Solve the Following Question.(3 Marks)

Question 513 Marks

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

Answer

Let sum of $n$ terms of two $A . . P$ be $s_n$ and $s^{\prime} n$. Then, $S_n=5 n+4$ and $s_n^{\prime} 9 n+16$ respectively.
Then, if ratio of sum of $n$ terms of $2 A . P$ is giben, then the ratio of there $n^{\text {th }}$ ther is obtained by replacing $n$ by ( $\left.2 n-1\right)$.
$\frac{\text{a}_\text{n}}{\text{a}_\text{n}}=\frac{5(2\text{n}-1)+1}{9(2\text{n}-1)+16}$
$\therefore$ Ratio of there 18th 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 523 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 533 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 543 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 553 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 563 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 573 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 583 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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Maths Solve the Following Question.(3 Marks)