3 Marks Question - Page 2 - MATHS STD 11 Science Questions - Vidyadip

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3 Marks Question

Question 513 Marks

The first and the last terms of an A.P. area and I respectively. Show that the sum of n^th term from the beginning and n^th term from the end is a + l.

Answer

The n^th term from starting
$=\text{a}_\text{n}=\text{aa}+(\text{n}-1)\text{d}\ .....(1)$
The n^th term from end
$=\text{l}-(\text{n}-1)\text{d}\ .....{(2)}$
Adding (1) and (2) we get
Sum of n^th term from begining and n^th term from the end
$=\text{a}+(\text{n}-1)\text{d}+\text{l}-(\text{n}-1)\text{d}$
$=\text{a}+\text{l}$
Hence proced.

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Question 523 Marks

How many numbers of two digit are divisible by 3?

Answer

The first two digit number divisible by 3 is 12.
and last two digit number divisible by 3 is 99.
So, the required series is 12, 15, 18, ...99.
Let there be n terms then n^th term = 99
$\Rightarrow99=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow99=12+(\text{n}-1)3$
$\Rightarrow\text{n}=30$
30 two digit number are divisible by 3.

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Question 533 Marks

If in A.P. is such that $\frac{\text{a}_4}{\text{a}_7}=\frac{2}{3},$ find $\frac{\text{a}_6}{\text{a}_8}.$

Answer

$\frac{\text{a}_4}{\text{a}_7}=\frac{2}{3}$ [Given]
$\Rightarrow\frac{\text{a}+3\text{d}}{\text{a}+6\text{d}}=\frac{2}{3}$
$\Rightarrow3\text{a}+9\text{d}=2\text{a}+12\text{d}$
$\Rightarrow\text{a}=3\text{d}\ .....(1)$
$\frac{\text{a}_6}{\text{a}_8}=\frac{\text{a}+5\text{d}}{\text{a}+7\text{d}}$ $[\because3\text{d from}(1)]$
$\Rightarrow\frac{8\text{d}}{10\text{d}}$
$\Rightarrow\frac{4}{5}$
$\frac{\text{a}_6}{\text{a}_8}=\frac{4}{5}$

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Question 543 Marks

The sum of first 7 terms of an A.P. is 10 and that next 7 terms is 17. find the progression.

Answer

Given,
$\text{a}_7=10$
$\text{s}_{14}-\text{s}_7=17\ .....{(1)}$
$\therefore\text{s}_{14}=17+\text{s}_7=17+10=27\ .....{(2)}$
From (1) and (2)
$\text{s}_7=\frac{7}{2}[2\text{a}+(7-1)\text{d}]$ $\Big[$Using $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\Big]$
$\Rightarrow10=7\text{a}+21\text{d}\ .....(3)$
and
$\text{s}_{14}=\frac{14}{2}[2\text{a}+13\text{d}]$
$\Rightarrow27=28\text{a}+182\text{d}\ .....{4}$
Solving (3) and (4)
$\text{a}=1$ and $\text{d}=\frac{1}{7}$
$\therefore$ The required A.P is
$1,\ 1+\frac{1}{7},\ 2+\frac{2}{7},\ 1+\frac{3}{7},...,\ +\infty$
or $1,\ \frac{8}{7},\ \frac{9}{7},\ \frac{10}{7},\ \frac{11}{7},\ ...,\ \infty$

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Question 553 Marks

Find the sum of n terms of the A.P. whose k^th terms is 5k + 1.

Answer

Here,
$\text{a}_\text{k}=5\text{k}+1$
$\text{a}_1=5+1=6$
$\text{a}_2=5(2)+1=11$
$\text{a}_3=5(3)+1=16$
$\text{d}=11-6=16-11=5$
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$=\frac{\text{n}}{2}[2(6)+(\text{n}-1)(5)]$
$=\frac{\text{n}}{2}[12+5\text{n}-5]$
$\text{s}_\text{n}=\frac{\text{n}}{2}(5\text{n}+7)$

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Question 563 Marks

Find the sum of all integers between 100 and 550, wgich are divisible by 9.

Answer

The series fromed bt all the integers between 100 and 550 which are divisible 9 is
108, 117, 123, ... , 549
Let there be n terms in the A.P then, the n^th term is 549
$549=\text{a}(\text{n}-1)\text{d}$
$549=108+(\text{n}-1)9$
$\Rightarrow\text{n}=50$
Then,
$\text{s}_\text{n}=\frac{\text{n}}{2}=[\text{a}+\text{l}]$
$\text{s}_{50}=\frac{50}{2}[108+549]$
$=16425$

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Question 573 Marks

If $\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big),\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big),\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)$ are in A.P., proved that a, b, c are in A.P.

Answer

$\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big),\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big),\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)$ are in A.P.
$\Rightarrow\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)+1,\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big)+1,\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)+1$ are in A.P.
$\Rightarrow\Big(\frac{\text{ac}+\text{ab}+\text{bc}}{\text{bc}}\Big),\ \Big(\frac{\text{ab}+\text{bc}+\text{ac}}{\text{ac}}\Big),\ \Big(\frac{\text{cd}+\text{ac}+\text{ab}}{\text{ab}}\Big)$ are in A.P.
$\Rightarrow\frac{1}{\text{bc}},\ \frac{1}{\text{ac}},\ \frac{1}{\text{ab}}$ are in A.P.
$\Rightarrow\frac{\text{abc}}{\text{bc}},\ \frac{\text{abc}}{\text{ac}},\ \frac{\text{abc}}{\text{ab}}$ are in A.P.
$\Rightarrow\text{a},\text{b},\text{c}$ are in A.P.

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Question 583 Marks

If $\text{s}_\text{n}=\text{n}^2\ \text{p}$ and $\text{s}_\text{m}=\text{m}^2\ \text{p},\ \text{m}\not=\text{n},$ in an A.P., prove that $\text{s}_\text{p}=\text{p}^3.$

Answer

Let a be the first term of the AP and d is the common difference. then
$\text{s}_\text{n}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$\text{n}^2\text{p}=\frac{\text{n}}{2}(2\text{a}+(\text{n}-1)\text{d})$
$\text{np}=\frac{1}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$2\text{np}=2\text{a}+(\text{n}-1)\text{d}\ .....{(1)}$
Again
$\text{s}_\text{m}=\frac{\text{m}}{2}(2\text{a}+(\text{m}-1)\text{d})$
$\text{m}^2\text{p}=\frac{\text{m}}{2}(2\text{a}+(\text{m}-1)\text{d})$
$\text{mp}=\frac{1}{2}[2\text{a}+(\text{m}-1)\text{d}]\ .....{(2)}$
Now subtract (1) from (2)
$2\text{p}(\text{m}-\text{n})=(\text{m}-\text{n})\text{d}$
$\text{d}=2\text{p}$
Therefore
$2\text{mp}=2\text{a}+(\text{m}-1)\times2\text{p}$
$2\text{a}=2\text{p}$
$\text{a}=\text{p}$
The sum up p terms will be:
$\text{s}_\text{p}=\frac{\text{p}}{2}(2\text{a}+(\text{p}-1)\text{d})$
$=\frac{\text{p}}{2}(2\text{p}+(\text{p}-1).2\text{p})$
$=\frac{\text{p}}{2}(2\text{p}+2\text{p}^2-2\text{p})$
$=\text{p}^3$

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Question 593 Marks

If the sum of n terms of an A.P. is $\text{np}+\frac{1}{2}\text{n}(\text{n}-1)$ Q, where P and Q are constants, find the common difference.

Answer

$\text{s}_\text{n}=\text{n}\text{p}+\frac{1}{2}(\text{n}-1)\text{Q}$ [Given]
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{p}+(\text{n}-1)\text{Q}]\ .....(1)$
We know,
$\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\ .....(2)$
Where a = first term and d = common diffrence (1) and (2) d = Q
$\therefore$ the common diffrence Q.

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Question 603 Marks

Find the sum of the following arithmetic progression:
$\frac{\text{x}-\text{y}}{\text{x}+\text{y}},\ \frac{3\text{x}-2\text{y}}{\text{x}+\text{y}},\ \frac{5\text{x}-3\text{y}}{\text{x}+\text{y}},\ ...$ to n terms.

Answer

$\frac{\text{x}-\text{y}}{\text{x}+\text{y}},\ \frac{3\text{x}-2\text{y}}{\text{x}+\text{y}},\ \frac{5\text{x}-3\text{y}}{\text{x}+\text{y}},\ ...$ to n terms.
n^th term is above sequence is $\frac{(2\text{n}-1)\text{x}-\text{ny}}{\text{x}+\text{y}}$
Sum of n terms is given by
$\frac{1}{\text{x}+\text{y}}[\text{x}+3\text{x}+5\text{x}+.....+(2\text{n}-1)\\\text{x}-(\text{y}+2\text{y}+3\text{y}...+\text{ny})]$
$=\frac{1}{\text{x}+\text{y}}\Big[\frac{\text{n}}{2}(2\text{x}+(\text{n}-1)2\text{x})-\frac{\text{n}(\text{n}+1)\text{y}}{2}\Big]$
$=\frac{1}{2(\text{x}+\text{y})}[2\text{n}^2\text{x}-2\text{n}^2\text{y}-\text{ny}]$

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Question 613 Marks

If $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P., Prove that:
$\frac{\text{b}+\text{c}}{\text{a}},\ \frac{\text{c}+\text{a}}{\text{b}},\ \frac{\text{a}+\text{b}}{\text{c}}$ are in A.P.

Answer

$\frac{\text{b}+\text{c}}{\text{a}},\ \frac{\text{c}+\text{a}}{\text{b}},\ \frac{\text{a}+\text{b}}{\text{c}}$ will be in A.p if $\frac{\text{c}+\text{a}}{\text{b}}-\frac{\text{b}+\text{c}}{\text{a}}=\frac{\text{a}+\text{b}}{\text{c}}-\frac{\text{c}+\text{a}}{\text{b}}$
if $\frac{\text{ca}+\text{a}^2-\text{b}^2-\text{cd}}{\text{ab}}=\frac{\text{ab}+\text{b}^2-\text{c}^2-\text{ac}}{\text{bc}}$
$\text{LHS}\Rightarrow\frac{\text{ca}+\text{a}^2-\text{b}^2-\text{cd}}{\text{ab}}$
$\Rightarrow\frac{\text{c}^2\text{a}+\text{a}^2\text{c}-\text{b}^2\text{c}-\text{c}^2\text{b}}{\text{abc}}$
$\Rightarrow\frac{\text{c}(\text{a}-\text{b})[\text{a}+\text{b}+\text{c}]}{\text{bc}}\ .....(1)$
$\text{RHS}\Rightarrow\frac{\text{ab}+\text{b}^2-\text{c}^2-\text{ac}}{\text{bc}}$
$\Rightarrow\frac{\text{a}^2\text{b}+\text{ab}^2-\text{ac}^2-\text{a}^2\text{c}}{\text{abc}}$
$\Rightarrow\frac{\text{a}(\text{b}-\text{c})[\text{a}+\text{b}+\text{c}]}{\text{abc}}\ .....(2)$
and since $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P
$\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$\text{c}(\text{b}-\text{a})=\text{a}(\text{b}-\text{c})\ .....(3)$
$\therefore$ LHS = RHS and the given terms are in A.P

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Question 623 Marks

A man saved ₹ 16,500 in ten years. In each year after the first he saved ₹ 100 more than he did in the receding year. How much did he save in the first year?

Answer

Let the amount saved by the man in the first year be x.
Then,
$\text{ATQ}$
$\text{x}+(\text{x}+100)+(\text{x}+200)+\ ...\ +(\text{x}+900)=16500$
As his saving increased by ₹ 100 every year.
$\therefore10\text{x}+100+200+300+\ ...\ +900=16500\ .....(1)$
Here,
$100+200+300+\ ...\ +(\text{x}+900)=16500$
$\text{a}=100,\ \text{d}=100$ and $\text{n}=9$
So,
$\text{S}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$
$\text{S}_9=\frac{\text{9}}{2}[100+900]=4500\ .....(2)$
From (1) and (2)
$10\text{x}+(4500)=16500$
$10\text{x}=12000$
or $\text{x}=1200$
The man saved ₹ 1200 in the frist year.

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Question 633 Marks

A carpenter was hired to build 192 window frames. The first day he made five frames and each day thereafter he made two more frames than he made the day before. How many days did it take him to finish the job?

Answer

Suppose carpenter took n days to finish his job.
First day carpenter made five frames $\text{a}_1=5$
Each day atter first day he made two more frames
$\text{d}=2$
$\therefore$ On n^th day frames made by carpenter are,
$\text{a}_\text{n}=\text{a}_\text{n}+(\text{n}-1)(\text{d})$
$\Rightarrow\text{a}_\text{n}5+(\text{n}-1)2$
Sum of all the frames till n^Ln day is
$\text{S}=\frac{\text{n}}{2}[\text{a}_1+\text{a}_\text{n}]$
$192=\frac{\text{n}}{2}[5+5+(\text{n}-1)2]$
$192=5\text{n}+\text{n}^2-\text{n}$
$\text{n}^2+4\text{n}-192=0$
$(\text{n}+16)(\text{n}-12)=0$
$\text{n}=-16$ or $\text{n}=12$
But number of days cannot be negative hence $\text{n}=12$
The carpenter took 12 days to finish his job.

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Question 643 Marks

If $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P., Prove that:
$\text{a}(\text{b}+\text{c}),\ \text{b}(\text{c}+\text{a}),\ \text{c}(\text{a}+\text{b})$ are in A.P.

Answer

$\text{a}(\text{b}+\text{c}),\ \text{b}(\text{c}+\text{a}),\ \text{c}(\text{a}+\text{b})$ are in A.P if $\text{b}(\text{c}+\text{a})-\text{a}(\text{b}+\text{c})=\text{c}(\text{a}+\text{b})=\text{c}(\text{a}+\text{b})-\text{b}(\text{c}+\text{a})$
$\text{LHS}=\text{b}(\text{c}+\text{a})-\text{a}(\text{b}+\text{c})$
$=\text{bc}+\text{ab}-\text{ab}-\text{ac}$
$=\text{c}(\text{b}-\text{a})$
$\text{RHS}=\text{c}(\text{a}+\text{b})-\text{b}(\text{c+a})$
$=\text{ca}+\text{cd}-\text{bc}-\text{ba}$
$=\text{a}(\text{c}-\text{d})\ .....(2)$
and $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P
$\therefore\frac{1}{\text{}a}-\frac{1}{\text{b}}=\frac{1}{\text{b}}-\frac{1}{\text{c}}$
or $\text{c}(\text{b}-\text{a})=\text{a}(\text{c}-\text{b})\ .....(3)$
From (1), (2) and (3)
$\text{a}(\text{b}+\text{c}),\ \text{b}(\text{c}+\text{a}),\ \text{c}(\text{a}+\text{b})$ are in A.P

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Question 653 Marks

How many numbers are there between 1 and 1000 which when divided by 7 leave remainder 4?

Answer

A number N is dividend by 7 leaves a remainder 4.
$\therefore\text{N}=7\text{q}+4$
N can take values 4, 11, 18, ..... 998
Now,
4, 11, 18, ..... 998 are arithmetic progression.
First term $\text{A}=4$
Common differnce $\text{d}=7$
Last term $\text{l}=998$
We know thet,
$\text{l}=\text{a}(\text{n}-1)\text{d}$
$\Rightarrow998=4+(\text{n}-1)7$
$\Rightarrow998=4+7\text{n}-1$
$\Rightarrow998=7\text{n}-3$
$\Rightarrow1001=7\text{n}$
$\Rightarrow\text{n}=\frac{1001}{7}$
$\Rightarrow\text{n}=143$
Hence, 143 numbers are there between 1 and 1000 which when divided by 7 leave remainder4.

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Question 663 Marks

Find the r^th term of an A.P., the sum of whose first n terme is 3n² + 2n.

Answer

Sum first n terms of the given AP is
$\text{s}_\text{n}=3\text{n}^2+2\text{n}$
$\text{s}_{\text{n}-1}=3(\text{n}-1)^2+2(\text{n}-1)$
$\text{a}_\text{n}=\text{s}_\text{n}-\text{s}_{\text{n}-1}$
$\text{a}_{\text{n}}=3\text{n}^2+2\text{n}-3(\text{n}-1)^2-2(\text{n}-1)$
$\text{a}_\text{n}=6\text{n}-1$
$\text{a}_\text{r}=6\text{r}-1$
$\text{r}^{\text{th}}\text{term is}\ 6\text{r}-1$

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Question 673 Marks

Which term of the sequence $12+8\text{i},\ 6\text{i},\ 10+4\text{i},\ ...$ is (a) real (b) purely imaginary?

Answer

The given sequence is $12+8\text{i},\ 6\text{i},\ 10+4\text{i},\ ...$
Here, $\text{a}=12+8\text{i}$
$\text{d}=-1-2\text{i}$
Then, $\text{a}_\text{n}=\text{a}(\text{n}-1)\text{d}$
$=12+8\text{i}+(\text{n}-1)(-1-2\text{i})$
$=(13-\text{n})+\text{i}(10-2\text{i})$
Let n^th term be purely real the $(10-2\text{n})=0$ or $\text{n}=5$
So, 5th term is purely real.
Let n^th term be purely im againarym than, $13-\text{n}=0$
$\therefore\text{n}=13$
So, 13th term is purely inaginary.

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Question 683 Marks

Find the A.M. between:
7 and 13

Answer

7 and 13
Let A be the arithem atic mean of 7 and 13
Then,
7, A, 13 are in A.P
$\Rightarrow\text{A}-7=13-\text{A}$
$\Rightarrow\text{A}=\frac{13-7}{2}=10$
$\therefore\text{A.M is 10}$

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Question 693 Marks

The sums of first n terms of two A.P.'s are in the ratio (7n + 2) : (n + 4). Find the ratio of their 5th terms.

Answer

Let sum of n term 1 A.P series are other s_n
The,
$\text{s}_\text{n}7\text{n}+2\ .....(1)$
$\text{s}_\text{n}=\text{n}+4\ .....(2)$
the ratio of sum of n terms of A.P is given, then the ratio of there n^th term is obtained by (2n - 1).
$\frac{\text{a}_\text{n}}{\text{a}_\text{n}}=\frac{7(2\text{n}-1)+2}{(2\text{n}-1)+4}$
Putting n = 5 to get the ratio of 5th term, we get
$\frac{\text{a}_5}{\text{a}_5}=\frac{7(2\times5-1)+1}{(2\times5-1)+4}=\frac{65}{13}=\frac{5}{1}$
The ratio is 5 : 1

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Question 703 Marks

There are 25 trees at equal distances of 5 metres in a line with a well, the distance of the well from the nearest tree being 10 metres. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees.

Answer

There are 25 trees at equal distance of 5 m in line with a will (w), and the distance of the well from the nearesst tree = 10 m.
Thus,
The total distance travelled by gardener to tree 1 and back is $2\times10\text{m}=20\text{m}$
Similarly for all the 25 trees.
The distance covred by gardener is
$=2[10+(10+5)+(10+2\times5)\\+(10+3\times5)+\ ...\ +(10+23\times5)]$
This froms a seroes of 1st term a = 10, common difference d = 5 and n = 25
$\therefore10+(10+5)+(10+2\times5)+\ ...\ +(10+23\times5)$
$\Rightarrow\text{S}_{25}=\frac{25}{2}[2\times10+(24)5]=25[10+60]=1750\text{m}$
From(1) and (2)
Total distance $=2\times1750\text{m}=3500\text{m}$

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Question 713 Marks

If $\text{a}^2,\ \text{b}^2,\ \text{c}^2$ are in A.P., prove that $\frac{\text{a}}{\text{b}+\text{c}},\frac{\text{b}}{\text{c}+\text{a}},\frac{\text{c}}{\text{a}+\text{b}}$ are in A.P.

Answer

$\frac{\text{a}}{\text{b}+\text{c}},\frac{\text{b}}{\text{c}+\text{a}},\frac{\text{c}}{\text{a}+\text{b}}$ are in A.P if $\frac{\text{b}}{\text{a}+\text{c}}-\frac{\text{a}}{\text{b}+\text{c}}=\frac{\text{c}}{\text{a}+\text{b}}-\frac{\text{b}}{\text{a}+\text{c}}$
$\text{LHS}=\frac{\text{b}}{\text{a}+\text{c}}-\frac{\text{a}}{\text{b}+\text{c}}$
$\Rightarrow\frac{\text{b}^2+\text{bc}-\text{a}^2-\text{ac}}{(\text{a}+\text{c})(\text{b}+\text{c})}$
$\Rightarrow\frac{(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})}{(\text{a}+\text{c})(\text{b}+\text{c})}\ .....(1)$
$\text{RHS}=\frac{\text{a}}{\text{a}+\text{b}}=\frac{\text{b}}{\text{a}+\text{c}}$
$\Rightarrow\frac{\text{ca}+\text{c}^2-\text{b}^2-\text{ab}}{(\text{a}+\text{b})(\text{b}+\text{c})}$
$\Rightarrow\frac{(\text{c}-\text{d})(\text{a}+\text{b}+\text{c})}{(\text{a}+\text{b})(\text{b}+\text{c})}\ .....(2)$
and
$\text{a}^2,\ \text{b}^2,\ \text{c}^2$ are in A.P
$\therefore\text{b}^2-\text{a}^2=\text{c}^2-\text{b}^2\ .....(3)$
Substituting $\text{b}^2-\text{a}^2$ with $\text{c}^2-\text{b}^2$
$(1)=(2)$
$\therefore\frac{\text{a}}{\text{b}+\text{c}},\ \frac{\text{b}}{\text{a}+\text{c}},\ \frac{\text{c}}{\text{a}+\text{b}}$ are in A.P

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Question 723 Marks

Insert six A.M.s between 15 and −13.

Answer

Let A₁, A₂, A₃, A₄, A₅, A₆, be the seven 6 A.M.s between 15 and -13.
Then, 15 A₁, A₂, A₃, A₄, A₅, A₆, -13 are in A.P. of 8 terms
Here,
-13 = 15 + 7d
⇒ d -4
= -4
A₁ = 15 + d = 15 + (−4) = 11
A₂ = 15 + 2d = 15 + (−8) = 7
A₃ = 15 + 3d = 15 + (−12) = 3
A₄ = 15 + 4d = 15 + (−16) = −1
A₅ = 15 + 5d = 15 + (−20) = −5
A₆ = 15 + 6d = 15 + (−24) = −9
The 6 A.M.S between 15 and -13are 11, 7, 3, -1, -5 and -9

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Question 733 Marks

A man is employed to count ₹ 10710. he count at the rate od ₹ 180 per minute for half an hour. after this he counts at the rate of ₹ 3 less every minute than the preceding minute. find the time takan by him to count the entire amount.

Answer

The man of counts at the rate of ₹ 180 per minute for half an hour. After this he counts at the rate of ₹ 3 less every minute than preceding minute.
Then, the amount counted in first 30 mitnute
$=₹\ 180\times30=₹\ 5400$
The amount left to be counted aftar 30 minute
$=₹\ 10710=5400 =₹\ 5310$
ATQ
A.p formed is $(180-3)+(180-2\times3)+\ ...=5310$
Let time takan to count 5310 be t
Then,
$\text{S}_\text{t}=\frac{\text{t}}{2}[200-3\text{t}]$
$5310=\frac{\text{t}}{2}[200-3\text{t}]$
or $\text{t}=59$ minute
Thus, the total time takan by the man to count ₹ 10710 is (59 - 30) = 89 minutes.

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Question 743 Marks

If $\frac{\text{b}+\text{c}}{\text{a}},\ \frac{\text{c}+\text{a}}{\text{b}},\ \frac{\text{a}+\text{b}}{b}$ are A.P., prove that:
bc, ca, ab are in A.P.

Answer

$\text{bc},\ \text{ca},\ \text{ab}$ are in A.P.
Then,
$\text{ca}-\text{bc}=\text{ab}-\text{ca}$
$\text{c}(\text{a}-\text{b})=\text{a}(\text{b}-\text{c})\ .....(1)$
If $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P
$\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$\Rightarrow\text{c}(\text{a}-\text{b})=\text{a}(\text{b}-\text{c})\ .....(2)$
Thus, the condition necessare to prove $\text{bc},\ \text{ca},\ \text{ab}$ iv A.P is full filled.
Thes,
$\text{bc},\ \text{ca},\ \text{ab}$ are in A.P

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Question 753 Marks

The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. find the first term, the first term, the common difference and sum of first 20 terms.

Answer

Given,
$\text{a}_3=7=\text{a}+2\text{d}\ .....{(1)}$
$\text{a}_7=3\text{a}_3+2$
$\therefore\text{a}_7=3(7)+2$ $[\because\text{a}_3=7]$
$=23=\text{a}+6\text{d}\ .....{}(2)$
Solving (1) and (2)
$\text{a}=-1,\text{d}=4$
Then, sum of 20 terms of this A.P
$\Rightarrow\text{s}_{20}=\frac{20}{2}[2+(20-1)4]$ $\Big[$ Using $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}\Big]$
$=10\times74$
$=740$

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Question 763 Marks

In an A.P., show that $\text{a}_{\text{m}+\text{n}}+\text{a}_{\text{m}-\text{n}}=2\text{a}_\text{m}.$

Answer

It is given that the sequence $<\text{a}_\text{n}>$ is an A.P.
$\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}\ .....(1)$
Similarly from (1)
$\text{a}_{\text{m+n}}=\text{a}+(\text{m}+\text{n}-1)\text{d}\ .....(2)$
$\text{a}_{\text{m+n}}=\text{a}+(\text{m}-\text{n}-1)\text{d}\ .....(3)$
Adding (2) and (3)
$\text{a}_{\text{m+n}}+\text{a}_{\text{m}-\text{n}}=(\text{a}+(\text{m+n}-1)\text{d})(\text{a}+(\text{m}-\text{n}-1)\text{d})$
$=2\text{a}+(\text{m+n}-1+\text{m}-\text{n}-1)\text{d}$
$=\text{2a}+\text{2d}(\text{m}-1)$
$=2(\text{a}+(\text{m}-1)\text{d})$
$=\text{2a}_\text{m}$
Hence proved.

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Question 773 Marks

Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.

Answer

Here,
$\text{s}_\text{n}=3\text{n}^2\ .....(1)$ [Given]
Where n is number of term
$\therefore\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]\ .....(2)$
From (1) and (2)
$3\text{n}^2=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$6\text{n}=2\text{a}+\text{nd}=\text{d}$
Equating both sides
$6\text{n}=\text{nd}$
$\therefore\text{d}=6\ .....(3)$
and
$0=2\text{a}-\text{d}$
or $\text{d}=2\text{a}\ .....{(4)}$
From (3) and (4)
$\text{a}=3$ and $\text{d}=6$
$\therefore$ The required A.P. is $3,\ 9,\ 15,\ 21,\ ...,\ \infty$

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Question 783 Marks

How many terms of the A.P. $-6,\ -\frac{11}{2},\ 5,\ ...$ are needed to give the sum -25?

Answer

Let the number of terms to be added the series is n.
Now,
$\text{a}=-6$ and $\text{d}=0.5.$
Therefore,
$-25=\frac{\text{n}}{2}[2(-6)+(\text{n}-1)(0.5)]$
$-50=\text{n}[-12+0.5\text{n}-0.5]$
$-12.5\text{n}+0.5\text{n}^2+50=0$
$\text{n}^2-25\text{n}+100=0$
$\text{n}=20,5$
Therefore the value of n will either 20 or 5.

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Question 793 Marks

A man saves ₹ 32 during the first year, ₹ 36 in the second year and in this way he increases his savings by ₹ 4 every year. Find in what time his saving will be ₹ 200.

Answer

Let the man save ₹ 200 in n numbers of year.
Then,
$\text{ATQ}$
$32+36+40+\ ...\ +=200$
It rorms a series of n terms, with $\text{a}=32$ and $\text{d}=4$
$\therefore\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}(\text{n}-1)\text{d}]$
$\Rightarrow200=\frac{\text{n}}{2}[2(32)+(\text{n}-1)4]$
$\Rightarrow400=60\text{n}+4\text{n}^2$
$\Rightarrow\text{n}^2+15\text{n}-100=0$
$\Rightarrow\text{n}=5$ or $-20$ [It can't be negative]
$\therefore\text{n}=5$
The man will save ₹ 200 in year

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Question 803 Marks

If the sum of a certain number of terms of the AP 25, 22, 19, ... is 116. Find the last term.

Answer

Sum of terms 25, 22, 19, ... is 116
$\frac{\text{n}}{2}[50+(\text{n}-1)(-\text{3})]=116$
$\frac{\text{n}}{2}[53-3\text{n}]=116$
$53\text{n}-3\text{n}^2=232$
$3\text{n}^2-53\text{n}+232=0$
$3\text{n}^2-29\text{n}-24\text{n+232=0}$
$\text{n}(3\text{n}-29)-8(3\text{n}-29)=0$
$(3\text{n}-29)(\text{n}-8)=0$
$\Rightarrow\text{n}=8$ or $\frac{29}{3}$
N connor be in fracti on, so $\text{n}=8$
Last term $=25-7\times3=4$

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Question 813 Marks

If S₁ be the sum of (2n + 1) terms of an A.P. and S₂ be the sum of its odd terms, the prove that: S₁ : S₂ = (2n + 1) : (n + 1)

Answer

$\text{s}_{(2\text{n}+1)}=\text{s}_1=\frac{(2\text{a}+1)}{2}[2\text{a(2)}\text{n}+1-1\text{d}]$
$\text{s}_1=\frac{(2\text{n}+1)}{2}[2\text{a}+2\text{nd}]$
$=(2\text{n}+1)(\text{a}+\text{nd})\ .....(1)$
sum of odd terms $=\text{s}_2$
$\text{s}_2=\frac{(\text{n}+1)}{2}[2\text{a}+(\text{n}+1-1)(2\text{d})]$
$=\frac{(\text{n}+1)}{2}[2\text{a}+\text{nd}]$
$\text{s}_2=(\text{n}+1)(\text{a}+\text{nd})\ .....(2)$
From equation (1) and (2),
$\text{s}_1:\text{s}_2=(2\text{n}+1)(\text{a}+\text{nd}):(\text{n}+1)(\text{a}+\text{nd})$
$\text{s}_1:\text{s}_2=(2\text{n}+1);(\text{n}+1)$

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