(Each question 4 marks) - MATHS STD 11 Science Questions - Vidyadip

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19 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 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?

Answer

Total 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 $=₹\ 16680$

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Question 24 Marks

If a, b, c are in A.P., prove that: $\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$

Answer

If $\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$ 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.

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Question 34 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.

Answer

First 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 $\text{S}=\frac{20}{2}[2.300000+(20-1)10000]$ $=10[600000+190000]$ $=7900000$

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Question 44 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.

Answer

Total 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 $=₹\ 39100$

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Question 54 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.

Answer

To 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

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Question 64 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.

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Question 74 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.

Answer

If $\frac{1}{\text{a}},\frac{1}{\text{b}},\frac{1}{\text{c}}$ are in A.P $\frac{1}{\text{b}}-\frac{1}{\text{a}}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$ $\text{L.H.S}=\frac{1}{\text{b}}-\frac{1}{\text{a}}$ $=\frac{\text{a}-\text{b}}{\text{ab}}$ $=\frac{\text{a}(\text{a}-\text{b})}{\text{abc}}\ ...(\text{i})$ $\text{R.H.S}=\frac{1}{\text{c}}-\frac{1}{\text{b}}$ $=\frac{\text{a}(\text{b}-\text{c})}{\text{abc}}\ ...(\text{ii})$ 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

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Question 84 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?

Answer

Suppose 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.

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Question 94 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?

Answer

20 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 $=1240+1240$ $=2480\text{m}.$

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Question 104 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.

Answer

Let 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.

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Question 114 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 10th year.

Answer

Let the number of ratio manufactured increase by x each year and number of ratio manufacture in first year be a. So, A.P fromed 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 10th year

$\text{a}_{10}=\text{a}+9\text{d}$
$=550+9(25)=775$
775 Ratio's were manufatured in the 10th year.

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Question 124 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.

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Question 134 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?

Answer

A man accepts a position with an initial salary of ₹ 5200 per month. $\text{a}=5200$ Man w i 11 receive an au tom ati c increase of ₹ 320. $\text{d}=320$ Man's sa I ary 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.

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Question 144 Marks

Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

Answer

sum 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 $=\frac{22}{2}[26+21\times4]=11\times110=1210$

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Question 154 Marks

If a, b, c are in A.P., prove that: $(\text{a}-\text{c})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$

Answer

If $(\text{a}-\text{c})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$ 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 $(\text{a}-\text{b})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$

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Question 164 Marks

If a, b, c are in A.P., prove that: $\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$

Answer

If $\text{a}^3+\text{c}^3+6\text{abc}=8\text{c}^3$ 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. $\text{a}^3+\text{c}^3+6\text{abc}=8\text{b}^3$

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Question 174 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.

Answer

T.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

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Question 184 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 30th instalment?

Answer

In $1^{st}$ 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 = 100and common difference d = 5 Therefore at the $30^{th}$ insallment the amount he will pay $\text{T}_{30}=100+(30-1)(5)$ $=100+145$ $=245$

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Question 194 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.

Answer

x, 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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