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 $\text{a}_5=9+(-2)(5-1)=1$ $\text{a}_6=9+(-2)(6-1)=-1$ $\text{a}_7=9+(-2)(7-1)=-3$
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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, $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$ $=\frac{28}{2}[101+47]$ $=14\times148$ $=2072$
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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 $\text{a}_5=-1+(5-1)\frac{5}{4}=4$ $\text{a}_6=-1+(6-1)\frac{5}{4}=\frac{21}{4}$ $\text{a}_5=-1+(5-1)\frac{5}{4}=\text{a}_{4}$ $=-1+(7-1)\frac{5}{4}=\frac{26}{4}=\frac{13}{2}$
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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 $\text{a}_2=\text{a}+(2-1)\text{d}$ $=-13+5$ $=-8$
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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.
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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.
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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 $\text{d}=2$
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If the nth term an 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.
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Insert A.M.s between 7 and 71 in such a way that the $5^{\text {th }}$ 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 $A_1, A_2, A_3, \ldots ., A_n$. So, $7, A_1, A_2, A_3, \ldots, A_n, 71$ are in $A . P$ of $(n+2)$ terms $A_5=A_6=a+5 d=27\left[\right.$ Given] $\Rightarrow a+5 d=27 \Rightarrow d=15[\therefore a=7]$ Theb $(n+2)^{\text {th }}$ 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.
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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.
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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, $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$ $\text{s}_{127}=\frac{127}{2}[85+715]$ $=50800$
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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 $\text{l}-(\text{n}-1)\text{d}$ $=88-(12-1)3$ $=88-33$ $=55$
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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.
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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.
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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, $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$ $=\frac{61}{2}[2+182]$ $=61\times92$ $=5612$
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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 $\therefore\text{a}_5=3+(5-1)(-4)=-13$ $\text{a}_6=3+(6-1)(-4)=-17$ $\text{a}_7=3+(7-1)(-4)=-21$
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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
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The nth team of a sequence is given by $\text{a}_\text{n}=2\text{n}+7.$ show that it an A.P. also, find its 7th 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.
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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. $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $=\frac{\text{n}}{2}\big[2(\text{x}^2+\text{y}^2-2\text{xy})+(\text{x}-1)(-2\text{xy})\big]$ $=\text{n}[(\text{x}-\text{y})^2+(\text{n}-1)\text{xy}]$
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Find: $n^{th}$ term of the A.P. 13, 8, 3, -2, ...
AnswerFind $n^{th}$ A.P. 13, 8, 3, -2, ... Here, $\text{a}_1=13$ $\text{d}=-5$ $\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$ $=13+(\text{n}-1)(-5)$ $=-5\text{n}+18$
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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 $\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$ $\therefore\text{a}_{10}=\text{a}_1+(10-1)\text{d}$ $=1+(10-1)3\Rightarrow28$
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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., $=253-(12-1)5$ $=253-55$ $=198$
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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.
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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.
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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.
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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
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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 $A_1, A_2, A_3, A_4 \ldots . . . A_n$ An be the $n$ arithmetic means between 3 and 17. Let $d$ be the common difference of the A.P. 3 , $A_1, A_2, A_3, A_4 \ldots . . A_n$ 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 $\text{A}_\text{n}=3+\text{nd}=3+\text{n}\Big(\frac{14}{\text{n}+1}\Big)=\frac{17\text{n}+3}{\text{n}+1}$
$\therefore\frac{\text{A}_\text{n}}{\text{A}_1}=\frac{3}{1}$
$\Rightarrow\frac{\Big(\frac{17\text{n}+3}{\text{n}+1}\Big)}{\Big(\frac{3\text{n}+3}{\text{n}+3}\Big)}=\frac{3}{1}$
$\Rightarrow\frac{17\text{n}+3}{3\text{n}+17}=\frac{3}{1}$
$\Rightarrow17\text{n}+3=9\text{n}+51$
$\Rightarrow8\text{n}=48$
$\Rightarrow\text{n}=6$
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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.
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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.
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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 $\frac{31}{8},\ \frac{23}{4},\ \frac{61}{8},\ \frac{19}{4},\ \frac{91}{8},\ \frac{53}{4},\ \frac{121}{8}.$
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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 $\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$ $254=\text{n}-1)5$
$\therefore\text{n}=51$
$\therefore$ 51st twrm of the given A. P. is 254.
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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 $1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5},...$
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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.
Answerx, 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.
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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.
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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.
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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
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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.
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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
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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, $A_1=4+d=4+3=7 A_2=4+2 d=4+6=10 A_3=4+3 d=4+9=13$ $A_4=4+4 d=4+12=16$ The 4 A.M.s between 4 and 19 are $7,10,13,16$.
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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 $\Rightarrow\text{A}-12=-8-\text{A}$ $\Rightarrow\text{A}=\frac{12+(-8)}{2}=2$ $\therefore\text{A.M is 2}$
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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.
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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 $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{n}+(\text{n}-1)\text{d}]$ $\text{s}_{3\text{n}}=\frac{3\text{n}}{2}[6+(3\text{n}-1)2]$ $=3\text{n}(2\text{n}+3)$
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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 $n^2$.
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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 $\text{a}_5-\sqrt{2}+2\sqrt{2}(5-1)=9\sqrt{2}$ $\text{a}_6-\sqrt{2}+2\sqrt{2}(6-1)=11\sqrt{2}$ $\text{a}_7-\sqrt{2}+2\sqrt{2}(7-1)=13\sqrt{2}$
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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 $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $\text{s}_{22}=\frac{22}{2}[2\text{a}+2\text{b}+21(-2\text{b})]$ $=22\text{a}-440\text{b}$
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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.
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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 $\text{s}_{25}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $\text{s}_{25}=\frac{25}{2}\big(2\times3\times24\times\frac{3}{2}\big)$ $=525$
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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.
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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
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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 $=2\sqrt{2}$ $\therefore\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$ $\text{a}_{18}=\sqrt{2}+2\sqrt{2}(17)=35\sqrt{2}$
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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, $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$ $\text{s}_6=\frac{6}{2}[\text{a}^2+\text{b}^2-2\text{ab}+\text{a}^2+\text{b}^2+\text{ab}+6\text{ab}]$ $=6[\text{a}^2+\text{b}^2+3\text{ab}]$
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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.
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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.
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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.
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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 $203-2(12)=179$
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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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