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CBSE Questions for Class 10 Maths Arithmetic Progressions Quiz 1 - MCQExams.com
CBSE
Class 10 Maths
Arithmetic Progressions
Quiz 1
In an arithmetic series, $$a_1 = -14$$ and $$a_{5}=50$$. Find the sum of the first 5 terms.
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0%
$$60$$
0%
$$70$$
0%
$$80$$
0%
$$90$$
Explanation
First term = a = -14
last term, l = 50
$$a_{n}=50$$
n = 5
$$S_{n}=\cfrac{n}{2}[a+l]$$
$$S_{5}=\cfrac{5}{2}[-14+50]$$
$$S_{5}=2.5[36]$$
$$S_{5}=90$$
$$-20, -16, -12, -8.....$$ is the sequence and each term is increased by $$4$$. Then which of the following could not be a term in the sequence?
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0%
$$762$$
0%
$$200$$
0%
$$440$$
0%
$$0$$
The sum of first $$20$$ natural numbers is
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0%
$$55$$
0%
$$210$$
0%
$$110$$
0%
$$215$$
Explanation
The sum of first '$$n$$' natural numbers is given by $$ \dfrac { n(n + 1) }{ 2 } $$.
So, the sum of first $$20$$ natural numbers is $$ \dfrac { 20(20 + 1) }{ 2 } $$.
$$= \dfrac { 20(21) }{ 2 } $$
$$ = 210$$.
Find first term 'a' and common difference 'd' for the following AP.
$$\sqrt{2}$$, $$\sqrt{8}$$, $$\sqrt{18}$$,
$$\sqrt{32}$$,.....
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$$a =$$ $$\sqrt{2}$$$$, d=$$ $$\sqrt{4}$$
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$$a =$$ $$\sqrt{2}$$$$, d=$$ $$\sqrt{2}$$
0%
Its not an AP
0%
None of these
Explanation
The sequence $$\sqrt2,\sqrt8,\sqrt{18},\sqrt{32},...$$ can be written as $$\sqrt2,2\sqrt2,3\sqrt2,4\sqrt{2},...$$
Clearly, it is an AP with first term $$a=\sqrt2$$ and common difference $$d=\sqrt2$$.
The sum of the first $$1000$$ positive integer is
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0%
$$500500$$
0%
$$650650$$
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$$300300$$
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$$700700$$
Explanation
Sum of '$$n$$' natural numbers $$=$$ $$ \dfrac { n(n+ 1) }{ 2 } \\ $$.
So, sum of first $$1000$$ natural numbers $$=$$ $$ \dfrac { 1000(1000+ 1) }{ 2 } = 500(1001)= 500500\\ $$.
Hence, option A is correct.
State the following statement is true or false:
Progression means increment of quantity in a particular pattern.
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0%
True
0%
False
Explanation
Progression means increment of quantity in a particular pattern.
For example: $$2,4,6,8,....$$ and $$3,6,12,24,....$$
Which term of the sequence $$-7, -2, 3, 8,.........$$ is $$73$$?
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0%
$$17$$
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$$18$$
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$$16$$
0%
none
Explanation
Here first term $$a=-7$$ , common difference $$d= -2-(-7) = 5$$ and last term $$l= 73$$
Since, $$l=a+(n-1)d $$
$$\Rightarrow 73= -7+(n-1)5 $$
$$\Rightarrow 80= (n-1)5 $$
$$\Rightarrow 16= n-1 $$
$$\Rightarrow n=17 $$
$$\therefore $$ Option A is correct.
Find the $$11$$th term of the sequence $$5,2,-1,-4,......$$
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$$-25$$
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$$-28$$
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$$-22$$
0%
None
Explanation
$$ a_{1} = 5;d = a_{2}-a_{1} = 2-5 = -3 $$
$$ \therefore a_{11} = a+(n-1)d = 5+10(-3) $$
$$ \Rightarrow {a_{11} = -25} $$
The nth term of the sequence $$2, 4, 6, 8....$$ is
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$$n$$
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$$n - 1$$
0%
$$2n$$
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$$2n - 4$$
Explanation
Clearly, the difference of successive terms of above sequence is constant which is 2
So given sequence is in AP with first term 2 and common difference $$2$$
Hence general term is, $$a_n = a+(n-1)d=2+(n-1)2=2n$$
Hence option 'C' is correct choice
Find the A.P. whose sum to $$n$$ terms is $$2n$$$$^2$$ $$+ n$$
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The required A.P. is $$2, 6, 10, 14,...$$
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The required A.P. is $$3, 7, 11, 15,...$$
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The required A.P. is $$4, 8, 12, 16,...$$
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The required A.P. is $$5, 10, 15, 20,...$$
Explanation
Given, $$S_n=2n^2+n$$
Now, $$a_1=S_1=2(1)^2+1=3$$
$$a_2=S_2-S_1=2(2)^2+2-3=7$$
$$a_3=S_3-S_2=[2(3)^2+3]-[2(2)^2+2]=21-10=11$$
$$a_4=S_4-S_3=[2(4)^2+4]-[2(3)^2+3]=36-21=15$$ and so on.
Therefore the required A.P. is $$3,7,11,15...$$
If the common difference of an AP is $$-6,$$ then what is $$\displaystyle a_{16}-a_{12}?$$
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0%
$$-24$$
0%
$$24$$
0%
$$-30$$
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$$30$$
Explanation
Given, $$d=-6$$
$$a_{16}=a+15d$$
$$a_{12}=a+11d$$
$$a_{16}-a_{12}=(a+15d)-(a+11d)=4d$$
$$\therefore a_{16}-a_{12}=4\times -6=-24$$
If 100 times the $$100^{\mathrm{t}\mathrm{h}}$$ term of an AP with non zero common difference equals the 50 times its $$50^{\mathrm{t}\mathrm{h}}$$ term, then the $$150^{\mathrm{t}\mathrm{h}}$$ term of this AP is:
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$$-150$$
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$$150$$ times its $$50^{th}$$ term
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$$150$$
0%
$$0$$
Explanation
Let say $$a$$ be the first term and $$d$$ be the common difference of the AP.
Given: $$100(a+99d)=50(a+49d)$$ [using $$n^{th}$$ term $$a_{n}= a+(n-1)d $$]
$$\Rightarrow 2a+198d=a+49d$$
$$\Rightarrow a+149d=0$$
Also, $$T_{150}= a+ 149d=0$$
Hence, the option 'D' is correct.
Sum of first $$15$$ terms of $$2 + 5 + 8 + ...$$ is?
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$$44$$
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$$42$$
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$$345$$
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$$386$$
Explanation
Difference between terms $$=5-2=8-5=3$$
We can observe that difference between two consecutive terms is equal, hence the terms are in A.P.
Sum of fist n terms in an A.P. is given by the formula: $$S_n=\dfrac{n}{2}\times[2a+(n-1)d] $$
Given $$n=15,a= 2, d=5-2=3 ,$$
$$S_n=\dfrac{15}{2}\times[2\times2+(15-1)3] $$
$$S_n=\dfrac{15}{2}\times 46$$
$$S_n=345 $$
Check whether the following form an AP
$$\sqrt{3} , \sqrt{12} , \sqrt{27} , \sqrt{48}$$ , ...
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0%
True
0%
False
Explanation
We can write the given series as
$$\sqrt { 3 },2\sqrt { 3 },3\sqrt { 3 },4\sqrt { 3 } $$
Compare the given series with $${a}_{1},{a}_{2},{a}_{3}......$$
we get
$${ a }_{ 2 }-{ a }_{ 1 }=2\sqrt { 3 }-\sqrt { 3 }=\sqrt { 3 }$$
$${ a }_{ 3 }-{ a }_{ 2}=3\sqrt { 3 }-2\sqrt { 3 }=\sqrt { 3 }$$
and so on.
Since common difference is same and equal to $$\sqrt { 3 }$$, therefore given series is AP.
Which of the following is not an A.P.?
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$$13, 8, 3, - 2, - 7, 12$$
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$$10.8, 11.2, 11.6, 12, 12.4$$
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$$8\dfrac{1}{7},18\dfrac{2}{7},28\dfrac{3}{7},48\dfrac{4}{7},58\dfrac{5}{7}$$
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$$8\dfrac{3}{23},11\dfrac{6}{23},14\dfrac{9}{23},17\dfrac{12}{23}$$
Explanation
$$(A)$$ The last term of this sequence is not in AP
$$(B)$$ The sequence is in AP with common difference $$0.4$$
$$(C)$$ $$\cfrac { 57 }{ 7 } ,\cfrac { 128 }{ 7 } ,\cfrac { 199 }{ 7 } ,\cfrac { 340 }{ 7 } ,\cfrac { 411 }{ 7 } $$
The last two terms of this sequence are not in AP
$$(D)$$ $$\cfrac { 187 }{ 23 } ,\cfrac { 259 }{ 23 } ,\cfrac { 331 }{ 23 } ,\cfrac { 403 }{ 23 } $$
The sequence is in AP with common difference $$\cfrac{72}{23}$$
The list of numbers $$10, 6, 2, -2,\cdots$$ is
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0%
an $$A.P.$$ with $$d = 16$$
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an $$A.P.$$ with $$d = -4$$
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an $$A.P.$$ with $$d = 4$$
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not an $$A.P.$$
Explanation
Given series: $$10, 6, 2, -2...$$
Let, $$a_1 = 10$$,
$$a_2 = 6$$,
$$a_3 = 2$$,
$$a_4 = -2$$
$$d_1 = a_2 - a _1 = 6-10 = -4$$
$$d_2 = a_3 - a _2 = 2-6 = -4$$
$$d_3 = a_4 - a _3 = -2-2 = -4$$
The differences are equal, hence common difference of this A.P. is $$d = -4$$
The common difference of the AP $$\dfrac{1}{P}, \dfrac{1-P}{P}, \dfrac{1-2 P}{P}$$, .......... is:
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0%
P
0%
-1
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-P
0%
1
Explanation
Let $$a_{1}=\dfrac{1}{P}$$ ,
$$a_{2}=\dfrac{1-P}{P}$$ ,
$$a_{3}=\dfrac{1-2P}{P}$$
$$\therefore$$ Common difference = $$a_{2}-a_{1} or \hspace{2 mm} a_{3}-a_{2}; $$
=
$$\dfrac{(1-P)-(1)}{P} ; \hspace{2 mm} \dfrac{(1-2P)-(1-P)}{P}$$
= $$-1$$
What is the sum of all natural numbers from 1 to 100?
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5050
0%
50
0%
4550
0%
5150
Explanation
We know,
Sum of n natural numbers $$=\dfrac{n(n+1)}{2}$$
$$\therefore$$ Sum from 1 to 100 $$=\dfrac{100.101}{2}=5050$$
$$(Ans \to A)$$
If the first term of an A.P. is $$-1$$ and common difference is $$-3$$, then its $$12^{th}$$ term is
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0%
$$34$$
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$$32$$
0%
$$-32$$
0%
$$-34$$
Explanation
Given that,
The first term of an A.P. $$(a)=-1$$,
The common difference $$(d)=-3$$
To find out: $$12^{th}$$ term of the A.P.
We know that, $$ n^{th}$$ term of any AP is given by:
$$T_n=a+(n-1)d $$
Hence,
$$T_{12}=(-1)+(12-1)(-3)$$
$$\Rightarrow T_{12}=(-1)+(11)(-3)=-1-33$$
$$\therefore \ T_{12}= -34$$
Hence, the $$12^{th}$$ term of the given A.P. is $$-34$$.
Calculate $$10^{th}$$ term of an AP: $$4, 6, 8, ........ $$
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$$18$$
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$$20$$
0%
$$22$$
0%
$$26$$
Explanation
Let $$S=4+6+8+...\infty$$
The above series is an arithmetic progression with common difference of $$2$$.
We know that,
$$a_{n}=a_{1}+(n-1)d$$
$$\therefore a_{10}=4+(10-1)(2)$$
$$=4+9(2)$$
$$=18+4$$
$$=22$$
Write the sum of first five terms of the following Arithmetic Progressions where, the common difference $$d$$ and the first term $$a$$ are given,
$$a = 2, d = 2.5$$
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0%
$$32$$
0%
$$33$$
0%
$$34$$
0%
$$35$$
Explanation
Here the first term $$a=2$$ and common difference $$d=2.5$$.
Now $$a_n=a+(n-1)d$$
$$\therefore a_1=a=2$$
$$a_2=a+d=2+2.5=4.5$$
$$a_3=a+2d=2+2\times2.5=7$$
$$a_4=a+3d=2+3\times2.5=9.5$$
$$a_5=a+4d=2+4\times2.5=12$$
$$\therefore $$ series in A.P. is $$2, 4.5, 7, 9.5, 12,....$$
Then, the sum of first five terms is $$=2+4.5+7+9.5+12=35$$
Or
We can use the formula for sum of A.P $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$\therefore S_5$$ =$$\dfrac{5}{2} \times [2(2)+(5-1)(2.5)]=35$$
A property valued at Rs. 30000 will depreciate Rs. 1380 the first year, Rs. 1340 the second year, Rs. 1300 the third year, and so on. What will the depreciation during the eight year?
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Rs. 1150
0%
Rs. 1200
0%
Rs. 1100
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Rs. 1250
Explanation
Every year depreciation decrease by 40, so the given problem can be solved through the AP: 1380, 1340, 1300, ........
So $$a = 1380, d = 40, n = 8$$
$$\therefore$$ Depreciation during eight year
$$=1380 + (8-1) \times (-40) = Rs. 1100$$
For an A.P., the seventh term is 19 and the thirteenth term isFind its twenty second term.
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0%
60
0%
54
0%
64
0%
78
Explanation
For the given A.P., $$T_7=19$$ and $$ T_{13}=37$$
Now $$T_n = a+(n-1)d$$
$$\therefore T_7 = a + (7-1)d$$
$$\therefore 19=a+6d$$
$$\therefore a+6d = 19$$ ............ (1)
Again, $$T_{13} = a + (13-1)d$$
$$\therefore 37 = a + 12d$$
$$\therefore a+12 d =37$$ ............. (2)
Subtracting eq. (1) from eq. (2)
$$a + 12 d = 37$$
$$a + 6 d = 19$$
$$6d = 18$$
$$\therefore d=13$$
Substituting $$d=13$$ in $$a+6d =19$$,
$$a+6(3)=19$$
$$\therefore a+18 = 19$$
$$\therefore a =1$$
Thus, the first term of the A.P. is 1 and the common difference is 3.
Now, $$T_{22} =a + (22-1) d=1 + (22-1) (3)$$
$$=1 +(21) (3) =1+63=64$$
Thus, the twenty second term of the given A.P. is 64.
In a flower bed, there are $$23$$ rose plants in the first row, $$21 $$ in the second, $$19$$ in the third, and so on. There are $$5$$ rose plants in the last row. The number of rows in the flower bed are......
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0%
$$11$$
0%
$$41$$
0%
$$10$$
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$$9$$
Find the sum of all odd natural numbers from 1 to 150.
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0%
5625
0%
5635
0%
5645
0%
5655
Explanation
Clearly, the odd natural numbers from $$1$$ to $$150$$ are $$1,3,5,...,149$$.
This is an AP with first term $$a=1$$, common difference $$d=2$$ and last term $$a_n=l = 149$$.
Let there be $$n$$ terms in this AP.
Then,
$$a_n=149$$
$$\Rightarrow a_n = a+(n-1)d$$
$$149= 1+(n-1)\times2$$
$$149=1+ 2n-2$$
$$2n=150$$
$$\therefore n=75$$
$$\therefore$$ required sum $$=S_n=\dfrac n2[a+l]=\displaystyle \frac{75}2[1+149]=5625$$
The sum of 24 terms of the following series : $$\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} +$$ ..................
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$$300$$ $$\sqrt{2}$$
0%
$$150$$ $$\sqrt{2}$$
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$$250$$ $$\sqrt{2}$$
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$$325$$ $$\sqrt{2}$$
Explanation
We have
$$\sqrt{2} + \sqrt{8} + \sqrt{18} + \sqrt{32} +$$ ..................
We can rewrite the series as $$1\sqrt{2}+2\sqrt{2}+3\sqrt{2} + 4 \sqrt{2}+........$$
Taking $$\sqrt{2}$$ common, we get:
$$\sqrt{2} \ [1+2+3+4+....... \text{upto} \ 24 \ \text{terms}] $$
The $$24^{th}$$ term will be $$24$$.
Hence, $$\sqrt{2} \ [1+2+3+4+....... +24] $$
We know that, the sum of $$n$$ terms of an AP is given by
$$S_{n}=\frac {n}{2}(a+l)$$
Here, $$a = 1, l = 24, n= 24 $$
$$\therefore \ S_{24}=\sqrt{2}\left[\frac{24}{2}(1+24)\right]$$
$$\Rightarrow \sqrt{2} \times \displaystyle \frac{24 \times 25}{2} = 300 \sqrt{2}$$
Hence, the sum of $$24$$ terms of the given series is $$300\sqrt{2}$$.
Write the sum of first five terms of the following Arithmetic Progressions where, the common difference $$d$$ and the first term $$a$$ are given:
$$a = 4, d = 0$$
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0%
$$20$$
0%
$$22$$
0%
$$24$$
0%
$$26$$
Explanation
Here the first term $$a=4$$ and common difference $$d=0$$.
Now $$a_n=a+(n-1)d$$
$$\therefore a_1=a=4$$
$$a_2=a+d=4+0=4$$
$$a_3=a+2d=4+2\times0=4$$
$$a_4=a+3d=4+3\times0=4$$
$$a_5=a+4d=4+4\times0=4$$
$$\therefore$$ series in A.P. is $$4,4,4,4,4,....$$
Thus, the sum of first five terms is $$4+4+4+4+4=20$$
Hence, the answer is $$20$$.
Or
We can use the formula for sum of A.P $$S_n=\dfrac{n}{2} ( 2a + ( n - 1 ) d )$$
$$\therefore S_5$$$$=$$$$\dfrac{5}{2} \times [2(4)+(5-1)(0)]$$
$$=20$$
Summation of $$n$$ terms of an A.P. is
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$$\dfrac {n}{2}(a+l)$$
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$$\dfrac {n}{2}[2a+(n-1)d]$$
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$$\dfrac {a(r^n-1)}{(r-1)}$$
0%
$$\dfrac {a(1-r^n)}{(1-r)}$$
Explanation
General equation for the sum of n terms of an AP is given by:
$${\dfrac{n}{2}}{(2a+(n-1)d)}$$ $$=$$ $${\dfrac{n}{2}}{(a+l)}$$
$$S_n=54+51+48+....n$$ terms $$=513$$. Value of $$n$$ is
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0%
$$18$$
0%
$$19$$
0%
$$15$$
0%
None of these above
Explanation
From the given series, we have
$$S_n=513$$
$$\Rightarrow \dfrac {n}{2}[2(54)+(n-1)(-3)]=513$$
$$\Rightarrow n(108-3n+3)=1026$$
$$\Rightarrow n^2-37n+342=0$$
$$\Rightarrow n^2-19n-18n+342=0$$
$$\Rightarrow n(n-19)-18(n-19)=0$$
$$\Rightarrow (n-18)(n-19)=0$$
Therefore, $$n=18$$ or $$n=19$$
The nth term of the A.P. $$3, 7, 11, 15, ....$$ is given by -
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$$T_n = 4n-1$$
0%
$$T_n=3n+4$$
0%
$$T_n=3n-4$$
0%
$$T_n = 2n+1$$
Explanation
Given A.P. is $$3,7,11,15,....$$
Here $$a=3, d=4$$
We know $$T_n = a+(n-1) d $$
$$= 3 + (n-1) 4$$
$$ = 4n-1$$
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