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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 4 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 4
The sum of 2 + 4 + 6 + 8 + .......... + 80 is :
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$$1540$$
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$$1640$$
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$$1740$$
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$$1440$$
Explanation
Given sequence is: $$2+4+6+.......+80$$
Here, $$\displaystyle a=2,d=4-2=2,$$$$\displaystyle { a }_{ n }=80$$
$$\displaystyle { a }_{ n }=a+\left( n-1 \right) d$$
$$\displaystyle \Rightarrow 80=2+\left( n-1 \right) 2$$
$$\displaystyle \Rightarrow 80=2n\Rightarrow n=40$$
$$\Rightarrow \displaystyle { S }_{ 40 }=\frac { 40 }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} $$
$$\displaystyle =20\left\{ 4+78 \right\}$$
$$ =20\times 82$$
$$=1640$$
The sum of $$2, 7, 12, ..............$$ to $$10$$ terms is :
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$$240$$
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$$248$$
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$$245$$
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$$250$$
Explanation
Given series is: $$\displaystyle 2+7+12+........ $$ to $$10$$ terms
Here, $$\displaystyle a=2,d=7-2=5,n=10$$
$$\displaystyle \Rightarrow { S }_{ n }=\frac { n }{ 2 } \left\{ 2a+\left( n-1 \right) d \right\} $$
$$\displaystyle \Rightarrow { S }_{ 10}=\frac { 10 }{ 2 } \left\{ 4+45 \right\} $$
$$=5\times 49$$
$$=245$$
Which term in the $$\text{A.P.}~~ 5,13,21,...$$ is $$ 181 $$?
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$${21}^{\text{st}}$$
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$${22}^{\text{nd}}$$
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$${23}^{\text{rd}}$$
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$${24}^{\text{th}}$$
Explanation
Given sequence is $$5, 13, 21, ...$$
Given: $$a = 5, ~d = 8, ~n^{th}\text{ term} = 181$$
To find: $$n$$
$$\because~~ { t }_{ n }~ = a +(n- 1)d\\ \Longrightarrow 181 =5 + (n - 1)8\\ \Longrightarrow 181 = 5+ 8n - 8\\ \Longrightarrow 181= 8n - 3\\ \Longrightarrow 8n= 184\\\implies ~n =23$$.
$$\therefore$$ $$181$$ is the $$23^{\text{rd}}$$ term.
If the sum of $$n$$ terms of AP. is $$476, l = 20, a = 36$$, then $$n$$ is equal to :
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$$14$$
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$$15$$
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$$16$$
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$$17$$
Explanation
Given, $$\displaystyle { S }_{ n }=476,l=20,a=36,n=?$$
We know $$\displaystyle { S }_{ n }=\frac { n }{ 2 } \left[ a+l \right] $$
$$\Rightarrow \displaystyle 476=\frac { n }{ 2 } \left[ 36+20 \right]$$
$$ =\dfrac { 56n }{ 2 } $$
$$=28n$$
$$\Rightarrow \therefore n=\dfrac { 476 }{ 28 } =17$$
If $$\displaystyle a=3,n=20$$ and $$\displaystyle { S }_{ n }=300$$, then $$l$$ is :
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$$30$$
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$$25$$
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$$27$$
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$$14$$
Explanation
Given, $$a=3, n=20, S_n=300$$
We need to find $$l$$
$$\displaystyle { S }_{ n }=\frac { n }{ 2 } \left( a+l \right) $$
$$\displaystyle \therefore 300=\frac { 20 }{ 2 } \left( 3+l \right) $$
$$\displaystyle \therefore l=27$$
If numbers $$a, b$$ and $$c$$ are in AP, then
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$$\displaystyle b-a=c-b$$
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$$\displaystyle b+a=c+b$$
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$$\displaystyle a-c=b-d$$
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None of these
Explanation
Since, given three numbers are in AP, then the common difference will be same.
$$\therefore$$ Common difference = $$ b - a = c - b$$
The $$4^{th}$$ term of an AP is $$14$$ and its $$12^{th}$$ term is $$70$$. What is its first term?
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$$-10$$
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$$-7$$
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$$7$$
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$$10$$
Explanation
As we know for an AP with first term $$a$$ and common difference $$d$$, genral term is $$a_n = a+(n-1)d$$
$$\therefore \displaystyle a+3d=14$$ and
$$\displaystyle a+ 11d=70$$
On subtracting both the above equations, we get
$$\displaystyle 8d=56$$
or $$\displaystyle d=7$$
$$\displaystyle \therefore \quad a=-7$$
The sum of $$n$$ natural number is :
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$$\displaystyle \frac { n\left( n+1 \right) }{ 2 } $$
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$$\displaystyle \frac { n(n-1) }{ 2 } $$
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$$\displaystyle \left( n+2 \right) $$
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$$\displaystyle \frac { n+1 }{ 2 } $$
Explanation
As the first $$n$$ natural numbers are
$$1,2,3,4,5,6,.........n$$
They form an AP with $$a=1$$ and $$d=1$$
The sum of AP for $$n$$ numbers
$$=\dfrac{n}{2}(2a+(n-1)d)$$
$$=\dfrac{n}{2}(2+n-1)$$
$$=\dfrac{n(n+1)}{2}$$
The $$3^{rd}$$ term of an A.P. is $$-40$$ and $$13^{th}$$ term is zero, then $$d$$ is equal to :
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$$-4$$
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$$4$$
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$$0$$
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$$-1$$
Explanation
Given $$3^{rd}$$ term $$=-40$$, $$13^{th}$$ term $$=0$$
Now using, $$a_n=a+(n-1)d$$
$$\therefore \displaystyle { a }_{ 3 }=a+2d=-40$$ ......(i)
and $$\displaystyle { a }_{ 13 }=a+12d=0$$ ......(ii)
Subtracting equations (i) and (ii), we get
$$\displaystyle d=4$$
If $$p, (p - 2)$$ and $$3 p$$ are in AP, then the value of $$p$$ is
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$$-3$$
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$$-2$$
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$$3$$
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$$2$$
Explanation
Since, given three numbers are in AP, then the common difference between two consecutive terms will be same.
$$\therefore$$
$$\displaystyle (p-2)-p=3p-(p-2)$$
$$\Rightarrow -2=2p+2$$
or $$\displaystyle p=-2$$
The common difference of the $$A.P:\;\;5, 3, 1, -1,\dots$$ is
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$$-2$$
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$$2$$
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$$-1$$
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$$3$$
Explanation
Common difference of an $$A.P$$ is equal to difference between the consecutive terms of the sequence,
$$d= 3 - 5$$
$$= -2$$
Hence, option $$A$$ is the correct answer.
If $$\displaystyle l=20,d=-1,n=17$$ , then the first term is :
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$$30$$
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$$32$$
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$$34$$
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$$36$$
Explanation
Given $$l = 20, d = -1, n = 17$$
We need to find $$a$$.
By using $$l=a+(n-1)d$$
$$\therefore 20=a+(17-1)(-1)$$
$$20=a-16$$
$$a=20+16$$
$$\therefore a=36$$
$$50^{th}$$ term of the AP. $$2, 5, 8, 11,.....$$ is :
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$$147$$
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$$149$$
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$$151$$
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$$153$$
Explanation
Given series $$\displaystyle 2,5,8,11,.....$$ is in A.P.
$$\therefore \displaystyle a=2,d=3$$
We need to find $$50^{th}$$ term.
$$\therefore n=50$$
$$\therefore \displaystyle { a }_{ 50 }=a+49d$$
$$=2+49\left( 3 \right) $$
$$=2+147$$
$$=149$$
Find the sum of the following APs:
$$-37, -33, -29, .....$$ to $$12$$ terms
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-180
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180
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200
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-200
Explanation
Here, $$a=-37, d=4$$ and $$n=12$$
Here we use the formula as given below:
$$S_n = \dfrac{n}{2}[2a+(n-1)d]$$
$$\therefore S_{12}=\dfrac {12}{2}(2\times -37+4\times 11)=6\times -30=-180$$
Find the sum of the following APs:
$$2, 7, 12, ......,$$ to $$10$$ terms
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245
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345
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276
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250
Explanation
$$S=\dfrac {n}{2}(2a+d(n-1))$$
Here, $$a=2, d=5$$ and $$n=10$$
$$Sum=\dfrac {10}{2}(2(2)+45)$$
$$=5\times 49=245$$
The value obtained by subtracting the $$10^{th}$$ term of an AP from the $$17^{th}$$ term is $$56$$. Find the common difference.
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$$7$$
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$$16$$
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$$9$$
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$$8$$
Explanation
In an Arithmetic Progression the g
eneral term of an A.P. is given by $$T_n$$:
$$ \boxed {T_{n} = a+(n-1)d} $$
According to the statement:
$$ T_{17}-T_{10} = 56 $$
$$ a+(17-1)d-[a+(10-1)d] = 56 $$
$$ a+16d-[a+9d] = 56 $$
$$ a+16d-a-9d = 56 $$
$$ 7a = 56 $$
$$\boxed {d = 8} $$
The common difference of the sequence $$5,8,11,14,$$ is
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$$3$$
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$$-3$$
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$$0$$
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$$1$$
Explanation
Given, $$5,8,11,14$$
$$\therefore 8-5=3$$
$$11-8=3$$
$$14-11=3$$
The common difference of the sequence $$5,8,11,14$$ is $$3$$
In an AP, the $$9th$$ term is $$-72$$. The $$10th$$ term is $$60$$ less than the $$4th$$ term. Find the first term of the AP
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$$8$$
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$$-8$$
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$$-152$$
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$$10$$
Explanation
Given that:
$$a_9=-72$$ and $$a_{10}=a_4-60$$
Let $$a$$ be the first term and $$d$$ be the common difference then
$$a+8d=-72$$ ......(1)
And
$$a+9d=a+3d-60$$
$$\Rightarrow 6d=-60\Rightarrow d=-10.$$
Substituting this value in equation (1), we get
$$a+8\times (-10)=-72$$
$$a=80-72$$
$$a=8$$
Hence,
The first term is $$8.$$
The first term of an AP is $$-50$$ and the $$50th$$ term is $$48$$. Find the common difference
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$$4$$
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$$1$$
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$$-3$$
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$$2$$
Explanation
Given,
First term $$ a = -50 $$
$$ 50^{th} $$ term, $$ T_{50} = 48 $$
$$ d = ? $$
$$ \boxed {T_{n} = a+(n-1)d} $$
$$ T_{50} = 48 $$
$$ 48 = -50+(50-1)d $$
$$ 48 = -50+49d $$
$$ 49d = 98 $$
$$ \boxed {d = 2} $$
Find the sum of the following APs:
$$0.6, 1.7, 2.8, ......$$ to $$100$$ terms
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0%
5475
0%
5505
0%
6589
0%
3844
Explanation
Here $$a=0.6, d=1.1$$ and $$n=100$$
As we know, $$S_n = \dfrac {n}{2}[2a+(n-1)d]$$
$$\therefore S_{100}=\dfrac {100}{2} (2(0.6)+1.1\times 99)=50\times 110.1=5505$$
In an AP, the $$4th$$ term is $$36$$. The $$21st$$ term is $$108$$ more than the $$9th$$ term. Find the common difference.
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$$12$$
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$$9$$
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$$4$$
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$$-3$$
Explanation
Given: $$ T_{4} = 36 $$
$$ T_{21} = 108+T_{9} $$
$$ d = ? $$
General term of an A.P.
$$=T_{n} = a+(n-1)d $$
$$T_{21} = 108+T_{9} $$
$$ a+(21-1)d = 108+a+(9-1)d $$
$$ 20d = 108 +8d $$
$$ 12d = 108 $$
$$ \boxed {d = 9} $$
Find the common difference in the series: 0.2, 0.9, 1.6 ..............
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0.5
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0.1
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0.7
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0.6
Explanation
First term, $$a_1 = 0.2$$
Second term, $$a_2 = 0.9$$
Common difference, $$d = a_2 - a_1$$
$$d= 0.9 - 0.2$$
$$\therefore d = 0.7$$
Find the sum of the arithmetic series $$6 + 12 + 18 + ....... 96$$.
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$$815$$
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$$816$$
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$$817$$
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$$819$$
Explanation
Given series is $$6+12+18+....96$$.
Here $$a = 6$$; last term, $$l = 96, d = 12-6 =6$$
We know that, last term, $$l = a + (n - 1) d$$
$$\Rightarrow 96 = 6 + (n - 1) 6$$
$$\Rightarrow 96 = 6 + 6n - 6$$
$$\Rightarrow 96 = 6n$$
$$\Rightarrow n = \dfrac{96}{6}$$
$$\Rightarrow n = 16$$ = total number of terms in the series
We know that,
Sum $$(S_n) = \dfrac{n}{2}$$ [First term $$+$$ Last term]
$$= \dfrac{16}{2} [6 + 96]$$
$$= 8 [102]$$
Therefore, $$S_{16} = 816$$
Find the common difference, if the sum of first n terms will be (n - 2) (n - 1).
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1
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2
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3
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4
Explanation
The sum of first n terms, $$S_n = \dfrac{n}{2} (2a + (n - 1)d)$$
On comparing the equations, we get
$$\dfrac{n}{2} (2a + (n - 1) d) = (n - 2)(n + 1)$$
$$(2a - d) n + dn^2 = n^2 - n - 2$$
On comparing the factors of $$n^2$$, we find the value of $$ d= 1$$
$$\therefore$$ The common difference is 1.
Find the common difference in the sequence 4, 8, 12, 16, .......... 20
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8
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3
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4
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2
Explanation
First term, $$a_1 = 4$$
Second term, $$a_2 = 8$$
Common difference, $$d = a_2 = a_1$$
$$d = 8 - 4 = 4$$
$$\therefore$$ The common difference is 4.
Find the number of terms in an A.P: $$-1,\ -5,\ -9,\ ..,\ - 197$$
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48
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49
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50
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47
Explanation
Given: AP: $$-1, \ -5,\ -9,\ ...,\ -197$$
Here, First term $$a = -1$$
Common difference $$d = a_2 - a_1 = -5 - (-1) = - 4$$
And last term $$a_n = - 197$$
We know that, $$a_n = a + (n - 1) d$$
$$\Rightarrow -197 = -1 + (n - 1) - 4$$
$$\Rightarrow -197 = -1 - 4n + 4$$
$$\Rightarrow -197 + 1 - 4 = -4n$$
$$\Rightarrow n = 50$$
$$\therefore$$ There are 50 terms in the A.P
Which term of the$$ A.P. 2, 9, 16, 23 .............$$ is $$100$$?
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$$15$$
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$$10$$
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$$11$$
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$$12$$
Explanation
$$a = 2$$
common difference, $$d = a_2 - a_1, a - 2 = 7$$
we know that, $$a_n = a + (n - 1) d$$
$$a_n = 100$$
$$100 = a + (n - 1) 7$$
$$100 = 2 + 7n - 7$$
$$100 - 2 + 7 = 7n$$
$$\dfrac{105}{7} = n$$
$$n = 15$$
Hence, $$15th$$ term of $$A.P$$. is $$100$$.
Find 11th term of the A.P. : -3, 1, 5, ........
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36
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37
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38
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39
Explanation
First term, $$a_1 = -3$$
Common difference, $$d = a_2 - a_1 = 1 - (-3) = 4$$
We know that, $$a_n = a_1 + (n - 1) d$$
$$a_{11} = - 3 + (11- 1) 4$$
$$= -3 + (10) 4$$
$$= - 3 + 40$$
$$a_{11} = 37$$
_____ is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
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Geometric value
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Geometric series
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Arithmetic progression
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Arithmetic mean
Explanation
$$\text {Arithmetic progression}$$ is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
Ex: $$1, 2, 3, 4 $$ ..... In this sequence, the first number is $$1$$ and the next number is $$2$$. The difference between the two numbers is $$1$$ and so on.
By adding $$1$$ to the preceding number, we are forming the next term and so on, and the series is in $$AP$$.
The sum of n terms of an arithmetic sequence can be calculated by .........................
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$$S_n = \dfrac{n}{2} [2a+ (n - 1) d]$$
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$$S_n = \dfrac{n}{2} [2a+ (n + 1) d]$$
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$$S_n = \dfrac{2n}{2} [2a+ (n - 1) d]$$
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$$S_n = \dfrac{2n}{2} [2a- (n - 1) d]$$
Explanation
The sum of n terms of an arithmetic sequence can be calculated by $$S_n = \dfrac{n}{2} [2a+ (n - 1) d]$$
Where, n
denotes the total number of terms
a = first term
d = common difference
$$S_n$$ = Sum of n terms
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