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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 10 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 10
Which term of the AP : $$3,8, 13, 18,.. $$is $$78?$$
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$$t_{16}$$
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$$t_{18}$$
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$$t_{14}$$
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None of these
Explanation
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. $$t_{16}$$
Find the common difference of an AP. whose first term is $$100$$ and the sum of whose first six terms is five times the sum of the next six terms.
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$$10$$
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$$-10$$
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$$5$$
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$$-5$$
Explanation
Here $$a=100$$
Let difference is $$d$$.
$$\Rightarrow a_1+a_2+a_3+a_4+a_5+a_6=5(a_7+a_8+a_9+a_{10}+a_{11}+a_{12})$$
So by the formula, $$S_n = \dfrac{n}{2} (a+l) $$, where $$a$$ &$$l$$ are the first and last term of an AP, we have
$$ 6\left(\dfrac{a_1+a_6}{2} \right)=5\times 6\left(\dfrac{a_7+a_{12}}{2}\right)$$
$$\Rightarrow a_1+a_6=5(a_7+a_{12})$$
$$\Rightarrow a+a+5d=5(a+6d+a+11d)$$
$$\Rightarrow 2a+5d=10a+85d$$
$$\Rightarrow 80d=-8a$$
$$\Rightarrow d=\dfrac{-a}{10}\Rightarrow \dfrac{-100}{10}\Rightarrow -10$$
If $$n^{th}$$ term of an AP is $$\dfrac {1}{3}(2n+1)$$, then the sum of its $$19$$ term is
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$$131$$
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$$132$$
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$$133$$
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$$134$$
Explanation
Let $$T_n$$ be the $$n^{th}$$ term of an AP
$$\Rightarrow T_n=a+(n-1)d\\\ \ \ \ \ \ \ \ \ =nd+a-d$$
On comparing with the given expression it is clear that $$d=\dfrac{2}{3}$$ and $$a-d=\dfrac{1}{3}$$
$$\Rightarrow a-\dfrac{2}{3}=\dfrac{1}{3}$$
$$\Rightarrow a=1$$
We know that s
um 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.
$$\therefore S_{19}=\dfrac {19}{2}\left [ 2+(18)\dfrac{2}{3} \right ]\\=133$$
Hence, option $$(C)$$ is correct.
Which term of AP : $$3, 15, 27, 39,..$$ will be $$132$$ more than its $$54th$$ term?
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t$$_{65}$$
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t$$_{64}$$
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t$$_{60}$$
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None of these
Explanation
Clearly, the given Sequence is an AP with first term $$a=3$$ and common difference $$d=15-3=12$$.
Let $$nth$$ term be the $$132$$ more than $$54th$$ term of the given AP.
$$\Rightarrow a_n=a_{54}+132$$
$$\Rightarrow a+(n-1)d=a+53d+132\Rightarrow (n-1)\times12=53\times12+132$$
$$\Rightarrow n-1=53+11$$
$$\Rightarrow n=65$$
Hence, $$a_{65}$$ is $$132$$ more than $$a_{54}$$
The sum of the first nineteen terms of an A.P. $$a_1,\, a_2,\, a_3$$ .......... if it is known that $$a_4\, +\, a_8\, +\, a_{12}\, +\, a_{16}\, =\, 224$$ is .......... .
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$$1064$$
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$$896$$
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$$532$$
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$$448$$
Explanation
We know that $$n^{th}$$ term of an AP is given as,
$$a_n = a+(n-1)d$$
And sum of first $$n$$ terms, $$S_n = \dfrac{n}{2} (2a+(n-1)d)$$, where $$a$$ and $$d$$ are the first term amd common difference of an AP.
Since,
$$ { a }_{ 4 }+{ a }_{ 8 }+{ a }_{ 12 }+{ a }_{ 16 }=224$$
$$\therefore a+3d+a+7d+a+11d+a+15d=224\\ \Rightarrow 4a+36d=224\\\Rightarrow a+9d=56$$
Now
$$\displaystyle { S }_{ 19 }=\frac { 19 }{ 2 } \left[ 2a+\left( 19-1 \right) d \right] \\ =19\left[ a+9d \right] \\=19\left[ 56 \right] \\ =1064$$
If $$9^{th}$$ and $$19^{th}$$ terms of an AP are $$35$$ and $$75$$,
then its $$20^{th}$$ term is
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$$78$$
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$$79$$
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$$80$$
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$$81$$
Explanation
$${ T }_{ 9 }=35\Rightarrow a+\left( 9-1 \right) d=35\Rightarrow a+8d=35$$
$$ { T }_{ 19 }=75\Rightarrow a+\left( 19-1 \right) d=75\Rightarrow a+18d=75$$
Solving these two we get
$$d=4,a=3$$
Therefore,
$${ T }_{ 20 }=a+\left( 20-1 \right) d=3+19\times 4=79$$
The sum of all integers between $$81$$ and $$719$$ which
are divisible by $$5$$ is
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$$51800$$
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$$50800$$
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$$52800$$
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None of these
Explanation
All integers between $$81$$ and $$719$$ which are divisible by $$5$$ form an AP with common difference $$5$$.
First term $$a=85$$ , last term $$l=715$$ and common difference is $$d=5$$
Then, apply the formula for $$n^{th}$$ term of an AP
$$715=85+\left( n-1 \right) 5$$
$$126=n-1$$
$$\Rightarrow n=127$$
To get the required sum, apply the formula for sum of $$n$$ terms of an AP
$$\displaystyle { S }_{ n }=\dfrac { n }{ 2 } \left( a+l \right) \\=\dfrac { 127 }{ 2 } \left( 85+715 \right)\\ =50800$$
The value of $$1 + 3 + 5 + 7 + 9 + ............. + 25$$ is:
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$$196$$
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$$625$$
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$$225$$
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$$169$$
Explanation
We take the first term $$a=1$$, last term $$l=25$$, common difference $$d=2$$ and the number of terms $$=n$$.
Then, since it is an A.P. series,
$$l=a+(n-1)d$$
$$ n=\dfrac { l-a }{ d } +1\\ =\dfrac { 25-1 }{ 2 } +1\\$$
$$=\dfrac{25-1+2}{2}$$
$$=\dfrac{25+1}{2}$$
$$=\dfrac{26}{2}$$
$$=13$$
Let
$${ S }_{ n }$$ be the sum of $$n=13$$ terms
$${ \therefore \quad S }_{ n }=\dfrac { n }{ 2 } \times \left( a+l \right) \\ =\dfrac { 13 }{ 2 } \times \left( 1+25 \right) \\ =13\times 13=169.$$
So the required sum $$=169$$.
Ramkali saved $$Rs. 5$$ in the first week of a year and then increased her weekly savings by $$Rs. 1.75.$$ If in the nth week, her weekly savings become $$Rs. 20.75,$$ find $$n.$$
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$$n=10$$
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$$n=12$$
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$$n=16$$
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$$n=20$$
Explanation
Given that
$$ a =5$$
$$d = 1.75$$
$$a_{n} =20.75$$
$$a_{n}=a+\left(n-1\right)d$$
$$20.75=5+\left(n-1\right)1.75$$
$$15.75=\left(n-1\right)1.75$$
$$\left(n-1\right)=\displaystyle \frac{15.75}{1.75}=\frac{1575}{175}=\frac{63}{7}=9$$
$$\therefore n-1 = 9$$
$$\therefore n=10$$
The sum of $$3^{rd}$$ and $$15^{th}$$ elements of an arithmetic progression is equal to the sum of $$6^{th},\, 11^{th}\, and\, 13^{th}$$ elements of the same progression. Then which element of the series should necessarily be equal to zero?
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$$1st$$
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$$9th$$
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$$12th$$
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None of the above
Explanation
Let the first term of AP be $$a$$ and difference be $$d$$
We know that $$n$$th term,
$$a_n = a+(n-1)d$$,
where $$a$$ & $$d$$ are the first term amd common difference of an AP.
$$\therefore$$ Then third term will be $$=a+2d$$
$$15^{th}$$ will be$$=a+14d$$
$$6^{th}$$ will be$$ =a+5d$$
$$11^{th}$$ will be$$ =a+10d$$
$$13^{th}$$ will be$$=a+12d$$
Then the equation will be
$$a+2d+a+14d = a+5d+a+10d+a+12d$$
$$2a+16d = 3a+27d$$
$$a+11d=0$$
We understand $$a+11d$$ will be the $$12^{th}$$ term of arithmetic progression.
So, the correct answer is $$12$$.
The sum of first $$n$$ terms of the $$A.P$$. $$\sqrt{2}\, +\, \sqrt{8}\, +\, \sqrt{18}\, +\, \sqrt{32}\, +\, ........$$ is ......
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$$\displaystyle \frac{n(n\, +\, 1)}{2}$$
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$$2n (n + 1)$$
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$$\displaystyle \frac{n(n\, +\, 1)}{\sqrt{2}}$$
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$$1$$
Explanation
The series can be rewritten as $$\sqrt{2} + 2 \sqrt{2} + 3\sqrt{2} + 4 \sqrt{2} + ..... $$
So the nth sum $$S_{n} = \sqrt{2} \times \left(1 + 2 + 3 + ..... + n \right)$$
$$= \displaystyle \frac{\sqrt{2}\times n\times \left(n+1\right)}{2} $$
$$=\displaystyle \frac{n\left(n+1\right)}{\sqrt{2}}$$
The $$9th$$ term of an AP is $$499$$ and $$499th$$ terms is $$9.$$ The term which is equal to zero is
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$$501 th$$
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$$502 th$$
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$$508 th$$
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None of these
Explanation
$$a_{9}=a+8d$$
$$a_9=499 (given)$$
$$\therefore a+8d=499.........(i)$$
$$ a_{499}=a+498d$$
$$a_{499}=9$$
$$\therefore a+498d=9.........(ii)$$
Subtract (i) from (ii)
$$\Rightarrow 490 d = - 490 $$
$$\Rightarrow d = -1 $$
substitute the value of $$d$$
$$\Rightarrow a + 8(-1) = 499$$
$$\Rightarrow a = 507 $$
$$\therefore$$ For $$ a_{n}=0$$
$$ \Rightarrow a+(n-1)d=0$$
$$ \Rightarrow 507+(n-1)(-1)=0$$
$$ \Rightarrow 507=n-1$$
$$ \Rightarrow n=508$$
So, $$508^{th}$$ term is equal to zero.
$$8^{th}$$ term of series $$\displaystyle 2\sqrt{2}+\sqrt{2}+0+......$$ will be
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$$\displaystyle -5\sqrt{2}$$
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$$\displaystyle 5\sqrt{2}$$
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$$\displaystyle 10\sqrt{2}$$
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$$\displaystyle -10\sqrt{2}$$
Explanation
Given series is : $$2\sqrt2+ \sqrt2+0+.......$$
Then, $$ a=2\sqrt{2},d=(\sqrt 2-2\sqrt 2)=-\sqrt{2}$$
$$ a_{8}=a+7d$$
$$\therefore 2\sqrt{2}+7\times (-\sqrt{2})$$
$$\therefore a_{8}=-5\sqrt{2}$$
Find the sum of all whole numbers divisible by $$5$$ but less than $$100$$.
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$$950$$
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$$925$$
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$$880$$
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$$1050$$
Explanation
All whole numbers divisible by $$5$$ but less then $$100$$ are
$$0,5,10,15,20........95$$
It is an A.P. with first term, common difference and last term as $$0, 5$$ and $$95$$ respectively.
$$a=0,d=5,a_n=95$$
$$a_n=a+(n-1)d$$
$$95=0+(n-1)5$$
$$5n=100$$
$$n=20$$
Hence total numbers are $$=20$$
Sum of numbers are
$$S_{n}=\dfrac{n}{2}[a+(n-1)d]$$
$$S_{20}=\dfrac{20}{2}[0+(20-1)5]$$
$$S_{20}=10\times 95\\\ \ \ \ \ =950$$
The sum of all natural number divisible by $$5$$ but less than $$100$$ is $$950$$.
The $$4th$$ term from the end of the AP
$$-11, -8, -5, ....................49$$ is
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$$37$$
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$$40$$
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$$43$$
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$$58$$
Explanation
Given series of AP:
$$- 11, -8, -5,.......49$$
New series : $$49,........-5,-8,-11$$
In new series of AP first term $$a = 49,$$ common difference $$d = (-8) - (-5) = - 3$$
$$\displaystyle a_{4}=a+3d=49+3(-3)=49-9=40$$
$$\therefore $$ Option B is correct.
What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?
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$$530, 820$$
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$$179, 587$$
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$$560, 758$$
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$$164, 850$$
Explanation
$$\textbf{Step1-Find sequence of an A.P according to question}$$
$$\text{The lowest 3 digit number is }$$$$101$$$$\text{ and the highest number is}$$ $$998$$.
$$\text{Therefore, the AP is }$$$$101,104,107, ....... , 998$$
$$\text{ where, the first term is }$$$$a=101$$,
$$\text{common difference is $$d=104 - 101 = 3$$ and }$$
$$\text{n$^{th}$ term is }$$$$a_n=998$$
$$\textbf{Step2-Find the total number of terms of sequence}$$
$$\text{Let us find the number of terms is n}$$
$$\text{We know that the n$^{th}$}$$
$$\text{term of AP is given by}$$
$$a_{n} = $$
$$a+(n-1)d$$
$$\text{Now, substituting the values, we get:}$$
$$998=101+(n-1)3\\ \Rightarrow (n-1)3=998-101\\ \Rightarrow (n-1)3=897\\ \Rightarrow n-1=\dfrac { 897 }{ 3 } \\ \Rightarrow n-1=299\\ \Rightarrow n=299+1\\ \Rightarrow n=300$$
$$\textbf{Step3-Find sum of 300 terms of an A.P}$$
$$\text{We know }$$
$$S=\dfrac {n}{2}$$$$\text{(1st term + last term) }$$
$$\text{Therefore, we have:}$$
$$S=\dfrac { 300 }{ 2 } (101+998)$$
$$S=150\times 1099$$
$$S=164850$$
$$\textbf{Hence, the sum of all 3 digit numbers that leave a remainder of 2 when divided by 3 is 164,850.}$$
The sum of all odd integers between $$2$$ and $$50$$ divisible by $$3$$ is
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$$192$$
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$$216$$
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$$168$$
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$$240$$
Explanation
The odd intergers between 2 and 50 divisible by 3 are $$3, 9, 15,......45$$
This sequence forms an AP with first term $$a=3$$ and common difference $$d=9-3=6$$
Since, $$l=a+(n-1)d $$
$$\Rightarrow 45=3+(n-1)6 $$
$$\Rightarrow 42=(n-1)6 $$
$$\Rightarrow 7=n-1 $$
$$\Rightarrow n=8 $$
Number of terms in given sequence are 8.
$$ S_n=\dfrac n2(a+l) $$
$$\Rightarrow S_n=\dfrac{8}{2}(3+45)$$
$$\Rightarrow S_n=4(48)$$
$$\Rightarrow S_n= 192$$
$$\therefore $$ Option A is correct.
If the $$5$$th term of an A.P is eight times the first term and $$8$$th term exceeds twice the $$4$$th term by $$3$$, then the common difference is
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$$7$$
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$$5$$
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$$6$$
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$$3$$
Explanation
$$n^{th}$$ term of an AP is given as,
$$t_n=a+(n-1)d$$
Where $$a$$ and $$d$$ are first term and common difference respectively.
According to the given condition,
Thus,
$${ t }_{ 5 } = a + 4d$$
$${ t }_{ 1 } = a$$
$${ t }_{ 8 } = a + 7d$$
$$ { t }_{ 4 } = a + 3d$$
Given, $${ t }_{ 5 } = 8{ t }_{ 1 }$$
$$\Longrightarrow a+ 4d= 8a\\ \Longrightarrow \quad 4d= 7a\\ \Longrightarrow d= \dfrac { 7a }{ 4 } $$
Also given, $${ t }_{ 8 }$$ $$-2{ t }_{ 4 } = 3$$
$$\Longrightarrow a+ 7d- 2(a+ 3d)= 3\\ \Longrightarrow d - a= 3\\ $$
Substituting $$d= \dfrac { 7a }{ 4 } $$ in the above equation,
$$\dfrac{7a}{4}-a=3$$
$$\dfrac{3a}{4}=3$$
we get $$a = 4$$ and hence, $$d = 7$$.
If $$9^{th}$$ term of an A.P. is $$47$$ and $$16^{th}$$ term is $$82$$, then the general term is
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$$9n+2$$
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$$6n-7$$
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$$5n+2$$
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none
Explanation
$$a_9=a+8d=47$$........(1)
$$a_{16}=a+15d=82$$.....(2)
Subtract eq (1) by (2)
$$7d=35$$
$$\therefore d=5$$
$$a+8d=47$$
$$a+8\times 5=47$$
$$\therefore a=7$$
Hence general term is $$ a_n = a+(n-1)d = 5n+2 $$
Find $${a}_{25}-{a}_{15}$$ for the A.P $$-6, -10, -14, -18,..........$$
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$$-40$$
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$$-48$$
0%
$$40$$
0%
$$-44$$
Explanation
Given series is: $$-6,-10,-14,-18.....$$
$$a=-6, d=-4$$
As we know $$nth$$ term, $$a_n = a+(n-1)d$$
$$\therefore$$ $$a_{25}=a+24d$$
& $$a_{15}=a+14d$$
$$a_{25}-a_{15}=(a+24d)-(a+14d)=10d$$
In the following series $$d=-4$$
$$\therefore a_{25}-a_{15}=10\times (-4)=-40$$
If an A.P is given by $$17, 14, ......-40$$, then $$6$$th term from the end is
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0%
$$-22$$
0%
$$-28$$
0%
$$-25$$
0%
$$26$$
Explanation
Given an AP is 17,14...........................-40
Then we get AP is -40,-37,.............14,17
Then d=3,a=-40 n=6
Therefore, $$a_{6}=a+(n-1)d=(-40)+(6-1)3=-40+15=-25$$
,
If three times the $$9$$th term of an A.P is equal to five times its $$13$$th term, then which of the following term will be zero.
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$$26$$th
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$$19$$th
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$$21$$st
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$$36$$th
Explanation
According to the question $$3(a+8d)=5(a+12d)$$
$$\Rightarrow 3a+24d=5a+60d$$
$$\Rightarrow 2a+36d=0$$
$$\Rightarrow a+18d=0$$
$$\Rightarrow a+(19-1)d=0$$
Hence $$19^{th}$$ term is zero
The total two-digit numbers which are divisible by $$5$$, are
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$$19$$
0%
$$20$$
0%
$$17$$
0%
$$18$$
Explanation
$$10, 15, 20, ........ 95$$
$$n$$th term of AP$$=a+(n-1)d$$
$$\Rightarrow 95=10+(n-1)5$$
$$\Rightarrow 95=10+5n-5$$
$$\Rightarrow 95=5n+5$$
$$\Rightarrow n=18$$
Hence, required numbers $$=18$$
How many terms are there in the sequence $$15\cfrac{1}{2}, 13, ......-47$$?
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$$27$$
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$$26$$
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$$28$$
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None
Explanation
Given series is $$15 \dfrac {1}{2}, 13, ....., -47$$
$$a_n=a+(n-1)d$$
$$\therefore -47=\dfrac{31}{2}+(n-1)\left(\dfrac{-5}{2}\right)$$
$$-47=\dfrac{31}{2}-\dfrac{5n}{2}+\dfrac{5}{2}$$
$$-47=18-\dfrac{5n}{2}$$
$$\dfrac{5n}{2}=65$$
$$\therefore n=\dfrac{65\times 2}{5}=26$$
Hence number of terms are $$26$$
Find the $$n^{th}$$ term and, hence, the $$25^{th}$$ term of the following AP: $$2, 5, 8, 11$$,.........
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$$3n-1, 74$$
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$$2n, 73$$
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$$3n, 75$$
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$$3n+1, 72$$
Explanation
Given series, 2, 5, 8, 11, ... is in AP
Here
First term $$a = 2$$
Common difference $$d = 5 - 2$$
$$= 8 - 5 $$
$$= 3$$
We know that, general term
$$\displaystyle { t }_{ n }=a+\left( n-1 \right) d$$
$$\displaystyle =2+\left( n-1 \right) 3$$
$$\displaystyle =2+3n-3$$
$$\displaystyle =3n-1$$
$$\displaystyle { t }_{ 25 }=3\times 25-1$$
$$=75-1=74$$
If the first, second and last term of an A.P. are $$x,y$$ and $$2x$$ respectively, its sum is
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$$\cfrac{xy}{2(y-x)}$$
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$$\cfrac{3xy}{2(y-x)}$$
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$$\cfrac{2xy}{y-x}$$
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$$\cfrac{xy}{y-x}$$
Explanation
Here $$a_1=x,$$ $$a_2=y$$ and $$a_n=2x$$ .
$$\therefore \ $$ First term, $$a=x$$ and common difference, $$d=y-x$$
Since, $$a_n=a+(n-1)d $$
$$\Rightarrow 2x=x+(n-1)(y-x) $$
$$\Rightarrow x=(n-1)(y-x) $$
$$\displaystyle \Rightarrow \frac {x}{y-x}=n-1 $$
$$\displaystyle \Rightarrow \frac {x}{y-x}+1=n $$
$$\displaystyle \Rightarrow n=\frac {y}{y-x}$$
Now, we know that, $$S_n= \dfrac n2(a+a_n) $$
$$\therefore \ S_n= \displaystyle \dfrac {\frac {y}{y-x}}2(x+2x) $$
$$\Rightarrow S_n= \displaystyle \dfrac {\frac {y}{y-x}}2(3x) $$
$$\Rightarrow S_n=\displaystyle \dfrac {3xy}{2(y-x)} $$
Hence, option B is correct.
If in an A.P, $${a}_{9}=0, {a}_{19}=k$$, then $${a}_{29}$$ is
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0%
$${k}^{2}$$
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$$10k$$
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$$9k$$
0%
$$2k$$
Explanation
Let $$a$$ and $$d$$ be the first term and common difference of given AP.
Since, $$a_9 = 0 \Rightarrow a+8d=0$$ ....(1) [using $$a_n = a+(n-1)d$$]
Similarly,
$$a_{19} = k \Rightarrow a+18d=k$$ ....(2)
On subtracting (1) from (2). we get
$$a+18d-a-8d = k$$
$$\Rightarrow 10d=k $$
$$\Rightarrow d=\dfrac k{10} $$
By substituting $$d=\dfrac k{10} $$ in (1), we get
$$a+\dfrac {8k}{10}=0$$
$$\Rightarrow a= -\dfrac {8k}{10}$$
Now, $$a_{29}=a+28d = -\dfrac {8k}{10}+\dfrac {28k}{10} = 2k$$
Hence, option D is correct.
The common difference of $$-2, -4, -6, -8,...........$$ is :
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$$-2$$
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$$-1$$
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$$2$$
0%
$$3$$
Explanation
$$\textbf{Step-1: Write the property of Common difference of an A.P.}$$
$$\text{Given series: -2,-4,-6,-8...}$$
$$\text{We know that Common difference = Second term - Fiirst term = Third term - Second term}$$
$$\therefore$$ $$\text{Common difference=}$$ $$ - 4 - ( - 2) = - 4 + 2 = - 2$$ $$\text{or - 6 - ( - 4) = - 6 + 4 = - 2}$$
$$\textbf{Hence, In the given sequence -2, -4, -6, ......, the common difference is -2, option - A.}$$
The AP whose first term is 10 and common difference is 3 is
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10, 13, 16, 19, ...
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5,7,9, 11, ...
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8, 12, 16,20, ...
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All the above
Explanation
If $$a_1$$ is the first term, then the nth term is $$ a_1 + (n-1)d $$, where $$d$$ is the common difference
Here $$ a_1 = 10, d = 3,$$
Hence,
$$ a_2 = a_1 + (2-1)d = 10 + 3 = 13 $$
$$ a_3 = a_1 + (3-1)d = 10 + 6 = 16 $$ and so on.....
Which term of the A.P $$5,15, 25 -----$$ will be $$100$$ more than its $$31^{st}$$ term?
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$$41^{st}$$
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$$42^{nd}$$
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$$40^{th}$$
0%
$$43^{rd}$$
Explanation
The given A.P is $$5,15,25.......$$ where the first term is $$a=5$$ and the common difference $$d$$ will be calculated by subtracting the first term from the second term as shown below:
$$d=15-5=10$$
Now, $$31$$st term$$=a+30d$$, substituting the values of $$a$$ and $$d$$, we get:
$$31$$st term
$$=5+(30\times10)=5+300=305$$
It is given that the A.P of the given terms is $$100$$ more than its
$$31$$st term, thus, we have:
$$a_{ n }=305+100=405\\ \Rightarrow a+(n-1)d=405\\ \Rightarrow 5+(n-1)10=405\\ \Rightarrow (n-1)10=405-5\\ \Rightarrow (n-1)10=400\\ \Rightarrow n-1=\dfrac { 400 }{ 10 } \\ \Rightarrow n-1=40\\ \Rightarrow n=40+1\\ \Rightarrow n=41$$
Hence, the $$41$$st term of the given A.P will be $$100$$ more than its $$31$$st term.
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Practice Class 10 Maths Quiz Questions and Answers
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