$$
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First term: $$-1$$ and Common difference: $$-3$$
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First term: $$1$$ and Common difference: $$3$$
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First term: $$3$$ and Common difference: $$-2$$
Explanation
Given series is $$3,1,-1,-3,......$$
First term $$a=3$$
And common difference $$d=a_2-a_1$$
$$=1-3$$
$$=-2$$.
Write first four terms of the AP, when the first term $$a$$ and the common difference $$d$$ are given as follows:
$$a = 10, d =10$$
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0%
The
first four terms of the AP are
$$20, 40, 60, 80.$$
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The first four terms of the AP are $$10, 20, 30, 40.$$
0%
The first four terms of the AP are $$15, 30, 45, 60.$$
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Data insufficient
Explanation
Here $$a=10$$ and $$d=10$$.
Now, $$n^{th}$$ term $$=a_n=a+(n-1)d$$
$$\therefore t_1=a=10$$
$$t_2=a+d=10+10=20$$
$$t_3=a+2d=10+2\times10=30$$
$$t_4=a+3d=10+3\times10=40$$
Thus, the first four terms of given AP are $$10,20,30,40$$.
For the following AP, write the first term and the common difference
$$-5, -1, 3, 7$$
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First term: $$7$$ and Common difference: $$7$$
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First term: $$3$$ and Common difference: $$-1$$
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First term: $$-5$$ and Common difference: $$4$$
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First term: $$3$$ and Common difference: $$-2$$
Explanation
Given series is $$-5,-1,3,7,....$$
First term $$a=-5$$
And common difference $$d=a_2-a_1$$
$$=-1-(-5)$$
$$=-1+5$$
$$=4$$
Write first four terms of the AP, when the first term $$a$$ and the common difference $$d$$ are given as follows:
$$a = -2, d =0$$
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$$-2, -2, -2, -2$$
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$$-2, -4, -6, -2$$
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$$-2, -4, 4, -2$$
0%
$$-2, 4, 4, -2$$
Explanation
Here $$a=-2$$ and $$d=0$$.
Now, $$n^{th}$$ term $$=a_n=a+(n-1)d$$
$$\therefore t_1=a=-2$$
$$t_2=a+d=-2+0=-2$$
$$t_3=a+2d=-2+2\times0=-2$$
$$t_4=a+3d=-2+3\times0=-2$$
Thus, the first four terms of given AP are $$-2,-2,-2,-2$$.
For the following AP, write the first term and the common difference
$$0.6, 1.7, 2.8, 3.9$$
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First term: $$0.6$$ and Common difference: $$1.1$$
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First term: $$1.7$$ and Common difference: $$2.8$$
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First term: $$1.7$$ and Common difference: $$-1.1$$
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First term: $$0.6$$ and Common difference: $$-2.8$$
Explanation
Given series is $$0.6, 1.7, 2.8, 3.9,....$$
First term $$a=0.6$$
And common difference $$d=a_2-a_1$$
$$=1.7-0.6$$
$$=1.1$$
Is it an AP?
$$1, 4, 7, 10, 13, 16, 19, 22, 25, ...$$
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0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient
Explanation
Given series is $$1,4,7,10,13,16,19,22,25,....$$
Since, $$4-1 = 7-4=10-7,.....$$
i.e $$3=3=3,......$$
Given the common difference is $$3$$.
Therefore, the given sequence is an AP.
Hence, option A is correct.
Check if the sequence is an AP $$1, 3, 9, 27,....$$
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0%
Yes, it is an AP
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No
0%
Data Insufficient
0%
Ambiguous
Explanation
Given series is $$1,3,9,27,....$$
We have, $$3-1=2$$
$$9-3=6$$
$$27-9=18$$
This shows that the difference of a term and the preceding term is now always same. Hence, the given sequence is not an AP.
$$11^{th}$$ term of the AP: $$-3,$$ $$-\dfrac{1}{2}, 2,...$$ is
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$$28$$
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$$22$$
0%
$$-38$$
0%
$$-48$$
Explanation
Clearly, the given AP has first term $$a=-3$$ and common difference $$d=\displaystyle \frac{-1}2-(-3)=2-\dfrac{(-1)}{2}\Rightarrow\frac52$$.
Also, $$n=11$$
$$\therefore a_{11}=a+10d$$
$$=-3+10\times\displaystyle \frac52$$
$$=-3+25$$
$$=22$$
Hence, Option B is correct.
Find the sum of
$$2, 7, 12, ...$$to $$10$$ terms
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$$160$$
0%
$$245$$
0%
$$290$$
0%
$$300$$
Explanation
Clearly, the given sequence is an AP with first term $$a=2$$, common difference $$d=5$$ and $$n=10$$.
Now,
$$S_{10}=\dfrac {10}2[2a+(10-1)d]$$
$$=5[2\times2+9\times5]$$
$$=245$$
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