Question 13 Marks
Find the sum of odd integers from 1 to 2001.
AnswerLet the number of terms is n. Now the sum of the series is: 1 + 3 + 5 + ... + 2001 Here, $\text{l}=2001$ and $\text{d}=2$ Therefore, $\text{l}=\text{a}+(\text{n}-1)\text{d}$ $2001=1+(\text{n}-1)\text{d}$ $2(\text{n}-1)=2000$ $\text{n}-1=1000$ $\text{n}=1001$ Therefore the sum of the series is: $\text{s}=\frac{1001}{2}[2+(1001-1)2]$ $=1001^2$ $=10021001$
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In an A.P. the first term is $2$ and the sum of the first five terms is one fourth of the next five terms. Show that $20th$ term is $−112$.
AnswerHere the first term $a = 2$. Let the common difference is d. Now, $\frac{5}{2}[2\text{a}+(5-1)\text{d}]=\frac{1}{4}\Big[\frac{5}{2}[2(\text{a}+\text{d}5)+(5-1)\text{d}]\Big]$
$\frac{5}{2}[2.2+4\text{d}]=\frac{5}{8}[2.2+14\text{d}]$
$10+10\text{d}=\frac{5}{2}+\frac{35}{4}\text{d}$
$\frac{5}{4}\text{d}=-7.5$ $\text{d}=-6$ The $20^{th}$ term will be: $\text{a}+(\text{n}-1)\text{d}=2+(20-1)(-6)$ $=-112$ Hence it is shown.
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In a cricket team tournament 16 teams participated. A sum of ₹ 8000 is to be awarded among themselves as prize money. If the last place team is awarded ₹ 275 in prize money and the award increases by the same amount for successive finishing places, then how much amount will the first place team receive?
AnswerSuppose the award increases by ₹ x.$\text{d}=\text{x}$
In cricket team tournament 16 teams participated. $\text{n}=16$ The last place team is awarded ₹ 275 in prize money$\text{a}_1=275$
Sum of ₹ 8000 is to be awarded as prize money$\text{S}=8000$
$\frac{16}{2}[\text{a}_1+\text{a}_1+(16-1)\times]=8000$
$2\text{a}_1+15\text{x=1000}$
$15\text{x}=450$
$\text{x}=30$
The amount received by first place team$=\text{a}_{16}$
$=\text{a}_1+(16-1)\text{d}$
$=275+15\times30$
$=725$
The amount received by first place team is ₹ 725.
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In an A.P., show that $\text{a}_{\text{m}+\text{n}}+\text{a}_{\text{m}-\text{n}}=2\text{a}_\text{m}.$
AnswerIt 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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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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We know that the sum of the interior angles of a triangle is 180°. Show that the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Find the sum of the interior angles for a 21 sided polygon.
AnswerWe know that sum of interior angles of a polygon with n sides is given by, $\text{a}_\text{n}=180^\circ(\text{n}-2)$ Sum of interior angles of a polygon with 3 sides is given by, $\text{a}_3=180^\circ(3-2)=180^\circ\ .....(1)$ Sum of interior angles of a polygon with 7 sides is given by, $\text{a}_4=180^\circ(4-2)=360^\circ\ .....(2)$ Sum of interior angles of a polygon with 5 sides is given by, $\text{a}_5=180^\circ(5-2)=540^\circ\ .....(3)$ From $eq^n$ (2), $eq^n$ (2) and $eq^n$ (3) we get, $\text{a}_4=360^\circ=180^\circ+180^\circ=\text{a}_4+180^\circ=\text{a}_4+\text{d}$ $\text{a}_5=540^\circ=180^\circ+360^\circ=\text{a}_5+2\text{d}$ Hence the sums of the interior angles of polygons with 3, 4, 5, 6, ... sides form an arithmetic progression. Sum of interior angles of 21 sided polygon $=180^\circ(21-2)$ $=3420^\circ$
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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.
AnswerLet 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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The sum of first 7 terms of an A.P. is 10 and that next 7 terms is 17. find the progression.
AnswerGiven, $\text{a}_7=10$ $\text{s}_{14}-\text{s}_7=17\ .....{(1)}$ $\therefore\text{s}_{14}=17+\text{s}_7=17+10=27\ .....{(2)}$ 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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The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 34. Find the first term and the common difference of the A.P.
Answer$\text{a}_4+\text{a}_8=24$ [Given] $\Rightarrow(\text{a+3d})+(\text{a}+7\text{d})=24$ $\Rightarrow\text{a}+5\text{d}=12\ .....(1)$ $\text{a}_6+\text{a}_10=34$ $\Rightarrow(\text{a}+5\text{d})+(\text{a}+9\text{d})=34$ $\Rightarrow\text{a}+7\text{d}=17\ .....(2)$ From (1) amd (2) $\text{a}=\frac{-1}{2}$ and $\text{d}=\frac{5}{2}$ $\therefore$ 1st term is $\frac{-1}{2}$ and common diffrence is $\frac{5}{2}.$
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Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
AnswerLet $A_1, A_2, A_3, A_4, A_5$ be five numbers between 8 and 26 . Let $d$ be the common difference. Then, we have: $26=A_7 \Rightarrow$ $26=8+(7-1) d \Rightarrow 26=8+6 d \Rightarrow d=3 A_1=8+d=8+3=11 A_2=8+2 d=8+6=14 A_3=8+3 d=8+9=17$ $A_4=8+4 d=8+12=20 A_5=8+5 d=8+15=23$ Therefore, the five numbers are $11,14,17,20,23$.
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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.
AnswerThe $n^{th}$ 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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How many numbers of two digit are divisible by 3?
AnswerThe 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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Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.
AnswerLet the 3 nimber in A.P are $\text{a}-\text{d}+\text{a}+\text{a}+\text{d}=27$ $3\text{a}=27$ $\therefore\text{a}=9\ .....{(1)}$ and $(\text{a}-\text{d})(\text{a})(\text{a}+\text{d})=648$ $(9-\text{d})9(9-\text{d})=648\ [\because\text{a}=9]$ $9^2-\text{d}^2=72$ $\therefore\text{d}=3\ .....(2)$ $\therefore$ the given sequeance is 6, 9, 12.
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If the 5th and 12th terms of an A.P. are 30 and 65 respectively, what is the sum of first 20 terms?
Answer$\text{a}_{5}=\text{a}+4\text{d}=30\ .....(1)$ [Given] $\text{a}_{12}=\text{a}+11\text{d}=65\ .....(2)$ [Given] From (1) and (2) $\text{d}=5$ and $\text{a}=10$ Then, Sum of irst 20 terms is $\text{s}_{\text{n}}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $\Rightarrow\text{s}_{\text{20}}=\frac{\text{20}}{2}[2\times10+(\text{20}-1)\text{5}]$ $=1150$ Sum of first 20 tems is 1150.
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Find the sum of n terms of the A.P. whose $k^{th}$ terms is 5k + 1.
AnswerHere, $\text{a}_\text{k}=5\text{k}+1$ $\text{a}_1=5+1=6$ $\text{a}_2=5(2)+1=11$ $\text{a}_3=5(3)+1=16$ $\text{d}=11-6=16-11=5$ $\text{s}_\text{n}=\frac{\text{n}}{2}[2\text{a}+(\text{n}-1)\text{d}]$ $=\frac{\text{n}}{2}[2(6)+(\text{n}-1)(5)]$ $=\frac{\text{n}}{2}[12+5\text{n}-5]$ $\text{s}_\text{n}=\frac{\text{n}}{2}(5\text{n}+7)$
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An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find 32nd term.
AnswerGiven, $\text{n}=60$ $\text{a}=7$ $\text{l}=125$ $\therefore\text{a}+(\text{n}-1)\text{d}=125$ $7+(59)\text{d}=125$ $\text{d}=2$ $\therefore\text{a}_{32}=\text{a}+(32-1)\text{d}$ $=7+(31)2$ $=67$
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Find the sum of all integers between 100 and 550, wgich are divisible by 9.
AnswerThe 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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The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term.
AnswerGiven: $\text{a}_{10}=41=\text{a}+9\text{d}\ .....(1)$ $\text{a}_{18}=73=\text{a}+17\text{d}\ .....(2)$ Solving (1) and (2) $\text{a}+9\text{d}=41$ $\text{a}+17\text{d}=73$ we get $\text{a}=5$ and $\text{d}=4$ $\therefore\text{a}_{26}=\text{a}+(26-1)\text{d}$ $=5+25(4)$ $=105$ 26th term od the given A.P. is 105.
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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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The first term of an A.P. is 2 and the last term is 50. The sum of all these terms is 442. Find the common difference.
AnswerGiven, $\text{a}=2$ $\text{l}=50$ $\therefore\text{l}=\text{a}+(\text{n}-1)\text{d}$ $50=2+(\text{n}-1)\text{d}$ $(\text{n}-1)\text{d}=48\ .....(1)$ $s_n$ of all n terms is given 442 $\therefore\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a}+\text{l}]$ $442=\frac{\text{n}}{2}[2+50]$ or $\text{n}=17\ .....{(2)}$ From (1) and (2) $\text{d}=\frac{48}{\text{n}-1}=\frac{48}{16}=3$ The common difference is 3.
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Solve:25 + 22 + 19 + 16 + ... + x = 115
Answer25 + 22 + 19 + 16 + ... + x = 115 Here, sum of the given series of say n terms is 115 So, the $n^{th}$ term = x Here, $\text{a}=25$ and $\text{d}=22-25=-3$ $\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}$ $\Rightarrow\text{x}=25-3(\text{n}-1)$ $\Rightarrow\text{x}=28-3\text{n}\ ...(1)$ The sum of n terms $\text{s}_{\text{n}}=\frac{\text{n}}{2}[\text{a}+\text{l}]$ $\Rightarrow115=\frac{\text{n}}{2}[25+28-3\text{n}]$ $\Rightarrow230=53\text{n}-3\text{n}^2$ $\Rightarrow3\text{n}^2-53\text{n}-3\text{n}^2$ $\Rightarrow3\text{n}^2-30\text{n}-23\text{n}-230=0$ $\Rightarrow\text{n}=10$ or $\frac{23}{3}$ But n can't be function $\therefore\text{n}=10\ .....(2)$ From (1) and (2) $\text{x}=28-3\text{n}$ $=28-3(10)$ $=-2$ $\text{x}=-2$
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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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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 $\text{s}_\text{p}=\text{p}^3.$
AnswerLet 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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Solve:1 + 4 + 7 + 10 + ... + x = 590.
Answer1 + 4 + 7 + 10 + ... + x = 590. Here, $\text{a}=1$ $\text{d}=4-1=3$ Let there be n terms so the $n^{th}$ term = x $\Rightarrow\text{x}=1+(\text{n}-1)3$ $[\therefore\text{a}_\text{n}=\text{a}+(\text{n}-1)\text{d}]$ $\Rightarrow\text{x}=3\text{n}-2\ .....(1)$ and $\text{s}_{\text{n}}=590$ [given] $\Rightarrow\frac{\text{n}}{2}[\text{a}+\text{l}]=590$ $\Rightarrow\frac{\text{n}}{2}[1+3\text{n}-2]$ $[\because\text{l}=\text{x}=\text{3n}-2]$ $\Rightarrow3\text{n}^2-\text{n}-1080=0$ $\Rightarrow3\text{n}^2-60\text{n}+59(\text{n}-20)=0$ $\Rightarrow3\text{n}(\text{n}-20)+59(\text{n}-20)=0$ $\Rightarrow\text{n}=2 .....(2)$ from(1) and (2) $\text{x}=3\text{n}-2$ $=3(20)-2$ $=58$ $\text{x}=58$
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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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The number of terms of an A.P. is even; the sum of odd terms is 24, of the even terms is 30, and the last term exceeds the first by $10\frac{1}{2},$ find the number of terms and the series.
AnswerLet no. of term be 2n Odd terms sum $=24=\text{T}_1+\text{T}_3+...+\text{T}_{2\text{n}-1}$ Even terms sum $=30=\text{T}_2+\text{T}_4+...+\text{T}_{2\text{n}}$ Subtract above two equtions $\text{nd}=6$ $\text{T}_{2\text{n}}=\text{T}_1+\frac{21}{2}$ $\text{T}_{2\text{n}}-\text{a}=\frac{21}{2}$ $(2\text{n}-1)\text{d}=\frac{21}{2}$ $12-\frac{21}{2}=\text{d}=\frac{3}{2}$ $\Rightarrow\text{n}=6\times\frac{2}{3}=4$ Total terms $=2\text{n}=8$ Subtite above values in equation of Sum of even terms or add terms, we get $\text{a}=\frac{3}{2}$ So series is $\frac{3}{2},3\frac{9}{2}......$
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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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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?
AnswerLet 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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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.
AnswerGiven: $10\text{a}_{10}=15\text{a}_{15}$ $10(\text{a}+(10-1)\text{d})=15(\text{a}+(15-1)\text{d})$ $10\text{a}+90\text{d}=15\text{a}+210\text{d}$ $5\text{a}+120\text{d}=0$ $\text{a}+24\text{d}=0\ .....(1)$ $\text{a}_{25}=\text{a}+(25-1)\text{d}$ $=\text{a}+24\text{d}$ $=0\ [\because\text{from}(1)\text{a}+24\text{d}=0]$ Hence proved.
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Find an A.P. in which the sum of any number of terms is always three times the squared number of these terms.
AnswerHere, $\text{s}_\text{n}=3\text{n}^2\ .....(1)$ [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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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), $\text{s}_1:\text{s}_2=(2\text{n}+1)(\text{a}+\text{nd}):(\text{n}+1)(\text{a}+\text{nd})$
$\text{s}_1:\text{s}_2=(2\text{n}+1);(\text{n}+1)$
View full question & answer→Question 323 Marks
The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.
AnswerLet 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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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.
AnswerThe 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.
View full question & answer→Question 343 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.
AnswerGiven: $\text{a}_{\text{m+1}}=2\text{a}_{\text{n+1}}$ $\Rightarrow\text{a}+(\text{m}+1-1)\text{d}=2(\text{a}+(\text{n}+1-1)\text{d})$ $\Rightarrow\text{a}+\text{md}=\text{2a}+2\text{nd}$ $\Rightarrow\text{a}(\text{m}2-\text{n})\text{d}\ .....(1)$ 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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Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
AnswerLet 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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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.
AnswerGiven, $\text{a}_3=7=\text{a}+2\text{d}\ .....{(1)}$ $\text{a}_7=3\text{a}_3+2$ $\therefore\text{a}_7=3(7)+2$ $[\because\text{a}_3=7]$ $=23=\text{a}+6\text{d}\ .....{}(2)$ 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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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.
View full question & answer→Question 383 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?
AnswerSuppose 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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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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Is 68 a them of the A.P. 7, 10, 13, ...?
AnswerIs 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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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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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?
AnswerThe 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 $=₹\ 600000-405000$
$=105000$
View full question & answer→Question 433 Marks
How many numbers are there between 1 and 1000 which when divided by 7 leave remainder 4?
AnswerA 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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Insert six A.M.s between $15$ and $−13$.
AnswerLet $A_1, A_2, A_3, A_4, A_5, A_6$, 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, $-13=15+7 d \Rightarrow d-4=-4 A_1=15+d=15+(-4)=11 A_2=15+2 d=15+(-8)=7 A_3=15+3 d=$ $15+(-12)=3 A_4=15+4 d=15+(-16)=-1 A_5=15+5 d=15+(-20)=-5 A_6=15+6 d=15+(-24)=-9$ The 6 A.M.S between 15 and -13 are $11,7,3,-1,-5$ and -9
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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.
AnswerLet 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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Find the $r^{th}$ term of an A.P., the sum of whose first n terme is $3n^2 + 2n$.
AnswerSum 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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If the sum of a certain number of terms of the AP 25, 22, 19, ... is 116. Find the last term.
AnswerSum 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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Find the sum of all those integers between $100$ and $800$ each of which on division by $16$ leaves the remainder $7$.
AnswerThe 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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Which term of the sequence $12+8\text{i},\ 6\text{i},\ 10+4\text{i},\ ...$ is (a) real (b) purely imaginary?
AnswerThe 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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Find the A.M. between: 7 and 13
Answer7 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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The sums of n terms of two arithmetic progressions are in the ratio $5n + 4 : 9n + 6$. Find the ratio of their 18th terms.
AnswerLet 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 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 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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The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.
AnswerLet 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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Find the sum of all even integers between $101$ and $999$.
AnswerAll 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 $=\text{s}_\text{n}\frac{\text{n}}{2}[\text{a}+\text{l}]$
$=\frac{449}{2}[102+998]$
$=\frac{449}{2}[102+998]$
$=449\times550$
$=346950$
View full question & answer→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.
AnswerLet 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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If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.
AnswerLet 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,
View full question & answer→Question 563 Marks
How many terms of the A.P. $-6,\ -\frac{11}{2},\ 5,\ ...$ are needed to give the sum -25?
AnswerLet 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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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.
AnswerThere 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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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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