Question 13 Marks
Find the general term ($n^{th}$ term) and $23^{rd}$ term of the sequence$ 3, 1, -1, -3, ….. .$
AnswerThe given sequence is $1, -1, -3, …..$
Now,
$1 - 3 = -1 - 1 = -3 - (-1) = -2$
Hence, the given sequence is an $A.P.$ with first term $a = 3$ and common difference $d = -2.$
The general term ($n^{th}$ term) of an $A.P$. is given by
$t_n = a + (n - 1)d$
$= 3 + (n - 1)(-2)$
$= 3 - 2n + 2$
$= 5 - 2n$
Hence, $23^{rd} term = t_{23} = 5 - 2(23) = 5 - 46 = -41$
View full question & answer→Question 23 Marks
An A.P. consists of 50 terms of which $3^{\text {rd }}$ term is 12 and the last term is 106 . Find the $29^{\text {th }}$ term of the A.P.
AnswerFor a given $A.P.,$
Number of terms, $n = 50$
$3^{rd}$ term,$ t_3 = 12$
$\Rightarrow a + 2d = 12 ….(i)$
Last term, $l = 106$
$\Rightarrow t_{50} = 106$
$\Rightarrow a + 49d = 106 ….(ii)$
Subtracting (i) from (ii), we get
$47d = 94$
$\Rightarrow d = 2$
$\Rightarrow a + 2(2) = 12$
$\Rightarrow a = 8$
Hence, $t_{29} = a + 28d = 8 + 28(2) = 8 + 56 = 64$
View full question & answer→Question 33 Marks
If the third and the $9^{th}$ term of an A.P. be $4$ and $-8$ respectively, find which term is zero?
AnswerFor an A.P., a
$t_3=4$
$\Rightarrow a+2 d=4 \ldots \ldots (i)$
$t_9=-8$
$\Rightarrow a+8 d=-8 \ldots \ldots (ii)$
subtracting $(i)$ from $(ii),$ we get
$6d = -12$
$\Rightarrow d=-2$
Substituting $d = -2$ in $(i),$ we get
$a + 2(-2) = 4$
$\Rightarrow a -4=4$
$ \Rightarrow a =8 $
$ \Rightarrow \text { General term }= t _{ n }=8+( n -1)(-2)$
Let $p ^{\text {th }}$ term of this A.P be $0$
$\Rightarrow 8+(p-1) \times(-2)=0$
$\Rightarrow 8-2 p+2=0 $
$ \Rightarrow 10-2 p=0$
$ \Rightarrow 2 p=10 $
$ \Rightarrow p=5$
Thus, $5^{th}$ term of this A.P is $0$
View full question & answer→Question 43 Marks
Find the sum of last 8 terms of the A.P. $-12, -10, -8, ……, 58.$
AnswerFirst we will reverse the given A.P. as we have to find the sum of last 8 terms of the A.P.
$58, …., -8, -10, -12.$
Here $a = 58 , d = -2$
$S_n=\frac{n}{2}(2 a+(n-1) d)$
$\Rightarrow S_8=\frac{8}{2}(2 \times 58+(8-1)(-2)) $
$ =4(116-14)$
$=4 \times 102$
$ =408$
Therefore the sum of last 8 terms of the A.P.$ -12, -10, -8, ……, 58$ is $408.$
View full question & answer→Question 53 Marks
Can $2 n^2-7$ be the $n^{\text {th }}$ term of an $A.P?$ Explain.
AnswerWe have $2 n^2-7$
Substitute $n =1,2,3, \ldots$, we get
$2(1)^2-7,2(2)^2-7,2(3)^2-7,2(4)^2-7, \ldots$
$-5,1,11, \ldots$
Difference between the first and second term $=1-(-5)=6$
And Difference between the second and third term $=11-1=10$
Here, the common difference is not same.
Therefore the $n ^{\text {th }}$ term of an A.P can't be $2 n^2-7$.
View full question & answer→Question 63 Marks
Divide 216 into three parts which are in A.P. and the product of the two smaller parts is 5040.
AnswerLet the three parts of 216 in A.P be (a - d), a, (a + d).
⇒a - d + a + a + d = 216
⇒ 3a = 216
⇒ a = 72
Given that the product of the two smaller parts is 5040.
⇒a(a - d ) = 5040
⇒ 72(72 - d) = 5040
⇒ 72 - d = 70
⇒ d = 2
∴ a - d = 72 - 2 = 70, a = 72 and a + d = 72 + 2 = 74
Therefore the three parts of 216 are 70, 72 and 74.
View full question & answer→Question 73 Marks
How many three digit numbers are divisible by 7?
AnswerThe first three digit number which is divisible by $7$ is $105$ and the last digit which is divisible by $7$ is $994 .$
This is an A.P. in which $a=105, d=7$ and $t_n=994$.
We know that $n^{\text {th }}$ term of A.P is given by
$t_n = a + (n - 1)d.$
$\Rightarrow 994 = 105 + (n - 1)7$
$\Rightarrow 889 = 7n - 7$
$\Rightarrow 896 = 7n$
$\Rightarrow n = 128$
$\therefore$ There are $128$ three digit numbers which are divisible by $7 .$
View full question & answer→Question 83 Marks
The $6^{\text {th }}$ term of an $A.P.$ is $16$ and the $14^{\text {th }}$ term is $32$ . Determine the $36^{\text {th }}$ term.
AnswerLet 'a' be the first term and ' $d$ ' be the common difference of the given A.P.
Now, $t _6=16$ (given)
$\Rightarrow a+5 d=16$
And,
$t_{14} = 32 (given)$
$\Rightarrow a + 13d = 32 ….(ii)$
Subtracting (i) from (ii), we get
$8d = 16$
$\Rightarrow d = 2$
$\Rightarrow a + 5(2) = 16$
$\Rightarrow a = 6$
Hence, $36^{\text {th }}$ term $= t _{36}= a +35 d=6+35(2)=76$
View full question & answer→Question 93 Marks
Which term of the A.P. $105, 101, 97 …$ is the first negative term?
AnswerHere $a = 105$ and $d = 101 - 105 = -4$
Let $a_n$ be the first negative term.
$\Rightarrow a_n < 0$
$\Rightarrow a + (n - 1)d < 0$
$\Rightarrow 105 + (n - 1)(-4)<0$
$\Rightarrow 105 - 4n + 4 <0$
$\Rightarrow 109 - 4n < 0$
$\Rightarrow 109 <4n$
$\Rightarrow 27.25 < n$
The value of $n = 28$.
Therefore $28^{th}$ term is the first negative term of the given A.P.
View full question & answer→Question 103 Marks
If the$ n^{th}$ term of the $A.P. 58, 60, 62,....$ is equal to the $n^{th}$ term of the $A.P. -2, 5, 12, ….,$ find the value of n.
AnswerIn the first $A.P. 58, 60, 62,....$
$a = 58$ and $d = 2$
$t_n = a + (n - 1)d$
$\Rightarrow t_n = 58 + (n - 1)2 …. (i)$
In the first $A.P. -2, 5, 12, ….$
$a = -2$ and $d = 7$
$t_n = a + (n - 1)d$
$\Rightarrow t_n= -2 + (n - 1)7 …. (ii)$
Given that the $n^{th}$ term of first A.P is equal to the $n^{th}$ term of the second A.P.
$\Rightarrow 58 + (n - 1)2 = -2 + (n - 1)7 …$ from (i) and (ii)
$\Rightarrow 58 + 2n - 2 = -2 + 7n - 7$
$\Rightarrow 65 = 5n$
$\Rightarrow n = 13$
View full question & answer→Question 113 Marks
The $25^{\text {th }}$ term of an A.P. exceeds its $9^{\text {th }}$ term by 16 . Find its common difference.
Answer$n^{\text {th }}$ term of an A.P. is given by $t_n=a+(n-1) d$.
$\Rightarrow t_{25}=a+(25-1) d=a+24 d \text { and }$
$t_9=a+(9-1) d=a+8 d$
According to the condition in the question, we get
$t_{25}=t_9+16$
$\Rightarrow a+24 d=a+8 d+16$
$\Rightarrow 16 d=16$
$\Rightarrow d=1$
View full question & answer→Question 123 Marks
How many two digit numbers are divisible by $3?$
AnswerThe two digit numbers divisible by $3$ are as follows:
$12,15,18,21,............,99$
Clearly, this forms an A.P with first term, $a = 12$
and common difference, $d = 3$
Last term $= n^{th}$ term $= 99$
The general term of an A.P is given by
$t_n = a + (n - 1)d$
$\Rightarrow 99=12+(n-1)(3)$
$\Rightarrow 99=12+3 n-3$
$\Rightarrow 90=3 n$
$\Rightarrow n =30$
Thus, $30$ two digit numbers are divisible by $3.$
View full question & answer→Question 133 Marks
Is $-150$ a term of $11, 8, 5, 2,$ .......?
AnswerThe given sequence is 11, 8, 5, 2, …..
Now,
8 - 11 = 5 - 8 = 2 - 5 = -3
Hence, the given sequence is an A.P. with first term a = 11 and common difference d = -3.
The general term of an A.P. is given by
$t_n = a + (n - 1)d$
$\Rightarrow -150 = 11 + (n - 1)(-3)$
$\Rightarrow -161 = -3n + (-3)$
$\Rightarrow 3n = 164$
$\Rightarrow n=\frac{164}{3}$
The number of terms cannot be a fraction.
So, clearly, -150 is not a term of the given sequence.
View full question & answer→Question 143 Marks
Which term of the sequence $3, 8, 13, ........$ is $78$?
AnswerThe given sequence is $3, 8, 13, …..$
Now,
$8 - 3 = 13 - 8 = 5$
Hence, the given sequence is an A.P. with first term $a = 3$ and common difference $d = 5.$
Let the $n^{th}$ term of the given A.P. be 78.
$\Rightarrow 78 = 3 + (n - 1)(5)$
$\Rightarrow 75 = 5n - 5$
$\Rightarrow 5n = 80$
$\Rightarrow n = 16$
Thus, the $16^{th}$ term of the given sequence is $78$.
View full question & answer→Question 153 Marks
Mrs. Gupta repays her total loan of Rs.$1,18,000$ by paying installments every month. If the installments for the first month is Rs. $1,000$ and it increases by Rs. $100$ every month, What amount will she pays as the $30^{th}$ installments of loan? What amount of loan she still has to pay after the $30^{th}$ installments
AnswerTotal amount of loan $=$ Rs. $1,18,000$
First installment $= a =R s .1000$
Increase in instalment every month $= d = Rs. 100$
$30^{t h} \text { installment } =t_{30}$
$ =a=29 d$
$ =1000+29 \times 100$
$ =1000+2900$
$ =R s .3900$
Now, amount paid in 30 installments $=s_{30}$
$=\frac{30}{2}(2 \times 1000+29 \times 100)$
$=15(2000+2900)$
$=15 \times 4900$
$=\text { Rs. } 73,500$
$\therefore$ Amount of loan to be paid after the $30^{\text {th }}$ installent
$=R s .(1,18,0000-73,500)$
$=R s .44,500$
View full question & answer→Question 163 Marks
Two cars start together in the same direction from the same place. The first cargoes at uniform speed of $10\ km h^{-1}$. The second car goes at a speed of $8\ km h^{-1}$ in the first hour and thereafter increasing the speed by $0.5\ km h^{-1}$ each succeeding hour. After how many hours will the two cars meet?
AnswerLet the two cars meet after n hours.
That means the two cars travel the same distance in n hours.
Distance travelled by the $1^{\text {st }}$ car in $n$ hours $=10 \times n km$ Distance travelled by the $2^{\wedge}$ nd car in $n$ hours $=\frac{n}{2}(2 \times 8+(n-1) \times 0.5) km$
$\Rightarrow 10 \times n=\frac{n}{2}(2 \times 8+(n-1) \times 0.5)$
$\Rightarrow 20=16+0.5 n-0.5$
$\Rightarrow 0.5 n=4.5$
$\Rightarrow n=9$
Thus , the two cars will meet after $9$ hours.$d08$
View full question & answer→Question 173 Marks
Split $207$ into three parts such that these partsare in A.P. and the productof the two smaller parts in $4623.$
AnswerLet the tree parts in A.P nbe $(a-d)$, $a$ and $(a+d)$ Then, $(a-d)+a+(a+d)=207$
$\Rightarrow 3 a=207 $
$ \Rightarrow a=69$
It is given that
$(a-d) \times a=4623$
$ \Rightarrow(69-d) \times 69=4623$
$ \Rightarrow 69-d=67 $
$\Rightarrow d=2$
$\Rightarrow a=69$ and $d=2$
Thus, We have
$a-d=69-2=67$
$a=69$
$a+d=69+2=71$
Thus, the tree parts in A.P are $67,69$ and $71$
View full question & answer→Question 183 Marks
The angle of a quadrillateral are A.P. with common difference 20°.find its angles.
AnswerLet the four angle of a quadrilateral in AP.be a,
$a+20^{\circ}, a+40^{\circ}$ and $a+60^{\circ}$
$\therefore a+\left(a+20^{\circ}\right)+\left(a+40^{\circ}\right)+\left(a+60^{\circ}\right)=360^{\circ}.......$ (Angle sum property)
$\Rightarrow 4 a+120^{\circ}=360^{\circ}$
$\Rightarrow 4 a=240^{\circ}$
$\Rightarrow a=60^{\circ}......(1)$
Thus, angles of a quarilateral are $=a, a+20^{\circ}, a+40^{\circ}$ and $60^{\circ}$
$=60^{\circ}, 80^{\circ}, 100^{\circ}$ and $120^{\circ}$
View full question & answer→Question 193 Marks
$\frac{1}{a}, \frac{1}{b}$ and $\frac{1}{c}$ are in A.P show that : $bc , ca$ and $ab$ also in A.P.
Answer$\frac{1}{a}, \frac{1}{b}$ and $\frac{1}{c}$ are in A.P.
$\Rightarrow \frac{1}{b}-\frac{1}{a}=\frac{1}{c}-\frac{1}{b}$
$\Rightarrow \frac{a-b}{a b}=\frac{b-c}{b c}$
$\Rightarrow \frac{a-b}{a}=\frac{b-c}{c}$
$\Rightarrow a c-b c=a b-a c$
$\Rightarrow a c+a c=a b+b c$
$\Rightarrow 2 a c=a b+b c$
$\Rightarrow 2 c a=a b+b c$
$\Rightarrow b c, c a$ and $a b$ are also in A.P
View full question & answer→Question 203 Marks
The first and the last terms of an A.P. are $34$ and $700$ respectively. If the common difference is $18,$ how many terms are there and what is their sum?
AnswerLet there be n terms in this A.P
First-term $a = 34$
Common difference $d = 18$
Last term $l = 700$
$\Rightarrow a+(n-1) d=700 $
$\Rightarrow 34+(n-1) \times 18=700 $
$\Rightarrow(n-1) \times 18=666$
$ \Rightarrow n -1=37 $
$ \Rightarrow n =38$
Sum of first $n$ terms $=$
$\frac{n}{2}[a+l]=\frac{38}{2}[34+700]=19 \times 734=13946$
View full question & answer→Question 213 Marks
Find the sum of all natural numbers between $250$ and $1000$ which are divisible by $9.$
AnswerNatural numbers between $250$ and $1000$ which are divisible by $9$ are as follows:
$252, 261, 270, 279, ......, 999$
Clearly, this forms an A.P with the first term $a = 252,$ common difference $d = 9$ and last term $l = 99$
$l = a + (n -1)d$
$\Rightarrow 999=252+(n-1) \times 9$
$\Rightarrow 747=(n-1) \times 9$
$\Rightarrow n-1=83$
$\Rightarrow n=84$
Sum of first $n$ terms $=S=\frac{n}{2}[a+1]$
$\Rightarrow$ Sum of nattural numbers between $250$ and $1000$ whihc are divisible by $9$
$=\frac{84}{2}[252+999] $
$ =42 \times 1251 $
$=52542$
View full question & answer→Question 223 Marks
Find the sum of first $51$ terms of an A.P. whose $2$nd and $3$rd terms are $14$ and $18$ respectively.
AnswerGiven $t_2=14$ and $t_3=18$
$\Rightarrow d=t_3-t_2=18-14=4$
Now $t_2=14$
$\Rightarrow a + d = 14$
$\Rightarrow a = 4 = 14$
$\Rightarrow a = 10$
Sum of $n$ terms of an A.P $=\frac{n}{2}[2 a+(n-1) d]$
$\therefore$ Sum of first 51 terms of an A.P $=\frac{51}{2}[2 \times 10+50 \times 4]$
$=\frac{51}{2}[20+200] $
$=51 \times 110$
$=5610$
View full question & answer→Question 233 Marks
Find the sum of $28$ terms of an A.P. whose nth term is $8n - 5.$
Answernth term of an A.P. $=t_n=8 n-5$
Let a be the first and d be the common difference of this A.P.
Then
$a=t-1=8 \times 1-5=8-5=3$
$ t-2=8 \times 2-5=16-5=11$
$ \therefore d=t_2-t_1=11-3=8$
The sum of $n$ terms of an A.P, $=S=\frac{n}{2}[2 a+(n-1) d]$
$\Rightarrow \text { Sum of } 28 \text { terms of AP }=\frac{28}{2}[2 \times 3+27 \times 8] $
$=14[6+216] $
$ =14 \times 222 $
$ =3108$
View full question & answer→Question 243 Marks
The sum of $n$ natural number is $5 n^2+4 n$. Find its $8^{\text {th }}$ term.
AnswerSum of n natural numbers $S_n=5 n^2+4 n$
$\Rightarrow$ Sum of $(n-1)$ natural numbers $=S_{n-1}=5(n-1)^2+4(n-1)$
$=5\left(n^2+1-2 n\right)+4 n-4 $
$ =5 n^2+5-10 n+4 n-4 $
$ =5 n^2-6 n+1$
$n ^{\text {th }}$ term $=S_n=S_{n-1}=5 n^2+4 n-5 n^2+6 n-1=10 n-1$
$\Rightarrow$ "$8^{\text {th }}$ term" $=$ t $_8=10 x \times 8-1=80-1=79 $
View full question & answer→Question 253 Marks
Determine the A.P. Whose $3$rd term is $16$ and the $7$th term exceeds the $5$th term by $12.$
AnswerFor given A.P
$t_3=a+2 d=16 \ldots$..(i)
Now
$t_7-t_5=12$
$\Rightarrow(a+6 d)-(a=4 d)=12$
$\Rightarrow 2 d =12$
$\Rightarrow d =6$
Substituting the value of din $(i)$ we get
$a+2 \times 6=16$
$\Rightarrow a+12=16$
$\Rightarrow a=4$
Thus the required A.P $= a, a + d, a + 2d, a + 3d,....$
$= 4, 10, 16, 22,.....$
View full question & answer→Question 263 Marks
For what value of n, the nth term of A.P $63, 65, 67, ……..$ and nth term of A.P. $3, 10, 17,……..$ are equal to each other?
AnswerFor an A.P $63, 65, 67, ..... $ we have $a = 63$ and $d = 65 - 63 = 2$
nth term $=t_n=63+(n-1) \times 2$
For an A.P $3, 10, 17, ....,$ we have $a' = 3$ and $d' = 10 - 3 = 7$
nth term $=t I_n=3+(n-1) \times 7$
The two A.Ps will have equal nth term is
$t_n=t \prime_n$
$\Rightarrow 63+(n-1) \times 2=3+(n-1) \times 7 $
$\Rightarrow 63+2 n=7 n-4$
$ \Rightarrow 5 n=65 $
$ \Rightarrow n=13$
View full question & answer→Question 273 Marks
How many three-digit numbers are divisible by $87?$
AnswerThe three-digit number divisible by $87$ are as follow $174, 261, ....., 957$
Clearly, this forms an A.P. with the first term $a= 174$ and common difference $d = 87.$
Last term $= n^{th}$ term $= 957$
The general term of an A.P is given by
$t_n=a+(n-1) d$
$\Rightarrow 957=174+(n-1)(87) $
$ \Rightarrow 783=(n-1) \times 87$
$\Rightarrow 9 = n - 1$
$\Rightarrow n = 10$
Thus $10$ three digit numbers are divisble by $87.$
View full question & answer→Question 283 Marks
Which term of A.P. $5, 15, 25 …………$ will be $130$ more than its $31st$ term?
AnswerThe given A.P is $5, 15, 25,...$
Here, $a = 5$ and $d = 15 - 5 = 10$
Now $t_{31}=a+30 d=5+30 \times 10=5+300=305$
Let the required term be nth term.
$\therefore t_n-t_{31}=130$
$\Rightarrow$ $5 + (n - 1)(10) = 435$
$\Rightarrow$ $n - 1 = 43$
$\Rightarrow$ $n = 44$
Thus reuired term $= 44^{th}$ term.
View full question & answer→Question 293 Marks
In an A.P., ten times of its tenth term is equal to thirty times of its $30$th term. Find its $40$th term.
AnswerThe general term of an A.P is given by
$t_n=a+(n-1) d$
Given
$10 x t_{10}=30 \times t_{30}$
$\Rightarrow 10 \times(a+9 d)=30 \times(a+29 d)$
$\Rightarrow a + 9d = 3a + 87d$
$\Rightarrow 2a + 78d = 0$
$\Rightarrow a + 39d = 0$
$\Rightarrow a = -39d$
Now $t_{40}=a+39 d=-39 d+39 d=0$
View full question & answer→Question 303 Marks
Which term of A.P $3, 10, 17, ....$ Will be $84$ more than its $13^{th}$ term?
AnswerThe given A.P is $3, 10, 17, ....$
Here $a = 3,$ and $d = 10 - 3 = 7$
Now
$t_{13}=a+12 d=3+12 \times 7=3+84=87$
Let the required term be nth term
$\therefore t_n-t_{13}=84$
$\Rightarrow [a + (n - 1)d] - 87 = 84$
$\Rightarrow 3+(n-1) \times 7=171$
$\Rightarrow(n-1) \times 7=168$
$\Rightarrow n -1=24$
$\Rightarrow n =25$
Thus required term $= 25^{th}$ term
View full question & answer→Question 313 Marks
If a, b and c are in A.P show that 4a, 4b and 4c are in A.P
Answera, b and c are in AP
=> b - a = c - b
=> 2b = a + c
Given terms are 4a, 4b and 4c
Now 4b - 4a = 2(2b - 2a)
= 2(a + c - 2a)
= 2(c - a)
And 4c - 4b = 2(2c - 2b)
= 2(2c - a - c)
= 2(c - a)
Since 4b - 4a = 4c - 4b, the given terms are in A.P
View full question & answer→Question 323 Marks
Determine the value of $k$ for which $k^2+4 k+8,2 k^2+3 k+6$ and $3 k^2+4 k+4$ are in A.P.
AnswerSince $\left(k^2+4 k+8\right),\left(2 k^2+3 k+6\right)$ and $\left(3 k^2+4 k+4\right)$ are in A.P, we have
$\left(2 k^2+3 k+6\right)-\left(k^2+4 k+8\right)=\left(3 k^2+4 k+4\right)-\left(2 k^2+3 k+6\right)$
$ \Rightarrow 2 k^2=3 k+6-k^2-4 k-8=3 k^2+4 k+4-2 k^2-3 k-6 $
$ \Rightarrow k^2-k-2=k^2+k-2$
$\Rightarrow 2 k=0$
$\Rightarrow k =0$
View full question & answer→Question 333 Marks
Find the common difference and $99^{th}$ term of the arthimetic progression:
$7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, \ldots$.
AnswerFind the common difference and $99^{th}$ term of the arithmetic progression:
The given A.P is $7 \frac{3}{4}, 9 \frac{1}{2}, 11 \frac{1}{4}, \ldots$
i.e $\frac{31}{4}, \frac{19}{2}, \frac{45}{4}, \ldots$.
Common difference $=d=\frac{19}{2}-\frac{31}{4}=\frac{38-31}{4}=\frac{7}{4}=1 \frac{3}{4}$
First term $=a=\frac{31}{4}$
The general term of an A.P is given by
$t_n=a+(n-1) d$
$\Rightarrow t_{99}=\frac{31}{4}+(99-1) \times \frac{7}{4}=\frac{31}{4}+98 \times \frac{7}{4}=\frac{31}{4}+\frac{686}{4}=\frac{717}{4}=179 \frac{1}{4}$
View full question & answer→Question 343 Marks
Find the $50^{th}$ term of the sequence:
$\frac{1}{n}, \frac{n+1}{n}, \frac{2 n+1}{n}, \ldots$
AnswerThe given sequence is $\frac{1}{n}, \frac{n+1}{n}, \frac{2 n+1}{n} \ldots$
Now
$\frac{n+1}{n}-\frac{1}{n}=\frac{n+1-1}{n}=\frac{n}{n}=1 $
$ \frac{2 n+1}{n}-\frac{n+1}{n}=\frac{2 n+1-n-1}{n}=\frac{n}{n}=1 \text { etc }$
Hence the given sequence is an A.P. with first term $a = 1/n$ and common difference $d= 1$
The general term of an A.P is given by
$t_n=a+(n-1 d) $
$ \Rightarrow t_{50}=\frac{1}{n}+(50-1)(1)=\frac{1}{n}+49$
So the $50$ th term is $\frac{1}{n}+49$
View full question & answer→Question 353 Marks
Find the $100^{th}$ term of the sequence:
$\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3} \ldots \ldots$
AnswerThe given A.P is $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}$
Now
$2 \sqrt{3}-\sqrt{3}=\sqrt{3}$
$3 \sqrt{3}-2 \sqrt{3}=\sqrt{3}$ etc
hence the given sequence is an A.P with first term $a=\sqrt{3}$ and common difference $d=\sqrt{3}$.
The general term of an A.P is given by
$t_n=a+(n-1) d $
$ \Rightarrow t_{100}=\sqrt{3}+(100-1) \times \sqrt{3}=\sqrt{3}+996 \sqrt{3}=100 \sqrt{3}$
So the $100$ th term is $100 \sqrt{3}$
View full question & answer→Question 363 Marks
Find the 30th term of the sequence
$\frac{1}{2}, 1, \frac{3}{2}, \ldots$
AnswerThe given sequence is $\frac{1}{2}, 1, \frac{3}{2}$
Now
$1-\frac{1}{2}=\frac{1}{2}$
$\frac{3}{2}-1=\frac{1}{2}$, etc
hence the given sequence is an A.P with the first term $a=\frac{1}{2}$
and common difference $d =\frac{1}{2}$.
The general term of an A.P is given by
$t_n=a+(n+1) d$
$\Rightarrow t_{30}=\frac{1}{2}+(30-1)\left(\frac{1}{2}\right)=\frac{1}{2}+\frac{29}{2}=\frac{30}{2}=15$
So the 30th term is 15.
View full question & answer→Question 373 Marks
Find the 24th term of the sequence: 12, 10, 8, 6,…
AnswerThe given sequence is 12, 10, 8, 6, ....
Now
10 - 12 = -2
8 - 10 = -2
6 - 8 = -2, etc
Hence the given sequence is an A.P. with first term a = 12 and common difference d = -2
The general term of an A.P is given by
$t_n=a+(n-1) d$
$\Rightarrow t_{24}=12+(24-1)(-2)=12+23 \times(-2)=12-46=-34$
So the 24th term is -34
View full question & answer→Question 383 Marks
An A.P. consists of $60$ terms, If the first and the last terms be $7$ and $125$ respectively, find the $31^{st}$ term.
AnswerFor a given A.P.,
Number of terms, $n = 60$
First term, $a = 7$
Last term, $l = 125$
$\Rightarrow t_{60} = 125$
$\Rightarrow a + 59d = 125$
$\Rightarrow 7 + 59d = 125$
$\Rightarrow 59d = 118$
$\Rightarrow d = 2$
Hence, $t_{31} = a + 30d = 7 + 30(2) = 7 + 60 = 67$
View full question & answer→Question 393 Marks
Determine the arithmetic progression whose 3rd term is 5 and 7th term is 9.
AnswerFor an A.P
$t_3=5$
$\Rightarrow a+2 d=5$
And $t_7=9$
$\Rightarrow$ a + 6d = 9 ....(ii)
Substracting (i) from (ii) we get
4d = 4
$\Rightarrow$ d = 1
Substituting d = 1 in (i) we get
$a+2 \times 1=5$
$\Rightarrow$ a = 3
Thus the required A.P = a, a + d, a + 2d. a + 3d, ...
= 3, 4, 5, 6, .....
View full question & answer→Question 403 Marks
Find the 10th term from the end of the A.P 4, 9, 14, ..... 254
AnswerThe given A.P is 4, 9, 14, ... 254.
First term = 4
Common differenc3e = 9 - 4 = 5
Last term = l =254
For the reverse A.P first term = 254 and common difference = -5
Thus 10th term from the end of a given A.P
= 10th term from the end of an given A.P
$=254+(10-1) \times(-5)$ = 254 - 45
= 209
View full question & answer→Question 413 Marks
Which term of the A.P $1 + 4 + 7 + 10 + .....$ is $52?$
AnswerThe given A.P is $1 + 4 + 7 + 10 +.....$
Here first term $a = 1$ and common difference $d = 4 - 1 = 3$
Let nth term of the given A.P be $52$
$\Rightarrow 52=a+(n-1) d$
$\Rightarrow 52=1+(n-1) \times 3$
$\Rightarrow 51=(n-1) \times 3$
$\Rightarrow n -1=17$
$\Rightarrow n =18$
Thus the 18 th term of the given A.P is $52.$
View full question & answer→Question 423 Marks
How many terms are there in the series:
$\frac{3}{4}, 1,1 \frac{1}{4}, \ldots \ldots, 3$
AnswerThe given series is $\frac{3}{4}, 1,1 \frac{1}{4}, \ldots, 3 \Rightarrow \frac{3}{4}, 1 \cdot \frac{5}{4}, \ldots, 31-\frac{3}{4}=\frac{1}{4}, \frac{5}{4}-1=\frac{1}{4}$, etc
Thus the given series is an A.P with first term $a = 3/4$ and common difference $d = 1/4$
Last term $= l = 3$
$\frac{3}{4}+(n-1)\left(\frac{1}{4}\right)=3$
$\Rightarrow(n-1) \times \frac{1}{4}=3-\frac{3}{4} $
$ \Rightarrow(n-1) \times \frac{1}{4}=\frac{9}{4} $
$ \Rightarrow n-1=9 $
$ \Rightarrow n=10$
Thus there are $10$ terms in given series.
View full question & answer→Question 433 Marks
How many terms are there in the series:
0.5, 0.53, 0.56, ......, 1.1?
AnswerThe given series is 0.5, 0.53, 0.56, ...., 1.1
0.53 - 0.5 = 0.03, 0.56 - 0.53 = 0.03 etc.
Thus the given series is an A.P with the first term a = 0.5 and common difference d = 0.03.
Last term = l= 1.1
0.5 + (n - 1)(0.03) = 1.1
$\Rightarrow(n-1) \times 0.03=0.6$
$\Rightarrow$ n - 1 = 20
$\Rightarrow$ n = 21
Thus there are 21 terms in the given series.
View full question & answer→Question 443 Marks
How many terms are there in the series:
4, 7, 10, 13, ........,148?
AnswerThe given series is 4, 7, 10, 13, ........,148
7 - 4 = 3, 10 - 7 = 3, 13 - 10 = 3 etc
Thus the given series is an A.P with first term a =4 and common difference d= 3.
Last term = l= 148
4 + (n - 1)(3) = 148
$\Rightarrow(n-1) \times 3=144$
$\Rightarrow$ n - 1 = 48
$\Rightarrow$ n = 49
Thus there are 49 terms in the given series
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