2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip

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57 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

Which term of the sequence $24,\ 23\frac{1}{4},\ 22\frac{1}{2},\ 21\frac{3}{4}...$ is the first negative term?

Answer

The 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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Question 22 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.

Answer

Given,
$\text{a}=3\text{a}_1\ .....(1)$
$\text{a}_7=2\text{a}_3+1\ .....(2)$
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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Question 32 Marks

In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term.

Answer

Given:
$\text{a}_{24}=2\text{a}_{10}$
$\Rightarrow\text{a}+23\text{d}=\text{a2}(\text{a}+9\text{d})$
$\Rightarrow\text{a}=5\text{d}\ .....(1)$
$\text{a}_{72}=\text{a}+(72-1)\text{d}$
$=\text{a}+71\text{d}\ [\because\text{a}=5\text{d}\ \text{from}(1)]$
$\Rightarrow76\text{d}\ .....{(2)}$
$\text{a}_{34}=\text{a}+(34-1)\text{d}$
$=5\text{d}+33\text{d}\ [\because\text{a}=5\text{d}\ \text{from}(1)]$
$=38\text{ad}\ .....{(3)}$
From (2) and (3)
$\text{a}_{72}=2\text{a}_{34}$
Hence proved.

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Question 42 Marks

Find the 12th term from the end of the following arithmetic progressions,

1, 4, 7, 10, ..., 88

Answer

A.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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Question 52 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.

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Question 62 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.

Answer

Let A₁, A₂, A₃, A₄ ....A_n An be the n arithmetic means between 3 and 17.

Let d be the common difference of the A.P. 3, A₁, A₂, A₃, A₄ ....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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Question 72 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.

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Question 82 Marks

Insert 7 A.M.s between 2 and 17.

Answer

Let A₁, A₂, A₃, A₄, A₅, A₆, A₇ be the seven A.M.s between 2 and 17.
Then, 2, A₁, A₂, A₃, A₄, A₅, A₆, A₇ 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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Question 92 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

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Question 102 Marks

Find the sum of the series: 3 + 5 + 7 + 6 + 9 + 12 + 9 + 13 + 17 + ... to 3n terms.

Answer

In 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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Question 112 Marks

Find the sun of first n odd natural numbers.

Answer

The 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 n².

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Question 122 Marks

Show thet the sum of all odd integers between 1 and 1000 wich are divisibleby 3 is 83667.

Answer

The 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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Question 132 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.

Answer

Given: $\text{a}=5$
$\text{d}=3$
$\text{a}_\text{n}=$ 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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Question 142 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.

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Question 152 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.

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Question 162 Marks

Find the sun of all odd numbers between 100 and 200.

Answer

The 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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Question 172 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.

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Question 182 Marks

Find the 12th term from the end of the following arithmetic progressions,

3, 8, 13, ..., 253

Answer

A.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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Question 192 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
$\text{s}_{25}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{s}_{25}=\frac{25}{2}\big(2\times3\times24\times\frac{3}{2}\big)$
$=525$

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Question 202 Marks

Find the 12^th term from the end of the following arithmetic progressions,

3, 5, 7, 9, ... 201

Answer

A.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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Question 212 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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Question 222 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,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$=\frac{28}{2}[101+47]$
$=14\times148$
$=2072$

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Question 232 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
$\text{a}_5=-1+(5-1)\frac{5}{4}=4$
$\text{a}_6=-1+(6-1)\frac{5}{4}=\frac{21}{4}$
$\text{a}_5=-1+(5-1)\frac{5}{4}=\text{a}_{4}$
$=-1+(7-1)\frac{5}{4}=\frac{26}{4}=\frac{13}{2}$

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Question 242 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
$\text{a}_2=\text{a}+(2-1)\text{d}$
$=-13+5$
$=-8$

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Question 252 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.

Answer

The 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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Question 262 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.

Answer

Let there be n A.M between 7 and 71 let the A.M's be A₁, A₂, A₃, ...., A_n.
So,
7, A₁, A₂, A₃, ...., A_n, 71 are in A.P of (n + 2) terms
A₅ = A₆ = a +5d = 27 [Given]
⇒ a + 5d = 27
⇒ 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.

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Question 272 Marks

Find the sum of first n natural numbers.

Answer

A.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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Question 282 Marks

Find the sum of all integers between 84 and 719, which are multiples of 5.

Answer

The 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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Question 292 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,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$=\frac{61}{2}[2+182]$
$=61\times92$
$=5612$

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Question 302 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
$\therefore\text{a}_5=3+(5-1)(-4)=-13$
$\text{a}_6=3+(6-1)(-4)=-17$
$\text{a}_7=3+(7-1)(-4)=-21$

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Question 312 Marks

Find the sum of all natural numbers between q and 100, which are divisible by 2 or 5.

Answer

The 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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Question 322 Marks

Find the sum of the following arithmetic progression:
(x - y)², (x² + y²), (x + y)² ... to n terms.

Answer

(x - y)², (x² + y²), (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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Question 332 Marks

Find: n^th term of the A.P. 13, 8, 3, -2, ...

Answer

Find 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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Question 342 Marks

Find the sum of the following arithmetic progression:
41, 36, 31, ... to 12 terms.

Answer

41, 36, 31, ... to 12 terms
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\text{s}_{25}=\frac{25}{2}[2\times41+(11)(-5)]$
$=162$

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Question 352 Marks

How many terms are there in the A.P. 7, 10, 13, ...43?

Answer

The 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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Question 362 Marks

The 6th and 17th terms of an A.P. are 19 and 41 respectively, find the 40th term.

Answer

Given 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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Question 372 Marks

Find: 10th term of the A.P. 1, 4, 7, 10, ...

Answer

10th 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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Question 382 Marks

Is 302 a them of the A.P. 3, 8, , 13, ...?

Answer

Is 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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Question 392 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.

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Question 402 Marks

Which term of the A.P. 4, 9, 14, ... is 254?

Answer

Which 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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Question 412 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 $1,2,\frac{3}{2},\frac{5}{3},\frac{8}{5},...$

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Question 422 Marks

If x, y, z are in A.P. and A₁ is the A.M. of x and y and A₂ is the A.M. of y and z, then prove that the A.M. of A₁ and A₂ is y.

Answer

x, y, z are in A.P.
$\therefore\text{y}=\frac{\text{x}+\text{z}}{2}$
Now, A₁ 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, A₂is 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 A₃ be the arithmetic mean of A₁ and A₂.
$\text{A}_3=\frac{\text{A}_1+\text{A}_2}{2}$
$=\frac{\frac{3\text{x}+\text{z}}{4}+\frac{3\text{x}+\text{x}}{4}}{2}$
$=\frac{4\text{x}+4\text{z}}{2}$
$=\frac{\text{x}+\text{z}}{2}$
$=\text{y}$
Hence, proved.

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Question 432 Marks

How many teterms are there in the A.P. $-1,-\frac{5}{6},-\frac{2}{3},-\frac{1}{2},\ ...,\frac{10}{3}?$

Answer

The 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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Question 442 Marks

Which term of the A.P. 84, 80, 76, ... is 0?

Answer

Which 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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Question 452 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

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Question 462 Marks

Insert 4 A.M.s between 4 and 19.

Answer

Let A₁, A₂, A₃, A₄be the four A.M.s between 4 and 19.
Then,
4, A₁, A₂, A₃, A₄, 19 are A.P of 6 terms
A_n = a + (n - 1)d
a₆ = 19 = 4 + (6 - 1)d
or d = 3 .....(1)
Now,
A₁ = 4 + d = 4 + 3 = 7
A₂ = 4 + 2d = 4 + 6 = 10
A₃ = 4 + 3d = 4 + 9 = 13
A₄ = 4 + 4d = 4 + 12 = 16
The 4 A.M.s between 4 and 19 are 7, 10, 13, 16.

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Question 472 Marks

Find: 18th term of the A.P. $\sqrt{2},\ 3\sqrt{2},\ 5\sqrt{2},\ ...$

Answer

To 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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Question 482 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,
$\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\text{s}_6=\frac{6}{2}[\text{a}^2+\text{b}^2-2\text{ab}+\text{a}^2+\text{b}^2+\text{ab}+6\text{ab}]$
$=6[\text{a}^2+\text{b}^2+3\text{ab}]$

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Question 492 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.

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Question 502 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.

Answer

Let A₁, A₂......A_n be n A.M.s between two numbers a and b.
Then, a, A₁, A₂......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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Question 512 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
$\text{a}_5-\sqrt{2}+2\sqrt{2}(5-1)=9\sqrt{2}$
$\text{a}_6-\sqrt{2}+2\sqrt{2}(6-1)=11\sqrt{2}$
$\text{a}_7-\sqrt{2}+2\sqrt{2}(7-1)=13\sqrt{2}$

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Question 522 Marks

Find the sum of all integers between 50 and 500, which are divisible by 7.

Answer

The series of integers by 7 between 50 and 500 are
56, 63, 70, ..., 497
Let the number of terms be n then, n^thterm = 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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Question 532 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?

Answer

Given,
$\text{a}_1=-4=\text{a}+0\text{d}\ .....(1)$
$\text{a}=2=\text{a}+4\text{d}\ .....(2)$
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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Question 542 Marks

Find the sum of the following arithmetic progression:
a + b, a - b, a - 3b, ... to 22 terms.

Answer

a + 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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Question 552 Marks

Find the A.M. between:
12 and -8

Answer

12 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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Question 562 Marks

Which trem of the A.P. 3, 8, 13, ...os 248?

Answer

Let 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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Question 572 Marks

If 9th term of an A.P. is zero, prove that its 29th term is double the 19th term.

Answer

Given:
$\text{a}_9=0$
$\therefore\text{a}+8\text{d}=0$
$\text{a}=-8\text{d}\ .....(1)$
$\text{a}_{19}=\text{a}+(19-1)\text{d}$
$=\text{a}+18\text{d}$ $[\therefore\text{a}=-8\text{d}\ \text{from}(1)]$
$=10\text{d}\ .....(2)$
$\text{a}_{29}=\text{a}+(29-1)\text{d}$
$=-8\text{d}+28\text{d}$ $[\because\text{a}=-8\text{d}\ \text{from}(1)]$
$-20\text{d}\ .....(3)$
(2) and (3)
$\text{a}_{29}=2\text{a}_{19}$
Hence proved.

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