Question 15 Marks
Show that $\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2$ and $\text{y}^2+\text{yz}+\text{z}^2$ are consecutive terms of an A.P., if x, y and z are in A.P.
Answerx, y and z are in A.P.
Let d be the common difference then,
$\text{y}=\text{x}+\text{d}$ and $\text{x}=\text{x}+2\text{d}$
To show $\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2+\text{zx}+\text{x}^2$ and consecutive terms of an A.P., it is enough to show that,
$(\text{z}^2+\text{zx}+\text{x}^2)-(\text{x}^2+\text{xy}+\text{y}^2)\\=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$\text{LHS}=(\text{z}^2+\text{zx}+\text{x}^2)-(\text{x}^2+\text{xy}+\text{y}^2)$
$(\text{z}^2+\text{zx}-\text{zy}-\text{y}^2)$
$=(\text{x}2\text{d})^2+(\text{x}+2\text{d})\text{x}-\text{x}(\text{x}+\text{d})-(\text{x}+\text{d})^2$
$=\text{x}^2+4\text{xd}+4\text{d}^2+\text{x}^2+2\text{xd}\\-\text{x}^2-\text{xd}-\text{x}^2-2\text{xd}-\text{d}^2$
$=3\text{xd}+\text{d}3^2$
$\text{RHS}=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$=(\text{x}+\text{d})^2+(\text{x}+\text{d})(\text{x}+2\text{d})-(\text{x}+2\text{d})\text{x}=\text{x}^2$
$=\text{x}^2+2\text{d}\text{x}+\text{d}^2+\text{x}^2+2\text{dx}\\+\text{xd}+2\text{d}^2-\text{x}^2-2\text{dx}-\text{x}^2$
$=3\text{xd}+3\text{d}^2$
$\therefore\text{LHS}=\text{RHS}$
$\therefore\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2$ and $\text{y}^2+\text{yz}+\text{z}^2$ are consecutive terms of an A.P.
View full question & answer→Question 25 Marks
A farmer buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalments of ₹ 500 plus 12% interest on the unpaid amount. How much the tractor cost him?
Question 35 Marks
If a, b, c are in A.P., prove that:
$\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$
AnswerIf $\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$
Then,
$\text{a}^2+\text{c}^2+2\text{ac}-2\text{ab}=2(\text{ab}+\text{bc}+\text{ca})$
or $(\text{a}+\text{b}+-\text{c})^2-\text{b}^2=0$ $[\therefore(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{ac}+2\text{bc}]$
or $\text{b}=\text{a}+\text{c}-\text{b}$
or $2\text{b}=\text{a}+\text{c}$
$\text{b}=\frac{\text{a}+\text{b}}{2}$
and since,
$\text{a},\text{b},\text{c}$ are in A.P
$\text{b}=\frac{\text{a}+\text{c}}{2}$
Thus, $\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$
Hence proved.
View full question & answer→Question 45 Marks
The income of a person is ₹ 300,000 in the first year and he receives an increase of ₹ 10000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.
AnswerFirst year the person income is: 300,000
Second year his income will be: 300,000 + 10,000 = 310,,000
Thin way he receives the amount after 20 years will be:
300,000 + 310,000 + ... + 490,000
This is an AP with first term a = 300000and common difference d 10,000
Therefore
$\text{S}=\frac{20}{2}[2.300000+(20-1)10000]$
$=10[600000+190000]$
$=7900000$
View full question & answer→Question 55 Marks
Shamshad Ali buys a scooter for ₹ 22000. He pays ₹ 4000 cash and agrees to pay the balance in annual instalments of ₹ 1000 plus 10% interest on the unpaid amount. How much the scooter will cost him.
AnswerTotal cost of scooter
$=₹\ 4000+\begin{bmatrix}\{₹\ 1000+\text{S.I. on}\ ₹\ 18000\ \text{for}\ 1\ \text{year}\}\\+\{₹\ 1000+\text{S.I.}\ \text{on}\ ₹\ 17000\ \text{for}\ 1\ \text{year}\}\\+\ ....+18\text{times}\end{bmatrix}$
$=(4000+18000)+\text{S,I}\ \text{for}\ 1\ \text{year}\ \text{on}\$18000+17000\ +...\ +18\text{times})$
$=22000+\text{S.I}.\ \text{for}\ 1\ \text{year}\ \text{on}\ \$18000+17000+\ ...\text{to}\ 18\ \text{times})$
$=22000+\text{S.I.}\ \text{for}\ 1\ \text{year}\ \text{on}\ \big\{\frac{18}{2}(18000+1000)\big\}$
$=22000+9(19000)\times\frac{10}{100}$
$=22000+17100$
$=₹\ 39100$
Total cost of Scooter $=₹\ 39100$
View full question & answer→Question 65 Marks
If $\text{a},\ \text{b},\ \text{c}$ are in A.P., then show that:
$\text{bc}-\text{a}^2,\ \text{ca}-\text{b}^2,\ \text{ab}-\text{c}^2$ are in A.P.
AnswerTo prove $\text{bc}-\text{a}^2,\ \text{ca}-\text{b}^2,\ \text{ab}-\text{c}^2$ are in A.P
$(\text{ca}-\text{b}^2)-(\text{bc}-\text{a}^2)=(\text{ab}-\text{c}^2)-(\text{ca}-\text{b}^2)$
$\text{LHS}=(\text{a}-\text{b}^2-\text{ca}+\text{a}^2)$
$=(\text{a}-\text{b})[\text{a}+\text{b}+\text{c}]\ ......(1)$
$\text{RHS}=\text{ab}-\text{c}^2-\text{ca}+\text{b}^2$
$=(\text{b}-\text{c})[\text{a}+\text{b}+\text{c}]\ .....(2)$
and since a, b, c are in ab
$\text{b}-\text{c}=\text{a}-\text{b}$
$\therefore\text{LHS}=\text{RHS}$
and
Thus, $\text{bc}-\text{a}^2,\ \text{ca}-\text{b}^2,\ \text{ab}-\text{c}^2$ are in A.P
View full question & answer→Question 75 Marks
If $\text{a},\ \text{b},\ \text{c}$ are in A.P., then show that:
$\text{a}^2(\text{b}+\text{c}),\ \text{b}^2(\text{c}+\text{a}),\ \text{c}^2(\text{a}+\text{b})$ are also in A.P.
Answer$\text{a}^2(\text{b}+\text{c}),\ \text{b}^2(\text{c}+\text{a}),\ \text{c}^2(\text{a}+\text{b})$ are in A.P.
If $\text{b}^2(\text{c}+\text{a})-\text{a}^2(\text{b}+\text{c})=\text{c}^2(\text{a}+\text{c})$
$\Rightarrow\text{b}^2\text{c}+\text{b}^2\text{a}-\text{a}^2\text{b}-\text{a}^2\text{c}=\text{c}^2\text{a}+\text{c}^2\text{b}-\text{b}^2\text{a}-\text{b}^2\text{c}$
Given, $\text{b}-\text{a}=\text{c}-\text{b}$ $[\text{a},\ \text{b},\ \text{c}$ are inA.P.$]$
$\text{c}(\text{b}^2-\text{a}^2)+\text{ab}(\text{b}-\text{a})=\text{a}(\text{c}^2-\text{b}^2)+\text{bc}(\text{c}-\text{d})$
$(\text{b}-\text{a})(\text{ab}+\text{bc}+\text{ca})=(\text{c}-\text{b})(\text{ab}+\text{bc}+\text{ca})$
Cancelling $\text{ab}+\text{bc}+\text{ca}$ from both sides
$\text{b}-\text{a}=\text{c}-\text{b}$
$2\text{b}=\text{c}+\text{a}$ which is true
Hence, $\text{a}^2(\text{b}+\text{c}),(\text{c}+\text{a})\text{b}^2$ and $\text{c}^2(\text{a}+\text{b})$ are also in A.P.
View full question & answer→Question 85 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:
$\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P.
AnswerIf $\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}$ are in A.P
$\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$\text{L.H.S}=\frac{1}{\text{b}}-\frac{1}{\text{a}}$
$=\frac{\text{a}-\text{b}}{\text{ab}}$
$=\frac{\text{a}(\text{a}-\text{b})}{\text{abc}}\ ...(\text{i})$
$\text{R.H.S}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$=\frac{\text{a}(\text{b}-\text{c})}{\text{abc}}\ ...(\text{ii})$
Since, $\frac{\text{b}+\text{c}}{\text{a}},\frac{\text{c}+\text{a}}{\text{b}},\frac{\text{a}+\text{b}}{\text{c}}$ are in A.P
$\frac{\text{b}+\text{c}}{\text{a}}-\frac{\text{c}+\text{a}}{\text{b}}=\frac{\text{c}+\text{a}}{\text{b}}-\frac{\text{a}+\text{b}}{\text{c}}$
$\frac{\text{b}^2+\text{cd}-\text{ac}-\text{a}^2}{\text{ab}}=\frac{\text{c}^2+\text{ac}-\text{ab}-\text{b}^2}{\text{bc}}$
$\Rightarrow\frac{(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})}{\text{ab}}=\frac{(\text{c}-\text{d})(\text{a}+\text{b}+\text{c})}{\text{bc}}$
or $\frac{\text{a}(\text{a}-\text{c})}{\text{abc}}=\frac{\text{c(a}-\text{b})}{\text{abc}}\ .....(3)$
From (1), (2) and (3)
$\text{LHS=RHS}$
Hence, $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P
View full question & answer→Question 95 Marks
A man saved ₹ 66000 in 20 years. In each succeeding year after the first year he saved ₹ 200 more than what he saved in the previous year. How much did he save in the first year?
AnswerSuppose the man saved ₹ x in the first year
$\text{a}_1=\text{x}$
In each succeeding year after the first year man saved ₹ 200 more then what he saved in the previous year.
$\text{d}=200$
Man saved ₹ 66000 in 20 years.
$\text{S}=66000$
$\frac{20}{2}[\text{a}_1+\text{a}_2+(20-1)200]=66000$
$\text{a}_1+1900=3300$
$\text{a}_1=1400$
Man saved ₹ 1400 in the first year.
View full question & answer→Question 105 Marks
In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Answer20 potatoes are placed in a line at intervals of 4 meters
$\therefore\text{n}=20$ and $\text{d}=4$
The first potato 24 meters from the starting point.
$\text{a}_1=24$
$\text{a}_2=\text{a}_1+\text{d}=24+8=32$
$\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$
$\text{a}_2=24+19\times24+76=100$
$\text{S}=\frac{20}{2}[\text{a}_1+\text{a}_2]=10[24+100]=1240$
As contestant is required to bring the potatos back to the starting point.
The distanced contestant would run
$=1240+1240$
$=2480\text{m}.$
View full question & answer→Question 115 Marks
A man arranges to pay off a debt of ₹ 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment.
AnswerLet the annual instalments from an ali thmetic series of common difaerenced and instalment a,
Then, series so firmed is
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})+\ ...\ +=3600$
or $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+\text{(n}-1\text{d}]$
or $3600=20[2\text{a}+39\text{d}]$
$2\text{a}+39\text{d}=180\ .....(1)$
and sum of first 30 terms is $\frac{2}{3}$ of 3600
$=2400$
$\Rightarrow2400=\frac{30}{2}[2\text{a}+(29)\text{d}]$
or $2\text{a}+29\text{d}=160\ .....(2)$
From (1) and (2)
$\text{a}=51$
The first installment paid by this man is ₹ 51.
View full question & answer→Question 125 Marks
A manufacturer of radio sets produced $600$ units in the third year and $700$ units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find:
- The production in the first year.
- The total product in $7$ years and
- The product in the $10^{th}$ year.
AnswerLet the number of ratio manufactured increase by $x$ each year and number of ratio manufacture in first year be a.
So, $\text{A.P}$ fromed $\text{ATQ}$ is,
$\text{a}, \text{a}+\text{x},\ \text{a}+2\text{x},\ ...$
Here,
$\text{a}_3=\text{a}+2\text{x}=600\ ...(1)$
$\text{a}_7=\text{a}+6\text{x}=700\ ...(2)$
From $(1)$ and $(2)$
$\text{a}=550, \text{x}=25$
- $550$ Ratio's were manufactured in the first year,
- The total produce in $7$ years is sum of produce in the first $7$ year.
$\text{S}_7=\frac{\text{7}}{\text{2}}[550+700]$ $\big[\because\text{S}_\text{n}[\text{a}+\text{l}]\big]$
$=4375$
$4375$ Ratio's were $m$ anufactured in first $7$ year.
- The produc in $10^{th}$ year
$\text{a}_{10}=\text{a}+9\text{d}$
$=550+9(25)=775$
$775$ Ratio's were manufatured in the $10^{th}$ year. View full question & answer→Question 135 Marks
If $\theta_1,\ \theta_2,\ \theta_3,\ ...\theta_\text{n}$ are in AP. whose common difference is d, show thet $\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+...+\sec\theta_{\text{n}-1}\sec\theta_\text{n}=\frac{\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$
Answer$\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+...+\sec\theta_{\text{n}-1}\sec\theta_\text{n}=\frac{\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$
$\theta_2-\theta_1=\theta_3-\theta_2=......=\text{d}$
$\sec\theta-1\sec\theta_2=\frac{1}{\cos\theta_1\cos\theta_2}=\frac{\sin\text{d}}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{\sin(\theta_2-\theta_1)}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{\sin\theta_2\cos\theta_1-\cos\theta_2\sin\theta_1}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{1}{\sin\text{d}}\Big[\frac{\sin\theta_2\cos\theta_1}{(\cos\theta_1\cos\theta_2)}-\frac{\cos\theta_2\sin\theta_1}{(\cos\theta_1\cos\theta_2)}\Big]$
$=\frac{1}{\sin\text{d}}[\tan\theta_2-\tan\theta_1]$
Similaely, $\sec\theta_2\sec\theta_3=\frac{1}{\sin\text{d}}[\tan\theta_3\tan\theta-2]$
If we add up all terms, we get
$=\frac{1}{\sin\text{d}}[\tan\theta_2-\tan\theta_1+\tan\theta_3-\tan\theta-2+......+\tan\theta_\text{n}=\tan\theta_{\text{n}-1}]$
$=\frac{1}{\sin\text{d}}[\tan\theta_\text{n}-\tan\theta_1]$
Hence proved.
View full question & answer→Question 145 Marks
A man accepts a position with an initial salary of $₹ 5200$ per month. It is understood that he will receive an automatic increase of $₹ 320$ in the very next month and each month thereafter.
- Find his salary for the tenth month.
- What is his total earnings during the first year?
AnswerA man accepts a position with an initial salary of $₹ 5200$ per month.
$\text{a}=5200$
Man $w$ is $11$ receive an automati $c$ increase of $₹ 320.$
$\text{d}=320$
Man's saIary for the $n"$ month is given by,
$\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$
Total earnig of the man for the first year
$=\frac{12}{2}[\text{a}_1+\text{a}_{12}]$
$=6[5200+5200+(12-1)320]$
$=83520$
Total earnlg of the man for the first year Is $₹ 83,520.$
View full question & answer→Question 155 Marks
Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Answersum of all two digit numbers which when divided by 4,
yields 1 as remainder,
$\Rightarrow$ all $4\text{n}+1$ terms with $\text{n}\geq3$
$\text{n}=22,\text{a}=13,\text{d}=4$
Sumof terms $=\frac{22}{2}[26+21\times4]=11\times110=1210$
View full question & answer→Question 165 Marks
If a, b, c are in A.P., prove that:
$(\text{a}-\text{c})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$
AnswerIf $(\text{a}-\text{c})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$
Then,
$\text{a}^2+\text{c}^2-2\text{ac}=4(\text{ab})-\text{b}^2-\text{ac}+\text{bc}$
$\Rightarrow\text{a}^2+\text{c}^24\text{b}^2+2\text{ac}-4\text{ac}-4\text{bc}=0$
$\Rightarrow(\text{a}+\text{c}-2\text{b})^2=0$ $\big[$ Using $(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{ac}+2\text{bc}\big]$
$\therefore\text{a}+\text{c}-2\text{b}=0$
or $\text{a}+\text{c}=2\text{b}$
and since,
$\text{a},\ \text{b},\ \text{c}$ are in A.P [Given]
$\text{a}+\text{c}=2\text{b}$
Hence proved
$(\text{a}-\text{b})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$
View full question & answer→Question 175 Marks
If a, b, c are in A.P., prove that:
$\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$
AnswerIf $\text{a}^3+\text{c}^3+6\text{abc}=8\text{c}^3$
or $\text{a}^3+\text{c}^3-(2\text{b})^3+6\text{abc}=0$
or $\text{a}^3+(-2\text{b})^3+\text{c}^3+3\times\text{a}\times(-2\text{b})\times\text{c}=0$
$\therefore(\text{a}-2\text{b}+\text{c})=0$ $\begin{bmatrix}\therefore\text{x}^3+\text{y}^3+\text{z}^3+3\text{xyz}=0\\\text{or if}\ \text{x}+\text{y}+\text{z}=0\end{bmatrix}$
or $\text{a}+\text{c}=2\text{b}$
$\text{a}-\text{b}=\text{c}-\text{b}$
and since, a, b, c are in A.P
Thus, $\text{a}-\text{b}=\text{c}-\text{d}$
Hence proved. $\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$
View full question & answer→Question 185 Marks
If a, b, c are in A.P., then show that:
$\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P.
AnswerT.P $\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P.
$\text{b}+\text{c}-\text{c},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P only if $(\text{c}+\text{a}-\text{b})-(\text{b}+\text{c}-\text{a})=(\text{a}+\text{b}-\text{c})-(\text{c}+\text{a}-\text{b})$
$\text{LHS}=(\text{c}+\text{a}-\text{b})-(\text{b}+\text{c}-\text{a})$
$\Rightarrow2\text{a}-2\text{b}\ .....(1)$
$\text{RHS}=(\text{a}+\text{b}-\text{c})-(\text{c}+\text{a}-\text{b})$
$\Rightarrow2\text{b}-\text{2c}\ .....(2)$
since,
$\text{a},\ \text{b},\ \text{c}$ are in A.P
$\therefore\text{b}-\text{a}=\text{c}-\text{b}$
or $\text{a}-\text{b}=\text{b}-\text{c}\ .....(3)$
From (1), (2) and (3)
LHS = RHS
Thus, given numbers
$\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P
View full question & answer→Question 195 Marks
A man starts repaying a loan as first instalment of $₹\ 100 = 00.$ If he increases the instalments by $₹\ 5$ every month, what amount he will pay in the $30$th instalment?
AnswerIn $1^{st}$ installment the man paid $100$ rupees.
In $2^{nd}$ installment the man paid $(100 + 5) = 105$
Likewise he pays to the 30th insallment as follows:
$100+105+\ ...\ +(100+5\times29)$
This is an AP with $a = 100$ and common difference $d = 5$
Therefore at the $30^{th}$ insallment the amount he will pay
$\text{T}_{30}=100+(30-1)(5)$
$=100+145$
$=245$
View full question & answer→Question 205 Marks
A farmer buys a used tractor for ₹ 12000. He pays ₹ 6000 cash and agrees to pay the balance in annual instalments of ₹ 500 plus 12% interest on the unpaid amount. How much the tractor cost him?
AnswerTotal cost of tractor
$=6000+[(500+12\%\ \text{of}\ 1\ \text{year})\\+(500+12\%\ \text{of}\ 5500\ 1\text{year})+\ .....\ +12\ \text{times}]$
$=6000+6000+\frac{12}{100}(6000+5500+\ .....\ +12\ \text{times})$
$=12000+\frac{12}{100}\big[\frac{12}{100}(6000+5000)\big]$
$=12000+\frac{12}{100}\times\frac{12}{2}\times6500$
$=12000+(72\times65)$
$=12000+4680$
$=16680$
Total cost of tractor $=₹\ 16680$
View full question & answer→Question 215 Marks
If a, b, c are in A.P., prove that:
$\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$
AnswerIf $\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$
Then,
$\text{a}^2+\text{c}^2+2\text{ac}-2\text{ab}=2(\text{ab}+\text{bc}+\text{ca})$
or $(\text{a}+\text{b}+-\text{c})^2-\text{b}^2=0$ $[\therefore(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{ac}+2\text{bc}]$
or $\text{b}=\text{a}+\text{c}-\text{b}$
or $2\text{b}=\text{a}+\text{c}$
$\text{b}=\frac{\text{a}+\text{b}}{2}$
and since,
$\text{a},\text{b},\text{c}$ are in A.P
$\text{b}=\frac{\text{a}+\text{c}}{2}$
Thus, $\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$
Hence proved.
View full question & answer→Question 225 Marks
The income of a person is ₹ 300,000 in the first year and he receives an increase of ₹ 10000 to his income per year for the next 19 years. Find the total amount, he received in 20 years.
AnswerFirst year the person income is: 300,000
Second year his income will be: 300,000 + 10,000 = 310,,000
Thin way he receives the amount after 20 years will be:
300,000 + 310,000 + ... + 490,000
This is an AP with first term a = 300000and common difference d 10,000
Therefore
$\text{S}=\frac{20}{2}[2.300000+(20-1)10000]$
$=10[600000+190000]$
$=7900000$
View full question & answer→Question 235 Marks
Shamshad Ali buys a scooter for ₹ 22000. He pays ₹ 4000 cash and agrees to pay the balance in annual instalments of ₹ 1000 plus 10% interest on the unpaid amount. How much the scooter will cost him.
AnswerTotal cost of scooter
$=₹\ 4000+\begin{bmatrix}\{₹\ 1000+\text{S.I. on}\ ₹\ 18000\ \text{for}\ 1\ \text{year}\}\\+\{₹\ 1000+\text{S.I.}\ \text{on}\ ₹\ 17000\ \text{for}\ 1\ \text{year}\}\\+\ ....+18\text{times}\end{bmatrix}$
$=(4000+18000)+\text{S,I}\ \text{for}\ 1\ \text{year}\ \text{on}\$18000+17000\ +...\ +18\text{times})$
$=22000+\text{S.I}.\ \text{for}\ 1\ \text{year}\ \text{on}\ \$18000+17000+\ ...\text{to}\ 18\ \text{times})$
$=22000+\text{S.I.}\ \text{for}\ 1\ \text{year}\ \text{on}\ \big\{\frac{18}{2}(18000+1000)\big\}$
$=22000+9(19000)\times\frac{10}{100}$
$=22000+17100$
$=₹\ 39100$
Total cost of Scooter $=₹\ 39100$
View full question & answer→Question 245 Marks
If $\text{a},\ \text{b},\ \text{c}$ are in A.P., then show that:
$\text{bc}-\text{a}^2,\ \text{ca}-\text{b}^2,\ \text{ab}-\text{c}^2$ are in A.P.
AnswerTo prove $\text{bc}-\text{a}^2,\ \text{ca}-\text{b}^2,\ \text{ab}-\text{c}^2$ are in A.P
$(\text{ca}-\text{b}^2)-(\text{bc}-\text{a}^2)=(\text{ab}-\text{c}^2)-(\text{ca}-\text{b}^2)$
$\text{LHS}=(\text{a}-\text{b}^2-\text{ca}+\text{a}^2)$
$=(\text{a}-\text{b})[\text{a}+\text{b}+\text{c}]\ ......(1)$
$\text{RHS}=\text{ab}-\text{c}^2-\text{ca}+\text{b}^2$
$=(\text{b}-\text{c})[\text{a}+\text{b}+\text{c}]\ .....(2)$
and since a, b, c are in ab
$\text{b}-\text{c}=\text{a}-\text{b}$
$\therefore\text{LHS}=\text{RHS}$
and
Thus, $\text{bc}-\text{a}^2,\ \text{ca}-\text{b}^2,\ \text{ab}-\text{c}^2$ are in A.P
View full question & answer→Question 255 Marks
If $\text{a},\ \text{b},\ \text{c}$ are in A.P., then show that:
$\text{a}^2(\text{b}+\text{c}),\ \text{b}^2(\text{c}+\text{a}),\ \text{c}^2(\text{a}+\text{b})$ are also in A.P.
Answer$\text{a}^2(\text{b}+\text{c}),\ \text{b}^2(\text{c}+\text{a}),\ \text{c}^2(\text{a}+\text{b})$ are in A.P.
If $\text{b}^2(\text{c}+\text{a})-\text{a}^2(\text{b}+\text{c})=\text{c}^2(\text{a}+\text{c})$
$\Rightarrow\text{b}^2\text{c}+\text{b}^2\text{a}-\text{a}^2\text{b}-\text{a}^2\text{c}=\text{c}^2\text{a}+\text{c}^2\text{b}-\text{b}^2\text{a}-\text{b}^2\text{c}$
Given, $\text{b}-\text{a}=\text{c}-\text{b}$ $[\text{a},\ \text{b},\ \text{c}$ are inA.P.$]$
$\text{c}(\text{b}^2-\text{a}^2)+\text{ab}(\text{b}-\text{a})=\text{a}(\text{c}^2-\text{b}^2)+\text{bc}(\text{c}-\text{d})$
$(\text{b}-\text{a})(\text{ab}+\text{bc}+\text{ca})=(\text{c}-\text{b})(\text{ab}+\text{bc}+\text{ca})$
Cancelling $\text{ab}+\text{bc}+\text{ca}$ from both sides
$\text{b}-\text{a}=\text{c}-\text{b}$
$2\text{b}=\text{c}+\text{a}$ which is true
Hence, $\text{a}^2(\text{b}+\text{c}),(\text{c}+\text{a})\text{b}^2$ and $\text{c}^2(\text{a}+\text{b})$ are also in A.P.
View full question & answer→Question 265 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:
$\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P.
AnswerIf $\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}$ are in A.P
$\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$\text{L.H.S}=\frac{1}{\text{b}}-\frac{1}{\text{a}}$
$=\frac{\text{a}-\text{b}}{\text{ab}}$
$=\frac{\text{a}(\text{a}-\text{b})}{\text{abc}}\ ...(\text{i})$
$\text{R.H.S}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$
$=\frac{\text{a}(\text{b}-\text{c})}{\text{abc}}\ ...(\text{ii})$
Since, $\frac{\text{b}+\text{c}}{\text{a}},\frac{\text{c}+\text{a}}{\text{b}},\frac{\text{a}+\text{b}}{\text{c}}$ are in A.P
$\frac{\text{b}+\text{c}}{\text{a}}-\frac{\text{c}+\text{a}}{\text{b}}=\frac{\text{c}+\text{a}}{\text{b}}-\frac{\text{a}+\text{b}}{\text{c}}$
$\frac{\text{b}^2+\text{cd}-\text{ac}-\text{a}^2}{\text{ab}}=\frac{\text{c}^2+\text{ac}-\text{ab}-\text{b}^2}{\text{bc}}$
$\Rightarrow\frac{(\text{b}-\text{a})(\text{a}+\text{b}+\text{c})}{\text{ab}}=\frac{(\text{c}-\text{d})(\text{a}+\text{b}+\text{c})}{\text{bc}}$
or $\frac{\text{a}(\text{a}-\text{c})}{\text{abc}}=\frac{\text{c(a}-\text{b})}{\text{abc}}\ .....(3)$
From (1), (2) and (3)
$\text{LHS=RHS}$
Hence, $\frac{1}{\text{a}},\ \frac{1}{\text{b}},\ \frac{1}{\text{c}}$ are in A.P
View full question & answer→Question 275 Marks
A man saved ₹ 66000 in 20 years. In each succeeding year after the first year he saved ₹ 200 more than what he saved in the previous year. How much did he save in the first year?
AnswerSuppose the man saved ₹ x in the first year
$\text{a}_1=\text{x}$
In each succeeding year after the first year man saved ₹ 200 more then what he saved in the previous year.
$\text{d}=200$
Man saved ₹ 66000 in 20 years.
$\text{S}=66000$
$\frac{20}{2}[\text{a}_1+\text{a}_2+(20-1)200]=66000$
$\text{a}_1+1900=3300$
$\text{a}_1=1400$
Man saved ₹ 1400 in the first year.
View full question & answer→Question 285 Marks
In a potato race 20 potatoes are placed in a line at intervals of 4 meters with the first potato 24 metres from the starting point. A contestant is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing back all the potatoes?
Answer20 potatoes are placed in a line at intervals of 4 meters
$\therefore\text{n}=20$ and $\text{d}=4$
The first potato 24 meters from the starting point.
$\text{a}_1=24$
$\text{a}_2=\text{a}_1+\text{d}=24+8=32$
$\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$
$\text{a}_2=24+19\times24+76=100$
$\text{S}=\frac{20}{2}[\text{a}_1+\text{a}_2]=10[24+100]=1240$
As contestant is required to bring the potatos back to the starting point.
The distanced contestant would run
$=1240+1240$
$=2480\text{m}.$
View full question & answer→Question 295 Marks
A man arranges to pay off a debt of ₹ 3600 by 40 annual instalments which form an arithmetic series. When 30 of the instalments are paid, he dies leaving one-third of the debt unpaid, find the value of the first instalment.
AnswerLet the annual instalments from an ali thmetic series of common difaerenced and instalment a,
Then, series so firmed is
$\text{a}+(\text{a}+\text{d})+(\text{a}+2\text{d})+\ ...\ +=3600$
or $\text{S}_\text{n}=\frac{\text{n}}{2}[2\text{a}+\text{(n}-1\text{d}]$
or $3600=20[2\text{a}+39\text{d}]$
$2\text{a}+39\text{d}=180\ .....(1)$
and sum of first 30 terms is $\frac{2}{3}$ of 3600
$=2400$
$\Rightarrow2400=\frac{30}{2}[2\text{a}+(29)\text{d}]$
or $2\text{a}+29\text{d}=160\ .....(2)$
From (1) and (2)
$\text{a}=51$
The first installment paid by this man is ₹ 51.
View full question & answer→Question 305 Marks
A manufacturer of radio sets produced $600$ units in the third year and $700$ units in the seventh year. Assuming that the product increases uniformly by a fixed number every year, find:
- The production in the first year.
- The total product in $7$ years and
- The product in the $10^{th}$ year.
AnswerLet the number of ratio manufactured increase by $x$ each year and number of ratio manufacture in first year be a.
So, $\text{A.P}$ fromed $\text{ATQ}$ is,
$\text{a}, \text{a}+\text{x},\ \text{a}+2\text{x},\ ...$
Here,
$\text{a}_3=\text{a}+2\text{x}=600\ ...(1)$
$\text{a}_7=\text{a}+6\text{x}=700\ ...(2)$
From $(1)$ and $(2)$
$\text{a}=550, \text{x}=25$
- $550$ Ratio's were manufactured in the first year,
- The total produce in $7$ years is sum of produce in the first $7$ year.
$\text{S}_7=\frac{\text{7}}{\text{2}}[550+700]$ $\big[\because\text{S}_\text{n}[\text{a}+\text{l}]\big]$
$=4375$
$4375$ Ratio's were $m$ anufactured in first $7$ year.
- The produc in $10^{th}$ year
$\text{a}_{10}=\text{a}+9\text{d}$
$=550+9(25)$
$=775$
$775$ Ratio's were manufatured in the $10^{th}$ year. View full question & answer→Question 315 Marks
If $\theta_1,\ \theta_2,\ \theta_3,\ ...\theta_\text{n}$ are in AP. whose common difference is d, show thet $\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+...+\sec\theta_{\text{n}-1}\sec\theta_\text{n}=\frac{\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$
Answer$\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+...+\sec\theta_{\text{n}-1}\sec\theta_\text{n}=\frac{\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$
$\theta_2-\theta_1=\theta_3-\theta_2=......=\text{d}$
$\sec\theta-1\sec\theta_2=\frac{1}{\cos\theta_1\cos\theta_2}=\frac{\sin\text{d}}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{\sin(\theta_2-\theta_1)}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{\sin\theta_2\cos\theta_1-\cos\theta_2\sin\theta_1}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{1}{\sin\text{d}}\Big[\frac{\sin\theta_2\cos\theta_1}{(\cos\theta_1\cos\theta_2)}-\frac{\cos\theta_2\sin\theta_1}{(\cos\theta_1\cos\theta_2)}\Big]$
$=\frac{1}{\sin\text{d}}[\tan\theta_2-\tan\theta_1]$
Similaely, $\sec\theta_2\sec\theta_3=\frac{1}{\sin\text{d}}[\tan\theta_3\tan\theta-2]$
If we add up all terms, we get
$=\frac{1}{\sin\text{d}}[\tan\theta_2-\tan\theta_1+\tan\theta_3-\tan\theta-2+......+\tan\theta_\text{n}=\tan\theta_{\text{n}-1}]$
$=\frac{1}{\sin\text{d}}[\tan\theta_\text{n}-\tan\theta_1]$
Hence proved.
View full question & answer→Question 325 Marks
A man accepts a position with an initial salary of $₹ 5200$ per month. It is understood that he will receive an automatic increase of $₹ 320$ in the very next month and each month thereafter.
- Find his salary for the tenth month.
- What is his total earnings during the first year?
AnswerA man accepts a position with an initial salary of $₹ 5200$ per month.
$\text{a}=5200$
Man $w$ is $11$ receive an automatic increase of $₹ 320.$
$\text{d}=320$
Man's saIary for the $n"$ month is given by,
$\text{a}_\text{n}=\text{a}_1+(\text{n}-1)\text{d}$
Total earnig of the man for the first year
$=\frac{12}{2}[\text{a}_1+\text{a}_{12}]$
$=6[5200+5200+(12-1)320]$
$=83520$
Total earnlg of the man for the first year Is $₹ 83,520.$
View full question & answer→Question 335 Marks
Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.
Answersum of all two digit numbers which when divided by 4,
yields 1 as remainder,
$\Rightarrow$ all $4\text{n}+1$ terms with $\text{n}\geq3$
$\text{n}=22,\text{a}=13,\text{d}=4$
Sumof terms $=\frac{22}{2}[26+21\times4]=11\times110=1210$
View full question & answer→Question 345 Marks
If a, b, c are in A.P., prove that:
$(\text{a}-\text{c})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$
AnswerIf $(\text{a}-\text{c})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$
Then,
$\text{a}^2+\text{c}^2-2\text{ac}=4(\text{ab})-\text{b}^2-\text{ac}+\text{bc}$
$\Rightarrow\text{a}^2+\text{c}^24\text{b}^2+2\text{ac}-4\text{ac}-4\text{bc}=0$
$\Rightarrow(\text{a}+\text{c}-2\text{b})^2=0$ $\big[$ Using $(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{ac}+2\text{bc}\big]$
$\therefore\text{a}+\text{c}-2\text{b}=0$
or $\text{a}+\text{c}=2\text{b}$
and since,
$\text{a},\ \text{b},\ \text{c}$ are in A.P [Given]
$\text{a}+\text{c}=2\text{b}$
Hence proved
$(\text{a}-\text{b})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$
View full question & answer→Question 355 Marks
If a, b, c are in A.P., prove that:
$\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$
AnswerIf $\text{a}^3+\text{c}^3+6\text{abc}=8\text{c}^3$
or $\text{a}^3+\text{c}^3-(2\text{b})^3+6\text{abc}=0$
or $\text{a}^3+(-2\text{b})^3+\text{c}^3+3\times\text{a}\times(-2\text{b})\times\text{c}=0$
$\therefore(\text{a}-2\text{b}+\text{c})=0$ $\begin{bmatrix}\therefore\text{x}^3+\text{y}^3+\text{z}^3+3\text{xyz}=0\\\text{or if}\ \text{x}+\text{y}+\text{z}=0\end{bmatrix}$
or $\text{a}+\text{c}=2\text{b}$
$\text{a}-\text{b}=\text{c}-\text{b}$
and since, a, b, c are in A.P
Thus, $\text{a}-\text{b}=\text{c}-\text{d}$
Hence proved. $\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$
View full question & answer→Question 365 Marks
If a, b, c are in A.P., then show that:
$\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P.
AnswerT.P $\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P.
$\text{b}+\text{c}-\text{c},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P only if $(\text{c}+\text{a}-\text{b})-(\text{b}+\text{c}-\text{a})=(\text{a}+\text{b}-\text{c})-(\text{c}+\text{a}-\text{b})$
$\text{LHS}=(\text{c}+\text{a}-\text{b})-(\text{b}+\text{c}-\text{a})$
$\Rightarrow2\text{a}-2\text{b}\ .....(1)$
$\text{RHS}=(\text{a}+\text{b}-\text{c})-(\text{c}+\text{a}-\text{b})$
$\Rightarrow2\text{b}-\text{2c}\ .....(2)$
since,
$\text{a},\ \text{b},\ \text{c}$ are in A.P
$\therefore\text{b}-\text{a}=\text{c}-\text{b}$
or $\text{a}-\text{b}=\text{b}-\text{c}\ .....(3)$
From (1), (2) and (3)
LHS = RHS
Thus, given numbers
$\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P
View full question & answer→Question 375 Marks
A man starts repaying a loan as first instalment of $₹ 100 = 00$. If he increases the instalments by $₹ 5$ every month, what amount he will pay in the $30\ th$ instalment?
AnswerIn $1^{st}$ installment the man paid $100$ rupees.
In $2^{nd}$ installment the man paid $(100 + 5) = 105$
Likewise he pays to the 30th insallment as follows:
$100+105+\ ...\ +(100+5\times29)$
This is an AP with $a = 100$ and common difference $d = 5$
Therefore at the $30^{th}$ insallment the amount he will pay
$\text{T}_{30}=100+(30-1)(5)$
$=100+145$
$=245$
View full question & answer→Question 385 Marks
Show that $\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2$ and $\text{y}^2+\text{yz}+\text{z}^2$ are consecutive terms of an A.P., if x, y and z are in A.P.
Answerx, y and z are in A.P.
Let d be the common difference then,
$\text{y}=\text{x}+\text{d}$ and $\text{x}=\text{x}+2\text{d}$
To show $\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2+\text{zx}+\text{x}^2$ and consecutive terms of an A.P., it is enough to show that,
$(\text{z}^2+\text{zx}+\text{x}^2)-(\text{x}^2+\text{xy}+\text{y}^2)\\=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$\text{LHS}=(\text{z}^2+\text{zx}+\text{x}^2)-(\text{x}^2+\text{xy}+\text{y}^2)$
$(\text{z}^2+\text{zx}-\text{zy}-\text{y}^2)$
$=(\text{x}2\text{d})^2+(\text{x}+2\text{d})\text{x}-\text{x}(\text{x}+\text{d})-(\text{x}+\text{d})^2$
$=\text{x}^2+4\text{xd}+4\text{d}^2+\text{x}^2+2\text{xd}\\-\text{x}^2-\text{xd}-\text{x}^2-2\text{xd}-\text{d}^2$
$=3\text{xd}+\text{d}3^2$
$\text{RHS}=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$=(\text{y}^2+\text{yz}+\text{z}^2)-(\text{z}^2+\text{zx}+\text{x}^2)$
$=(\text{x}+\text{d})^2+(\text{x}+\text{d})(\text{x}+2\text{d})-(\text{x}+2\text{d})\text{x}=\text{x}^2$
$=\text{x}^2+2\text{d}\text{x}+\text{d}^2+\text{x}^2+2\text{dx}\\+\text{xd}+2\text{d}^2-\text{x}^2-2\text{dx}-\text{x}^2$
$=3\text{xd}+3\text{d}^2$
$\therefore\text{LHS}=\text{RHS}$
$\therefore\text{x}^2+\text{xy}+\text{y}^2,\ \text{z}^2+\text{zx}+\text{x}^2$ and $\text{y}^2+\text{yz}+\text{z}^2$ are consecutive terms of an A.P.
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