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

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3 Marks Question

Question 513 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 523 Marks

If 10 times the 10th term of an A.P. is equal to 15 times the 15th term, show that 25th term of the A.P. is zero.

Answer

Given:
$10\text{a}_{10}=15\text{a}_{15}$
$10(\text{a}+(10-1)\text{d})=15(\text{a}+(15-1)\text{d})$
$10\text{a}+90\text{d}=15\text{a}+210\text{d}$
$5\text{a}+120\text{d}=0$
$\text{a}+24\text{d}=0\ .....(1)$
$\text{a}_{25}=\text{a}+(25-1)\text{d}$
$=\text{a}+24\text{d}$
$=0\ [\because\text{from}(1)\text{a}+24\text{d}=0]$
Hence proved.

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Question 533 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 543 Marks

If $S_1$ be the sum of $(2n + 1)$ terms of an A.P. and $S_2 $ be the sum of its odd terms, the prove that: $S_1 : S_2 = (2n + 1) : (n + 1)$

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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Question 553 Marks

The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.

Answer

Let the angle be
$\text{a}-3\text{d},\ \text{a}-\text{d},\ \text{a}+3\text{d}$
Then,
Sum of all angles $=360^\circ$
$\text{a}-3\text{d}+\text{a}-\text{d}+\text{a}+\text{d}+\text{a}+3\text{d}=360^\circ$
$4\text{a}=360^\circ$
$\text{a}=90^\circ\ .....{ (1)}$
and
$(\text{a}-\text{d})-(\text{a}-3\text{d})=10$
$2\text{d}=10$
$\text{d}=5$
$\therefore$ The angle of the given quadrilateral are 75°, 85°, 95°, and 105°.

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Question 563 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 573 Marks

If (m+1)th term of an A.P. is twice the (n + 1)th term, prove that (3m + 1)th term is twice the (m +n + 1)th term.

Answer

Given:
$\text{a}_{\text{m+1}}=2\text{a}_{\text{n+1}}$
$\Rightarrow\text{a}+(\text{m}+1-1)\text{d}=2(\text{a}+(\text{n}+1-1)\text{d})$
$\Rightarrow\text{a}+\text{md}=\text{2a}+2\text{nd}$
$\Rightarrow\text{a}(\text{m}2-\text{n})\text{d}\ .....(1)$
Then,
$\text{a}_{3\text{m}+1}=\text{a}+(3\text{m}+1-1)\text{d}$
$=\text{a}+3\text{md}$
$=3\text{d}=2\text{nd}+3\text{md}$
$=2(2\text{m}-\text{n})\text{d}\ .....(2)$
$\text{a}_{\text{m+n+1}}=\text{a}+(\text{m+n+1}-1)\text{d}$
$=\text{md}-2\text{nd}+\text{md}+\text{nd}$
$=(2\text{m}-\text{n})\text{d}\ .....(3)$
From (2) and (3)
$\text{a}_{2\text{m+1}}=2\text{a}_{\text{m+n+1}}$
Hence proved,

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Question 583 Marks

Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Answer

Let the four numbers in A.P. be
$\text{a}=3\text{d},\ \text{a}-\text{d},\ \text{a}+\text{d},\ \text{a}+3\text{d}$
$(\text{a}-3\text{d})+(\text{a}-\text{d})+(\text{a}+\text{d})+(\text{a}+3\text{d})=50$
$4\text{a}=50$
$\text{a}=\frac{25}{2}\ .....{(1)}$
and
$(\text{a}+3\text{d})=4(\text{a}-3\text{d})$
$\frac{25+6\text{d}}{2}=50-12\text{d}$
$30\text{d}=75$
$\text{d}=\frac{25}{10}=\frac{5}{2}\ .....{(2)}$
$\therefore$ The required sequence is 5, 10, 15, 20.

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Question 593 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 603 Marks

If 12th term of an A.P. is -13 and the sum of the first four terms is 24, what is the sum of first 10 terms?

Answer

$\text{a}_{12}=\text{a}+11\text{d}=-13\ .....(1)$ [Given]
$\text{s}_4=\frac{4}{2}(2\text{a}+3\text{d})=24\ .....(2)$ [Given]
From (1) and (2)
$\text{d}=-2$ and $\text{a}=9$
Then,
Sum of irst 10 terms is
$\text{s}_{10}=\frac{10}{2}[2\times9+(9)(-2)]$ $\Big[$ using $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+{2\text{a}+(\text{n}-1)\text{d}}\Big]$
$=0$
Sum of first 10 tems is zero.

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Question 613 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 623 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 633 Marks

Is $68$ a them of the A.P. $7, 10, 13, ...?$

Answer

Is $168$ a term of A.P. $7, 10, 13, ...?$
Here, $\text{a}=7$
and $\text{x}=10-7=3$
$\therefore\text{a}_\text{n}$ term is $=\text{a}+(\text{n}-1)\text{d}$
Let $68$ be $n^{th}$ temr of A.P.
Then,
$68=7+3(\text{n}-1)$
$\Rightarrow68=7+3\text{n}-3$
$\Rightarrow68-4=3\text{n}$
$\Rightarrow64=3\text{n}$
$\Rightarrow\text{n}=\frac{64}{3}$
Which is note natural number.
$\therefore 68$ nota term of given A.P.

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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

A piece of equipment cost a certain factory ₹ 600,000. If it depreciates in value, 15% the first, 13.5% the next year, 12% the third year, and so on. What will be its value at the end of 10 years, all percentages applying to the original cost?

Answer

The piece of equipment deprecites 15% in first year i.e., $\frac{15}{100}\times600,000=₹\ 90,000$$\therefore$ value after 1st year $=600,000-90,000$
$=₹\ 510,000$
The equipment deprecites at the rate 135% in 2nd year i.e., $\frac{135}{1000}\times600,000=81000$
$\therefore$ value after 2 nd year =81000
The value after 3rd year $=\frac{12}{100}\times600000=72000$
The total depreciation in 10 years
$\Rightarrow\text{S}_{10}=\frac{10}{2}[2\times81000+(9)(-9000)]$
$=5[81000]$ $\big[$ using $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$
$\therefore$ The cost of machine aftar 10 years $=₹\ 600000-405000$
$=105000$

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Question 663 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 673 Marks

Insert six A.M.s between $15$ and $−13.$

Answer

Let$ A_1, A_2, A_3, A_4, A_5, A_6,$ be the seven $6$ A.M.s between $15$ and $-13.$
Then, $15 A_1, A_2, A_3, A_4, A_5, A_6, -13$ are in A.P. of $8$ terms
Here,
$-13 = 15 + 7d$
$\Rightarrow d -4$
$= -4$
$A_1 = 15 + d = 15 + (−4) = 11$
$A_2 = 15 + 2d = 15 + (−8) = 7$
$A_3 = 15 + 3d = 15 + (−12) = 3$
$A_4 = 15 + 4d = 15 + (−16) = −1$
$A_5 = 15 + 5d = 15 + (−20) = −5$
$A_6 = 15 + 6d = 15 + (−24) = −9$
The 6 A.M.S between $15$ and $-13$ are $11, 7, 3, -1, -5$ and $-9$

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Question 683 Marks

The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.

Answer

Let the 3rd term of A.P. be
$\text{a}-\text{d},\ \text{a},\ \text{a}+\text{d}$
then,
$\text{a}-\text{d}+\text{a}+\text{a}+\text{d}=21$
$\therefore\text{a}=7$
and
$(\text{a}-\text{d})(\text{a+d})=\text{a}+6$
$\text{a}^2-\text{d}^2=\text{a}+6$
$7^2=\text{d}^2=\text{a}+6$ $[\because\text{a}=7]$
$\text{d}^2=36$
$\text{d}=\pm6$
Since d can't be negative, therefore
$\therefore$ the A.P. is 1, 7, 13.

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Question 693 Marks

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

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 703 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 713 Marks

Find the sum of all those integers between $100$ and $800$ each of which on division by $16$ leaves the remainder $7$.

Answer

The fiest number between $100$ and $800$ each of which on division by $16$ leaves the remainder $7$ is $112$ and last number is $791.$
Thus, the series so series formed is $103, 119, ..., 791$
Let number of terms ben, then
$n^{th} $ term $= 791$
Then,
$\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\Rightarrow791=103+(\text{n}-1)16$
$\Rightarrow\text{n}=44$
Then, sum of all terms of the given series is
$\text{s}_{43}=\frac{44}{2}[103+791]$
$=\frac{44\times894}{2}$
$=19668$

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Question 723 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, $5$th term is purely real.
Let $n^{th}$ term be purely im againarym than, $13-\text{n}=0$
$\therefore\text{n}=13$
So, $13$th term is purely inaginary.

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Question 733 Marks

Find the A.M. between:
7 and 13

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Question 743 Marks

The sums of n terms of two arithmetic progressions are in the ratio $5n + 4 : 9n + 6.$ Find the ratio of their $18$th terms.

Answer

Let sum of $n$ terms of two A..P be $s_n $ and $s' n.$Then, $S_n = 5n + 4$ and $s'_n 9n +16$ respectively.
Then, if ratio of sum of n terms of 2A.P is giben, then the ratio of there $n^{th} $ ther is obtained by replacing $n$ by $(2n - 1).$
$\frac{\text{a}_\text{n}}{\text{a}_\text{n}}=\frac{5(2\text{n}-1)+1}{9(2\text{n}-1)+16}$
$\therefore$ Ratio of there $18$th term is
$\frac{\text{a}_{18}}{\text{a}_{18}}=\frac{5(2\times18-1)+4}{9(2\times18-1)+16}$
$=\frac{5\times35+4}{9\times35+16}$
$=\frac{179}{321}$

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Question 753 Marks

The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.

Answer

Let three numbers be $\text{a}-\text{d},\ \text{a},\ \text{a}+\text{d}$
Then,
$\text{a}-\text{d}+\text{a}+\text{a}+\text{d}=12$
$3\text{a}=12$
$\text{a}=4$
and
$(\text{a}-\text{d})^3+\text{a}^3+(\text{a}+\text{d})^3=\pm288$
$\text{a}^3+\text{d}^3+3\text{ad}(\text{a+d})+\text{a}^3+\text{a}^3-\text{a}^3-3\text{ad}(\text{a}-\text{d})-288$
$\Rightarrow2\text{a}^3+3\text{a}^2\text{d}+3\text{ad}^2-3\text{a}^2\text{d}+3\text{ad}^2=288$
$\Rightarrow2\text{a}^3+3\text{a}^2\text{d}^2=288$
$\Rightarrow128+48\text{d}^2=288$
$\therefore\text{d}=\pm2$
$\therefore$ The required sequence is 2, 4, 6, or 4, 2.

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Question 763 Marks

Find the sum of all even integers between $101$ and $999.$

Answer

All even integers will have common differnce $= 2$
$\therefore$ A.P. is $102,\ 104,\ 106,\ ...,\ 998$
$\text{t}_\text{n}=\text{a}+(\text{n}-1)\text{d}$
$\text{t}_\text{n}=998,\text{a}=102,\text{d}=2$
$998=102+(\text{n}-1)(2)$
$998=102+2\text{n}-2$
$998-100=2\text{n}$
$2\text{n}=898$
$\text{n}=449$
$s_{449}$ can be calculated by
$=\text{s}_\text{n}\frac{\text{n}}{2}[\text{a}+\text{l}]$
$=\frac{449}{2}[102+998]$
$=\frac{449}{2}[102+998]$
$=449\times550$
$=346950$

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Question 773 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 783 Marks

If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.

Answer

Let 3 number in A.P be
$\text{a}-\text{d},$ and $\text{a}+\text{d}$
$\Rightarrow(\text{a}-\text{d})+(\text{a})+(\text{a}+\text{d})=24$
$3\text{a}=24$
$\text{a}=8$
and
$(\text{a}-\text{d})(\text{a})(\text{a}+\text{d})=440$
$8^2-\text{d}^2=55$
$\text{d}=3$
$\therefore$ The required sequnce is 5, 8, 11,

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Question 793 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 803 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 813 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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