Question 12 Marks
Show thet the followiong sequences is an A.P. Also, find the common difference and write 3 more terms in each case.
$9,\ 7,\ 5,\ 3,\ ...$
Answer$9,7,5,3,...$
$\text{a}_4-\text{a}_3=\text{a}_3-\text{a}_2-\text{a}_1=-2$
$\therefore$ The comman difference is -2
and the given sequence is A.P
$\text{a}_5=9+(-2)(5-1)=1$
$\text{a}_6=9+(-2)(6-1)=-1$
$\text{a}_7=9+(-2)(7-1)=-3$
View full question & answer→Question 22 Marks
Find the sum of the following series:
$101 + 99 + 97 + ... + 47$
Answer$101 + 99 + 97 + ... + 47$
$a_n$ term of A.P. of n terms is $47$
$\therefore47=\text{a}+(\text{n}-1)\text{d}$
$47=101+(\text{n}-1)(-2)$
or $\text{n}=28$
Then,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$=\frac{28}{2}[101+47]$
$=14\times148$
$=2072$
View full question & answer→Question 32 Marks
Show thet the followiong sequences is an A.P. Also, find the common difference and write 3 more terms in each case.
$-1,\frac{1}{4},\frac{3}{2},\frac{11}{4},...$
Answer$-1,\frac{1}{4},\frac{3}{2},\frac{11}{4},...$
$\text{a}_1=-1,\text{a}_2=\frac{1}{4},\text{a}_3=\frac{3}{2},\text{a}_4=\frac{11}{4}$
$\text{a}_4-\text{a}_3=\text{a}_3-\text{a}_2=\text{a}_2-\text{a}_1=\frac{5}{4}$
$\therefore$ common differnce is $\text{d}=\frac{5}{4}$
$\therefore$ The given sequence is A.P
$\text{a}_5=-1+(5-1)\frac{5}{4}=4$
$\text{a}_6=-1+(6-1)\frac{5}{4}=\frac{21}{4}$
$\text{a}_5=-1+(5-1)\frac{5}{4}=\text{a}_{4}$
$=-1+(7-1)\frac{5}{4}=\frac{26}{4}=\frac{13}{2}$
View full question & answer→Question 42 Marks
Find the second term and nth term of an A.P. whose 6th term is 12 and the 8th term is 22.
Answer$\text{a}_6=\text{a}+5\text{d}=12\ .....(1)$
$\text{a}_8=\text{a}+7\text{d}=22\ .....{(2)}$
Solving (1) and (2)
$\text{a}=-13$ and $\text{d}=5$
then,
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$=-13+(\text{n}-1)5$
$=5\text{n}-18$
and
$\text{a}_2=\text{a}+(2-1)\text{d}$
$=-13+5$
$=-8$
View full question & answer→Question 52 Marks
If the nth term of the A.P. $9, 7, 5, ...$ is same as the $n^{th}$ term of the A.P. $15, 12, 9, ..$ find $n.$
AnswerThe given A.P. is $9, 7, 5, ...$ and $15, 12, 9$
Here,
$\text{a}=9,\ \text{A}=15$
$\text{d}=-2, \ \text{D}=3$
Let $\text{a}_\text{n}=\text{A}_\text{n}$ for same n.
$\Rightarrow+\text{a}(\text{n}-1)\text{d}=\text{A}+(\text{n}-1)\text{d}$
$\Rightarrow9+(\text{n}-1)(-2)=15+(\text{n}-1)3$
$\Rightarrow\text{n}=7$
$\therefore$ 7th term both the A.P. is same.
View full question & answer→Question 62 Marks
Find the sum of the following arithmetic progression:
41, 36, 31, ... to 12 terms.
Answer41, 36, 31, ... to 12 terms
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{s}_{25}=\frac{25}{2}[2\times41+(11)(-5)]$
$=162$
View full question & answer→Question 72 Marks
In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.
AnswerGiven:
$\text{a}_{24}=2\text{a}_{10}$
$\Rightarrow\text{a}+23\text{d}=\text{a2}(\text{a}+9\text{d})$
$\Rightarrow\text{a}=5\text{d}\ .....(1)$
$\text{a}_{72}=\text{a}+(72-1)\text{d}$
$=\text{a}+71\text{d}\ [\because\text{a}=5\text{d}\ \text{from}(1)]$
$\Rightarrow76\text{d}\ .....{(2)}$
$\text{a}_{34}=\text{a}+(34-1)\text{d}$
$=5\text{d}+33\text{d}\ [\because\text{a}=5\text{d}\ \text{from}(1)]$
$=38\text{ad}\ .....{(3)}$
From (2) and (3)
$\text{a}_{72}=2\text{a}_{34}$
Hence proved.
View full question & answer→Question 82 Marks
The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference.
AnswerGiven,
$\text{a}=3\text{a}_1\ .....(1)$
$\text{a}_7=2\text{a}_3+1\ .....(2)$
Expandind (1) and (2)
$\text{a}+3\text{d}=2\text{a}$
$\therefore2\text{a}=3\text{d}$ or $\text{a}=\frac{3\text{d}}{2}\ .....(3)$
$\text{a}+6\text{d}=2\text{a}+4\text{d}+1$
$\text{a}+1=2\text{d}\ .....{(4)}$
From (3) and (4)
$\text{a}=3$ and $\text{d}=2$
View full question & answer→Question 92 Marks
If the $n^{th}$ term $a_n$ of sequence is given by $\text{a}_\text{n}=\text{n}^2-\text{n}+1,$ write down its first five terms.
Answer$\text{a}_\text{n}=\text{n}^2-\text{n}+1$ is the given sequence
then, first 5 tems are $\text{a}_1,\ \text{a}_2,\ \text{a}_3,\ \text{a}_4$ and $\text{a}_5$
$\text{a}_1=(1)^2-1+1=1$
$\text{a}_2=(2)^2-2+1=3$
$\text{a}_2=(3)^2-3+1=7$
$\text{a}^4=(4)^2-4+1=13$
$\text{a}_5=(5)^2-5+1=21$
First 5 terms $1, 3, 7, 13$ and $21.$
View full question & answer→Question 102 Marks
Insert A.M.s between $7$ and $71$ in such a way that the $5^{th}$ A.M. is $27$. Find the number of A.M.s.
AnswerLet there be n A.M between $7$ and $71$ let the A.M's be $A_1, A_2, A_3, ...., A_n.$
So,
$7, A_1, A_2, A_3, ...., A_n, 71$ are in A.P of $(n + 2)$ terms
$A_5 = A_6 = a +5d = 27 [$Given$]$
$\Rightarrow a + 5d = 27$
$\Rightarrow d = 15 \big[\therefore a = 7 \big]$
Theb $(n + 2)^{th} $ term of A.P is $71$
$\therefore a_{n+2} = 7 a + (n + - 1)d$
or $n = 15$
There are $15$ AM's between $7$ and $71.$
View full question & answer→Question 112 Marks
Find the sum of first n natural numbers.
AnswerA.P. formed is 1, 2, 3, 4, ..., n.
Here,
$\text{a}=1$
$\text{d}=1$
$\text{l}=\text{n}$
So sum of n terms $=\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2+(\text{n}-1)1]$
$=\frac{\text{n}(\text{n}+1)}{2}$ is the sum of first n natural numbers.
View full question & answer→Question 122 Marks
Find the sum of all integers between $84$ and $719,$ which are multiples of $5.$
AnswerThe required series is $85, 90, 95, ..., 715$
let there be n terms in the A.P
Then,
$n^{th} $ term $= 715$
$715 = 85 (n - 1) 5$
$n = 127$
Then,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\text{s}_{127}=\frac{127}{2}[85+715]$
$=50800$
View full question & answer→Question 132 Marks
Find the 12th term from the end of the following arithmetic progressions,
1, 4, 7, 10, ..., 88
AnswerA.P. is 1, 4, 7, 10, ..., 88
Then, 12th term from end is $\text{l}-(\text{n}-1)\text{d}$
$=88-(12-1)3$
$=88-33$
$=55$
View full question & answer→Question 142 Marks
Which term of the sequence $24,\ 23\frac{1}{4},\ 22\frac{1}{2},\ 21\frac{3}{4}...$ is the first negative term?
AnswerThe given sequence is $24,\ 23\frac{1}{4},\ 22\frac{1}{2},\ 21\frac{3}{4}...$
Here, $\text{a}=24$
$\text{d}=23\frac{1}{4}-24=\frac{93-96}{4}=\frac{-3}{4}$
$\text{a}_\text{n}<0$
$\text{a}+(\text{n}-1)\text{d}<0$
$24-\frac{3}{4}(\text{n}-1)<0$
$96-3\text{n}+3<0$
$99<3\text{n}$
$33<\text{n}$ or $\text{n}>33$
$\therefore$ 34th term is 1st negetive term.
View full question & answer→Question 152 Marks
A sequence is defined by $\text{a}_\text{n}=\text{n}^3-6\text{n}^2-11\text{n}-6,\text{n}\in\text{N.}$ show that the first three terms of the sequence are zero and all other terms are positive.
Answer$\text{a}_\text{n}=\text{n}^3-6\text{n}^2-11\text{n}-6,\text{n}\in\text{N.}$The first three terms are $\text{a}_1,\text{ a}_2$ and $\text{a}_3$
$\text{a}_1=(1)^3-(1)^2+11(1)-6=0$
$\text{a}_2=(2)^3=(2)^2+11(2)-6=0$
$\text{a}_3(3)^2-(3)^2+11(3)-6=0$
$\therefore$ the $1^\text{st}\ 3$ terms are zero.
and
$\text{a}_\text{n}=\text{n}^3-6\text{n}^2-11\text{n}-6$
$=(\text{n}-2)^3-(\text{n}-2)$ is positive.
View full question & answer→Question 162 Marks
Find the sum of the following series:
$2 + 5 + 8 + ... + 182$
Answer$2 + 5 + 8 + ... + 182$
$a_n $ term of given A.P. is $182$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}=182$
$\Rightarrow182=2+(\text{n}-1)3$
Then,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$=\frac{61}{2}[2+182]$
$=61\times92$
$=5612$
View full question & answer→Question 172 Marks
Show thet the followiong sequences is an A.P. Also, find the common difference and write 3 more terms in each case.
$3,-1,-5,-9...$
Answer$3,-1,-5,-9...$
$\text{a}_1=3,\text{a}_2=-1,\text{a}_3=-5,\text{a}_4=-4$
$\text{a}_2-\text{a}_1=-1-3=-4$
$\text{a}_3-\text{a}_2=-5-(-1)=-4$
$\text{a}_4-\text{a}_3-=9(-5)=-4$
$\therefore$ common difirence is $=-4$
$\text{a}_4-\text{a}_3=\text{a}_3-\text{a}_2=\text{a}$
$\therefore$ The given sequerence is a A.P
$\therefore\text{a}_5=3+(5-1)(-4)=-13$
$\text{a}_6=3+(6-1)(-4)=-17$
$\text{a}_7=3+(7-1)(-4)=-21$
View full question & answer→Question 182 Marks
Find the sum of all natural numbers between q and 100, which are divisible by 2 or 5.
AnswerThe natural numbers which are divisible by 2 or 5 are:
2 + 4 + 5 + 6 + 8 + 10 + ... + 100 = (2 + 4 + 6 + ...+ 100) + (5 + 15 + 25 + ...+ 95) Now (2 + 4 + 6 + ... +100) and (5 + 15 + 25 + ... + 95) are AP with common difference 2 and 10 respopectively.
Therefore
$2+4+6+\ ...\ +100=2\frac{50}{2}(1+50)$
$=2550$
Again,
$5+15+25+\ ...\ +95=5(1+3+5+\ ...\ +19)$
$=5\Big(\frac{10}{2}\Big)(1+19)$
$=500$
Therefore the sum of the number divisible by 2 or 5 is:
2 + 4 + 5 + 6 + 8 + 10 + + ... + 100 = 2550 + 500
= 3050
View full question & answer→Question 192 Marks
The $n^{th}$ team of a sequence is given by $\text{a}_\text{n}=2\text{n}+7.$ show that it an A.P. also, find its $7^{th}$ term.
Answer$\text{a}_{\text{n}}=2\text{n}+7$
$\text{a}_1=2(1)+7=9$
$\text{a}_2=2(2)+7=11$
$\text{a}_3=2(3)+7=13$
Here, $\text{a}_3-\text{a}_2=\text{a}_2-\text{a}_1=2$
$\therefore$ The given sequence is A.P
$\text{a}_7=2(7)+7=21$
$7^{th}$ term is $21.$
View full question & answer→Question 202 Marks
Find the sum of the following arithmetic progression:
$(x - y)^2, (x^2 + y^2), (x + y)^2 ...$ to $n$ terms.
Answer$(x - y)^2, (x^2 + y^2), (x + y)^2 ...$ to $n$ terms.
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}\big[2(\text{x}^2+\text{y}^2-2\text{xy})+(\text{x}-1)(-2\text{xy})\big]$
$=\text{n}[(\text{x}-\text{y})^2+(\text{n}-1)\text{xy}]$
View full question & answer→Question 212 Marks
Find: $n^{th}$ term of the A.P. $13, 8, 3, -2, ...$
AnswerFind $n^{th}$ A.P. $13, 8, 3, -2, ...$
Here, $\text{a}_1=13$
$\text{d}=-5$
$\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$=13+(\text{n}-1)(-5)$
$=-5\text{n}+18$
View full question & answer→Question 222 Marks
Find: 10th term of the A.P. 1, 4, 7, 10, ...
Answer10th term of the A.P.1, 4, 7, 10, ...
Here, 1st term $=\text{a}_1=1$
and common difference d $=4-1=3$
we know $\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$
$\therefore\text{a}_{10}=\text{a}_1+(10-1)\text{d}$
$=1+(10-1)3\Rightarrow28$
View full question & answer→Question 232 Marks
Find the 12th term from the end of the following arithmetic progressions,
3, 8, 13, ..., 253
AnswerA.P. is 3, 8, 13, ..., 253.
then, 12th term from end is $\text{l}-(\text{n}-1)\text{d}$ i.e.,
$=253-(12-1)5$
$=253-55$
$=198$
View full question & answer→Question 242 Marks
Write the first five terms in the following sequences:$\text{a}_1=1,\text{a}_\text{n}=\text{a}_{\text{n}-1}+2,\text{n}>1$
Answer$\text{a}_1=1,\text{a}_\text{n}=\text{a}_{\text{n}-1}+2,\text{n}\geq2$
$\text{a}_2=\text{a}_{\text{2}-1}+2=\text{a}_{1+2}=3$ $[\because\text{a}_1=1]$
$\text{a}_3=\text{a}_{\text{3}-1}+2=\text{a}_{2}+2=5$ $[\because\text{a}_2=3]$
$\text{a}_4=\text{a}_{\text{4}-1}+2=\text{a}_{3}+2=7$ $[\because\text{a}_3=5]$
$\text{a}_5=\text{a}_{\text{5}-1}+2=\text{a}_{4}+2=9$ $[\because\text{a}_4=7]$
$\therefore$ The first 5 terms os series are 1, 3, 5, 7, 11.
View full question & answer→Question 252 Marks
Write the first five terms in the following sequences:$\text{a}_1=1=\text{a}_2,\text{a}_\text{n}=\ \text{a}_{\text{n}-1}+\text{a}_{\text{n}-2},\text{n}>2$
Answer$\text{a}_1=\text{a}_2=1$ $\text{a}_\text{n}=\text{a}_{\text{n}-1}+\text{a}_{\text{n}-2}\ \text{n}>2$ $\Rightarrow\text{a}_3=\text{a}_{3-1}+\text{a}_{3-2}$ $=\text{a}_2+\text{a}_1=1+1=2$ $\Rightarrow\text{a}_4=\text{a}_{4-1}+\text{a}_{4-2}$ $=\text{a}_3+\text{a}_2=2+1=3$ $\Rightarrow\text{a}_5=\text{a}_{5-1}+\text{a}_{5-2}$ $=\text{a}_4+\text{a}_3=5$$\therefore$ The given sequence is 1, 1, 3, 5.
View full question & answer→Question 262 Marks
The $n^{th}$ them of a sequence is given by $\text{a}_\text{n}=2\text{n}^2+\text{n}+1.$ Show that it is not an A.P.
Answer$\text{a}_\text{n}=2\text{n}^2+\text{n}+1.$
$\text{a}_1=2(1)^2+(1)+1=4$
$\text{a}_2=2(2)^2+(2)+1=11$
$\text{a}_3=2(3)^2+(3)+1=21$
$\text{a}_3-\text{a}_2\not=\text{a}_2-\text{a}_1$
$\therefore$ The given sequence is not as A.P. as consequtive term do not have common diffrence.
View full question & answer→Question 272 Marks
How many terms are there in the A.P. 7, 10, 13, ...43?
AnswerThe given A.P is 7, 10, 13, ... 43.
Let there be n terms,
then, n term = 43
or $43=\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow43=7+(\text{n}-1)3$
$\Rightarrow\text{n}=13$
Thus, there are 13 terms in the given sequence
View full question & answer→Question 282 Marks
There are n A.M.s between $3$ and $17$. The ratio of the last mean to the first mean is $3 : 1$. Find the value of $n.$
AnswerLet $A_1, A_2, A_3, A_4 ....A_n$ An be the n arithmetic means between $3$ and $17$.Let d be the common difference of the A.P. $3, A_1, A_2, A_3, A_4 ....A_n $ and $17.$ Then, we have:
$\text{d}=\frac{17-3}{\text{n}-1}=\frac{14}{\text{n}+1}$
Now, $\text{A}_\text{n}=3+\text{d}=3+\frac{14}{\text{n}+1}=\frac{3\text{n}+17}{\text{n}+1}$
and $\text{A}_\text{n}=3+\text{nd}=3+\text{n}\Big(\frac{14}{\text{n}+1}\Big)=\frac{17\text{n}+3}{\text{n}+1}$
$\therefore\frac{\text{A}_\text{n}}{\text{A}_1}=\frac{3}{1}$
$\Rightarrow\frac{\Big(\frac{17\text{n}+3}{\text{n}+1}\Big)}{\Big(\frac{3\text{n}+3}{\text{n}+3}\Big)}=\frac{3}{1}$
$\Rightarrow\frac{17\text{n}+3}{3\text{n}+17}=\frac{3}{1}$
$\Rightarrow17\text{n}+3=9\text{n}+51$
$\Rightarrow8\text{n}=48$
$\Rightarrow\text{n}=6$
View full question & answer→Question 292 Marks
The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.
AnswerGiven thet:
$\text{a}_6=19=\text{a}+(6-1)\text{d}\ .....(1)$
$\text{a}_{17}=41=\text{a}+(17-1)\text{d}\ .....(2)$
Solving (1) and (2), we get
$\text{a}=9$ and $\text{d}=2$
$\therefore\text{a}_{40}=\text{a}+(40-1)\text{d}$
$=9+(40-1)\text{d}$
$=9+39(2)$
$=87$
40th term of the given sequence is 87.
View full question & answer→Question 302 Marks
Is $302$ a them of the A.P. $3, 8, , 13, ...?$
AnswerIs $302$ a term of A.P $3, 8, 13$
Let $302$ be $n^{th}$ term of the given A.P.
Here, $302=3+(\text{n}-1)5$
$\frac{299}{5}=(\text{n}-1)$
$\text{n}=\frac{304}{5}$
Whivh is not a natural nimber.
$\therefore 302$ is not a term of given A.P.
View full question & answer→Question 312 Marks
Insert $7$ A.M.s between $2$ and $17$.
AnswerLet $A_1, A_2, A_3, A_4, A_5, A_6, A_7$ be the seven A.M.s between $2$ and $17$.
Then, $2, A_1, A_2, A_3, A_4, A_5, A_6, A_7$_ and $17$ are in A.P. whose common difference is as follows:
$\text{d}=\frac{17-2}{7+1}$
$=\frac{15}{8}$
$\text{A}_1=2+\text{d}=2+\frac{15}{8}=\frac{31}{8}$
$\text{A}_2=2+2\text{d}=2+\frac{15}{4}=\frac{23}{4}$
$\text{A}_3=2+3\text{d}=2+\frac{45}{8}=\frac{61}{8}$
$\text{A}_4=2+4\text{d}=2+\frac{15}{2}=\frac{19}{2}$
$\text{A}_5=2+5\text{d}=2+\frac{75}{8}=\frac{91}{8}$
$\text{A}_6=2+6\text{d}=2+\frac{45}{4}=\frac{53}{4}$
$\text{A}_7=2+7\text{d}=2+\frac{105}{8}=\frac{121}{8}$
Hence, the required A.M.S are $\frac{31}{8},\ \frac{23}{4},\ \frac{61}{8},\ \frac{19}{4},\ \frac{91}{8},\ \frac{53}{4},\ \frac{121}{8}.$
View full question & answer→Question 322 Marks
Which term of the A.P. $4, 9, 14, ...$ is $254$?
AnswerWhich term of A.P. is $4, 9, 14, ...$ is $254$?
Let $n^{th}$ term of A.P. be $254$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$254=\text{n}-1)5$
$\therefore\text{n}=51$
$\therefore$ $51$st twrm of the given A. P. is $254$.
View full question & answer→Question 332 Marks
The fibonacci sequence is defined by $\text{a}_1=1=\text{a}_2,\ \text{a}_\text{n}=\text{a}_{\text{n}-1}+\text{a}_{\text{n}-2}$ from $\text{n}>2.$ Find $\frac{\text{a}_{\text{n}+1}}{\text{a}_\text{n}}$ for $\text{n}=1,\ 2,\ 3,\ 4,\ 5.$
Answer$\text{a}_\text{n}=\text{a}_{\text{n}-1}+\text{a}_{\text{n}-2}$ For $\text{n}>2$
$\Rightarrow\text{a}_3=\text{a}_{3-1}+\text{a}_{3-2}=\text{a}_2+\text{a}_1=1+1=2$
$\Rightarrow\text{a}_4=\text{a}_{4-1}+\text{a}_{4-2}=\text{a}_3+\text{a}_2=2+1=3$
$\Rightarrow\text{a}_5=\text{a}_{5-1}+\text{a}_{5-2}=\text{a}_4+\text{a}_3=3+2=3$
$\Rightarrow\text{a}_6=\text{a}_{6-1}+\text{a}_{6-2}=\text{a}_5+\text{a}_4=5+31=8$
$\therefore$ for $\text{n}=1$
$\frac{\text{a}_{\text{n}-1}}{\text{a}_\text{n}}=\frac{\text{a}_2}{\text{a}_1}=\frac{1}{1}=1$
For $\text{n}=2$
$\frac{\text{a}^3}{\text{a}^2}=\frac{2}{1}=2$
For $\text{n}=3$
$\frac{\text{a}^3}{\text{a}^2}=\frac{3}{2}=1.5$
For $\text{n}=4$ and $\text{n}=5$
$\frac{\text{a}_5}{\text{a}_4}=\frac{5}{3}$ and $\frac{\text{a}_6}{\text{5}_5}=\frac{8}{5}$
$\therefore$ The reqquired series is $1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5},...$
View full question & answer→Question 342 Marks
If $x, y, z$ are in A.P. and $A_1$ is the A.M. of $x$ and $y$ and $A_2$ is the A.M. of $y$ and $z$, then prove that the A.M. of $A_1$ and $A_2$ is $y$.
Answer$x, y, z$ are in A.P.
$\therefore\text{y}=\frac{\text{x}+\text{z}}{2}$
Now, $A_1$ is the arithmetic mean of $x$ and $y$.
$\text{A}_1=\frac{\text{x}+\text{y}}{2}=\frac{\text{x}+\frac{\text{x}+\text{z}}{2}}{2}=\frac{3\text{z}+\text{x}}{4}$
And, $A_2$ is the arithmetic mean of $y$ and $z$.
$\text{A}_2=\frac{\text{y}+\text{z}}{2}=\frac{\text{x}+\frac{\text{z}}{2}+\text{z}}{2}=\frac{3\text{z}+\text{x}}{4}$
Let $A_3$ be the arithmetic mean of $A_1$ and $A_2$.
$\text{A}_3=\frac{\text{A}_1+\text{A}_2}{2}$
$=\frac{\frac{3\text{x}+\text{z}}{4}+\frac{3\text{x}+\text{x}}{4}}{2}$
$=\frac{4\text{x}+4\text{z}}{2}$
$=\frac{\text{x}+\text{z}}{2}$
$=\text{y}$
Hence, proved.
View full question & answer→Question 352 Marks
If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.
AnswerGiven:
$\text{a}_9=0$
$\therefore\text{a}+8\text{d}=0$
$\text{a}=-8\text{d}\ .....(1)$
$\text{a}_{19}=\text{a}+(19-1)\text{d}$
$=\text{a}+18\text{d}$ $[\therefore\text{a}=-8\text{d}\ \text{from}(1)]$
$=10\text{d}\ .....(2)$
$\text{a}_{29}=\text{a}+(29-1)\text{d}$
$=-8\text{d}+28\text{d}$ $[\because\text{a}=-8\text{d}\ \text{from}(1)]$
$-20\text{d}\ .....(3)$
(2) and (3)
$\text{a}_{29}=2\text{a}_{19}$
Hence proved.
View full question & answer→Question 362 Marks
How many teterms are there in the A.P. $-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2},\ ...,\frac{10}{3}?$
AnswerThe given A.P. is $-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2},\ ...,\frac{10}{3}?$
Let there be n terms
then, $n^{th}$ term $=\frac{10}{3}$
or $\frac{10}{3}=\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow\frac{10}{3}=-1+(\text{n}-1)\Big(\frac{-5}{6}+1\Big)$
$\Rightarrow\text{n}=27$
Thus, there are $27$ terms in the given sequence.
View full question & answer→Question 372 Marks
Write the first five terms in the following sequences:$\text{a}_1=\text{a}_2=2,\text{a}_\text{n}=\text{a}_{\text{n}-1}-1,\text{n}>2$
Answer$\text{a}_1=\text{a}_2=2$
$\text{a}_\text{n}=\text{a}_{\text{n}-1}-1\ \text{n}>2$
$\Rightarrow\text{a}_3=\text{a}_{3-1}-1$
$=\text{a}_2-1$
$=2-1=1$
$\Rightarrow\text{a}_4=\text{a}_{4-1}-1$
$=\text{a}_3-1=1-1=0$
$\Rightarrow\text{a}_5=\text{a}_{5-1}-1$
$=0-1=-1$
$\therefore$ The first 5 terms of the sequence are 2, 2, 1, 0, -1
View full question & answer→Question 382 Marks
Which term of the A.P. $84, 80, 76, ...$ is $0?$
AnswerWhich term of A.P. $84, 80, 76, is 0?$
Let $n^{th}$ term of A.P.be $0$
Then, $\text{a}_\text{n}=0=\text{a}+(\text{n}-1)\text{d}$
$\text{a}_\text{n}=0=84+(\text{n}-1)(-4)$
$\therefore\text{n}=22$
$\therefore$ 22nd term of the given A.P. is $0.$
View full question & answer→Question 392 Marks
Find the A.M. between:
(x - y) and (x + y)
Answer(x - y) and (x + y)
Let A be the arithem atic mean of (x - y) and (x + y)
Then,
$(\text{x}-\text{y}),\ \text{A},\ (\text{x}+\text{y})$ are in A.P
$\Rightarrow\text{A}-(\text{x}-\text{y})=(\text{x}+\text{y})-\text{A}$
$\Rightarrow\text{A}=\frac{(\text{x}-\text{y})+(\text{x}+\text{y})}{2}=\frac{2\text{x}}{2}=\text{x}$
$\therefore$ A.M is x
View full question & answer→Question 402 Marks
Insert $4$ A.M.s between $4$ and $19.$
AnswerLet $A_1, A_2, A_3, A_4$ be the four A.M.s between $4$ and $19.$
Then,
$4, A_1, A_2, A_3, A_4, 19$ are A.P of $6$ terms
$A_n = a + (n - 1)d$
$a_6 = 19 = 4 + (6 - 1)d$
or d = 3 .....(1)
Now,
$A_1 = 4 + d = 4 + 3 = 7$
$A_2 = 4 + 2d = 4 + 6 = 10$
$A_3 = 4 + 3d = 4 + 9 = 13$
$A_4 = 4 + 4d = 4 + 12 = 16$
The $4$ A.M.s between $4$ and $19$ are $7, 10, 13, 16.$
View full question & answer→Question 412 Marks
Find the A.M. between:
12 and -8
Answer12 and -8
Let A be the arithem atic mean of 12 and -8
Then,
12, A, -8 are in A.P
$\Rightarrow\text{A}-12=-8-\text{A}$
$\Rightarrow\text{A}=\frac{12+(-8)}{2}=2$
$\therefore\text{A.M is 2}$
View full question & answer→Question 422 Marks
Find the sun of all odd numbers between 100 and 200.
AnswerThe series so formed is 101, 103, 105, ... , 199
Let number of terms be n
then,
$\text{a}_\text{n}=\text{a}+(\text{n}-1)2$
$\Rightarrow199=101+(\text{n}-1)2$
$\Rightarrow\text{n}=50$
The sum of n terms $=\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\text{s}_{50}=\frac{50}{2}[101+199]$
$=7500$
The sum of odd numbers between 100 and 200 is 7500.
View full question & answer→Question 432 Marks
Find the sum of the series: 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + ... to 3n terms.
AnswerIn the given series 3 + 5 + 7 + 9 + ... to 3n
Here,
a = 3
d = 2
Number of terms = 3n
The sum of n terms is
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{n}+(\text{n}-1)\text{d}]$
$\text{s}_{3\text{n}}=\frac{3\text{n}}{2}[6+(3\text{n}-1)2]$
$=3\text{n}(2\text{n}+3)$
View full question & answer→Question 442 Marks
Find the sun of first $n$ odd natural numbers.
AnswerThe series of $n$ odd natural number are $1, 3, 5, ..., n$
Where $n$ is odd natural number
then, sum of $n$ terms is
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2(1)+(\text{n}-1)(2)]$
$=\text{n}^2$
The sum of n odd natural numbers is $n^2.$
View full question & answer→Question 452 Marks
Show thet the followiong sequences is an A.P. Also, find the common difference and write 3 more terms in each case.
$\sqrt{2},\ 3\sqrt{2},\ 5\sqrt{2},\ 7\sqrt{2},...$
Answer$\sqrt{2},3\sqrt{2},5\sqrt{2},7\sqrt{2},...$
$\text{a}_1=\sqrt{2},\text{a}_2=3\sqrt{2},\text{a}_3=5\sqrt{2},\text{a}_4=7\sqrt{2}$
$\text{a}_4-\text{a}_3=\text{a}_3-\text{a}_2=\text{a}_2-\text{a}_1=2\sqrt{2}$
$\therefore$ The comman dufference is $2\sqrt{2}$
and the given sequence is A.P
$\text{a}_5-\sqrt{2}+2\sqrt{2}(5-1)=9\sqrt{2}$
$\text{a}_6-\sqrt{2}+2\sqrt{2}(6-1)=11\sqrt{2}$
$\text{a}_7-\sqrt{2}+2\sqrt{2}(7-1)=13\sqrt{2}$
View full question & answer→Question 462 Marks
Find the sum of the following arithmetic progression:
a + b, a - b, a - 3b, ... to 22 terms.
Answera + b, a - b, a - 3b, ... to 22 terms
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{s}_{22}=\frac{22}{2}[2\text{a}+2\text{b}+21(-2\text{b})]$
$=22\text{a}-440\text{b}$
View full question & answer→Question 472 Marks
Show thet the sum of all odd integers between $1$ and $1000$ wich are divisibleby $3$ is $83667$.
AnswerThe odd numbers between $1$ and $100$ divisible by $3$ are $3, 9, 15, ..., 999$
Let the number of terms be $n$ then, $n^{th}$ term is $999$.
$\text{a}_\text{n}=\text{a}(\text{n}-1)\text{d}$
$999=3+(\text{n}-1)6$
$\Rightarrow\text{n}-167$
The sum of n terms
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\Rightarrow\text{s}_{167}=\frac{167}{2}[3+999]$
$=83667$
Hence proved.
View full question & answer→Question 482 Marks
Find the sum of the following arithmetic progression:
$9,\ \frac{9}{2},\ \frac{15}{2},\ ...$ to 25 terms.
Answer$9,\ \frac{9}{2},\ \frac{15}{2},\ ...$ to 25 terms
$\text{s}_{25}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{s}_{25}=\frac{25}{2}\big(2\times3\times24\times\frac{3}{2}\big)$
$=525$
View full question & answer→Question 492 Marks
How many terms are there in the A.P. whose first and fifth terms are -14 and 2 repectively and the sum of the terms is 40?
AnswerGiven,
$\text{a}_1=-4=\text{a}+0\text{d}\ .....(1)$
$\text{a}=2=\text{a}+4\text{d}\ .....(2)$
Solving (1) and (2)
$\text{a}_1=\text{a}=-14$ and $\text{d}=4$
Let ther be n terms then sum of there n terms = 40
$\therefore\text{s}_{\text{n}}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\Rightarrow40=\frac{\text{n}}{2}[-28+(\text{n}-1)4]$
$\Rightarrow4\text{n}^2-32\text{n}-80=0$
or $\text{n}=10$ or $-2$
But n can't be negative
$\therefore\text{n}=10$
The given A.P. has 10 terms.
View full question & answer→Question 502 Marks
Which trem of the A.P. $3, 8, 13, ...$ os $248$?
AnswerLet $n^{th}$ term of A.P. $= 248$
$\therefore\text{a}_\text{n}=248=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow248=3+(\text{n}-1)5$
$\therefore\text{n}=50$
$\therefore$ 50th term of the given A.P. is 248
View full question & answer→Question 512 Marks
Find: 18th term of the A.P. $\sqrt{2},\ 3\sqrt{2},\ 5\sqrt{2},\ ...$
AnswerTo find 18th term of A.P. $\sqrt{2},3\sqrt{2},5\sqrt{2},\ ...$
Here, 1st term $\text{a}_1=\sqrt{2}$
and d = cpmmon difference $=2\sqrt{2}$
$\therefore\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$
$\text{a}_{18}=\sqrt{2}+2\sqrt{2}(17)=35\sqrt{2}$
View full question & answer→Question 522 Marks
Find the sum of the following series:
$(\text{a}-\text{b})^2+(\text{a}^2+\text{b}^2)+(\text{a}+\text{b})^2+\ ...\ +[(\text{a}+\text{b}^2)+6\text{ab}]$
Answer$(\text{a}-\text{b})^2+(\text{a}^2+\text{b}^2)+(\text{a}+\text{b})^2+\ ...\ +[(\text{a}+\text{b}^2)+6\text{ab}]$
Let number of terms be n
Then,
$\text{a}_\text{n}=(\text{a}+\text{b}^2)+6\text{ab}$
$\Rightarrow(\text{a}+\text{b})^2+(\text{n}-1)(2\text{nd})=(\text{a}+\text{b})^2+6\text{ab}$
$\Rightarrow\text{a}^2+\text{b}^2-2\text{ab}+2\text{abn}-2\text{ab}=\text{a}^2+\text{b}^2+2\text{ab}+6\text{ab}$
then,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\text{s}_6=\frac{6}{2}[\text{a}^2+\text{b}^2-2\text{ab}+\text{a}^2+\text{b}^2+\text{ab}+6\text{ab}]$
$=6[\text{a}^2+\text{b}^2+3\text{ab}]$
View full question & answer→Question 532 Marks
The first term of an A.P. is 5, the common difference is 3 and last term is 80; find the number of terms.
AnswerGiven: $\text{a}=5$
$\text{d}=3$
$\text{a}_\text{n}=$ last there be n terms
$\therefore\text{a}_\text{n}=80=\text{a}+(\text{n}-1)\text{d}$
$80=5+(\text{n}-1)3$
$\Rightarrow\text{n}=26$
$\therefore$ Thus, thre are 26 term in the given sequence.
View full question & answer→Question 542 Marks
Find the first four terms of the sequence defined by $\text{a}_1=3$ and $\text{a}_\text{n}=3\text{a}_{\text{n}-1}+2$ for all $\text{n}>1.$
Answer$\text{a}_\text{n}=3\text{a}_{\text{n}-1}+2$
$\text{a}_1=3$
$\text{a}_1=\text{3a}_{2-1}+2=\text{3a}_{2-1}+2=3(3)+2=11$
$\text{a}_3=\text{3a}_{3-1}+2=3\text{a}_2+2=3(11)+2=35$
$\text{a}_4=\text{3a}_{4-1}+2=\text{3a}_3+2=3(35)+2=107$
First four terms of the sequence are 3, 11, 35 and 107.
View full question & answer→Question 552 Marks
If $n$ A.M.$s$ are inserted between two numbers, prove that the sum of the means equidistant from the beginning and the end is constant.
AnswerLet $A_1, A_2......A_n $ be n A.M.s between two numbers a and b.
Then, $a, A_1, A_2......A_n,$ b are in A.P. with common difference, $\text{d}=\frac{\text{b}-\text{a}}{\text{n}+1}.$
$\therefore\text{A}_1+\text{A}^2+\ ......+\text{A}_\text{n}=\frac{\text{n}}{2}[\text{A}_1+\text{A}_\text{n}]$
$=\frac{\text{n}}{2}[\text{A}_1-\text{d}+\text{A}_\text{n}+\text{d}]$
$=\frac{\text{n}}{2}[\text{a}+\text{b}]$
$=\text{n}\times\big[\frac{\text{a}+\text{b}}{2}\big]$
$=$ A.M. between $a$ and $b,$ which is constant.
View full question & answer→Question 562 Marks
Find the $12^{th}$ term from the end of the following arithmetic progressions,
$3, 5, 7, 9, ... 201$
AnswerA.p. is $3, 5, 7, 9, ..., 201.$
Here,
$\text{a}=3$
$\text{d}=2$
$n^{th}$ term from the end is $\text{l}-(\text{n}-1)\text{d}$
i.e. $201-(\text{n}-1)2$ or $203-2\text{n}\ .....{(1)}$
12th term from end is
$203-2(12)=179$
View full question & answer→Question 572 Marks
Find the sum of all integers between $50$ and $500$, which are divisible by $7.$
AnswerThe series of integers by $7$ between $50$ and $500$ are
$56, 63, 70, ..., 497$
Let the number of terms be n then, $n^{th}$ term $= 497$
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow497=56+(\text{n}-1)7$
$\Rightarrow\text{n}=64$
The sum $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\Rightarrow\text{s}_{64}=\frac{64}{2}[56+497]$
$=32\times553$
$=17696$
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