\(\text { In } \triangle A P C \text { and } \Delta B P D \text { , }\)
\(\angle A P C=\angle B P D\) [vertically opposite angles]
and \(\angle P A C=\angle P B D\) [alternate angles]
\(\therefore \Delta A P C \sim \Delta B P D\) [by AA similarity criterion]
\(\Rightarrow \frac{A P}{P B}=\frac{A C}{B D}=\frac{P C}{P D}\) ......(i)
[since, corresponding sides of similar triangles are proportional]
\(\text { In } \Delta A P E \text { and } \Delta B P F\)
\(\angle A P E=\angle B P F\) [vertically opposite angles]
and \(\angle P A E=\angle P B F\) [alternate angles]
\(\therefore \Delta A P E \sim \Delta B P F\) [by AA similarity criterion]
\(\Rightarrow \frac{A P}{P B}=\frac{A E}{B F}=\frac{P E}{P F}\) ......(ii)
[since, corresponding sides of similar triangles are proportional]
\(\text { In } \Delta P E C \text { and } \Delta P F D, \angle E P C=\angle F P D\) [vertically opposite angles]
and \(\angle P C E=\angle P D F\) [alternate angles]
\( \therefore \Delta P E C \sim \Delta P F D \) [by AA similarity criterion]
\(\Rightarrow \frac{P E}{P F}=\frac{P C}{P D}=\frac{E C}{F D}\) ........(iii)
[since, corresponding sides of similar triangles are proportional]
From Eqs, (i), (ii) and (iii), we get
\(\frac{A P}{P B}=\frac{A C}{B D}=\frac{A E}{B F}=\frac{P E}{P F}=\frac{E C}{F D}\)
\(\therefore \frac{A E}{B F}=\frac{A C}{B D}=\frac{C E}{F D}\)

n³-n = n n²-1)
= n (n+1)(n-1)
=( n-1) n(n+1)
= product of threeconsecutive positive integers
Now, we ave to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer a is of the form 3q, 3q + 1 or 3q + 2 for some integer q.
Let a, a + 1, a + 2 be any three consecutive integers.
Case I: if  a=3q
a(a + l)(a + 2) = 3q(3q + 1)(3q + 2)
 = 3q (2r)
= 6qr, which is divisible by 6.
(∵ Product of two consecutive integers (3q + 1) and (3q + 2) is an even integer, say 2r)
Case II: If   a=3q + 1
∴ a(a + l)(a + 2) = (3q + 1)(3q + 2)(3q + 3)
= (2r) (3)(q + 1)
= 6r(q + 1),
which is divisible by 6
Case III: If    a=3q + 2
ஃ a(a + l)(a + 2) = (3q + 2)(3q + 3)(3q + 4)
= multiple of 6 for ever
= 6r (say),
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.

In \(\triangle PDQ\), we have
PQ² = PD² + QD²
PD² = PQ² - QD²
= a² - c² .... (i)

In right \(\triangle PDR\), we have
PR² = PD² + DR²
PD² = PR² - DR² = b² - d²
From Eqs.(i) and (ii)
a² - c² = b² - d² \(\Rightarrow \)  a² - b² = c² - d²

By BPT,
PQ : QR : RS = AB : BC : CD = 6 : 9 : 12
Let PQ = 6x, QR = 9x and RS = 12x
Since, length of PS = 36km
\(\therefore \)  PQ + QR + RS = 36
\(\Rightarrow \)  6x + 9x + 12x = 36
\(\Rightarrow \)  27x = 36  \(\Rightarrow \) x = \(\frac{4}{3}\)
PQ = 8 cm, QR = 12 cm and RS = 16 cm

Yes, 16 : 1 ; unity of nation, fraternity and patriotism.

Rs.12000 in scheme A and Rs.10000 in scheme B

10 km/h, 40 km/h

Plot the lines as follows (0, 0), (4,4), (6,2)

The given system of equations is
\(\frac { 2xy }{ x+y } =\frac { 3 }{ 2 } \) and \(\frac { xy }{ 2x-y } =\frac { -3 }{ 10 } \)
\(\Rightarrow\) \(\frac { x+y }{ 2xy } =\frac { 4 }{ 3 } \) and  \(\frac { 2x-y }{ xy } =\frac { -10 }{ 3 } \)
\(\frac { 1 }{ y } +\frac { 1 }{ x } =\frac { 4 }{ 3 } \) and \(\frac { 2 }{ y } -\frac { 1 }{ x } =\frac { -10 }{ 3 } \)
Put \(\frac { 1 }{ x } =u\)  and \(\frac { 1 }{ y } =v\)  then the system of equations becomes u+v=\(\frac { 4 }{3 } \) and  -u+2v=\(\frac { -10 }{3 } \)
Now, solve these equation.
x=\(\frac {1}{2}\) and y=\(\frac {-3}{2}\)

Recall the identity a² - b² = (a - b)(a + b). Using it, we can write:
x² - 3 = (x - \(\sqrt{3}\))(x + \(\sqrt{3}\))
So, the value of x² - 3 is zero when x = \(\sqrt{3}\) or x = -\(\sqrt{3}\)
Therefore, the zeroes of x² - 3 are \(\sqrt{3}\) and -\(\sqrt{3}\)
Now,
sum of zeroes = \(\sqrt{3}\) - \(\sqrt{3}\) = 0 = \(\frac{-(Coefficient \quad of \quad x)}{Coefficient \quad of \quad x^{2}}\)
product of zeroes = (\(\sqrt{3}\))(-\(\sqrt{3}\)) = -3 = \(\frac{-3}{1}=\frac{Constant \quad term}{Coefficient \quad of \quad x^{2}}\)

Factors of 1 to 10 numbers
1 = 1
2 = 1 x 2
3 = 1 x 3
4 = 1 x 2 x 2
5 = 1 x 5
6 = 1 x 2 x 3
7 = 1 x 7
8 = 1 x 2 x 2 x 2
9 = 1 x 3 x 3
10 = 1 x 2 x 5
\(\therefore \)  LCM of numbers 1 to 10
= LCM (1, 2, 3, 4, 5, 6, 7, 8, 9, 10)
= 2 x 2 x 2 x 3 x 3 x 5 x 7 = 2520

\(a={ x }^{ 3 }{ y }^{ 2 }=x\times x\times x\times y\times y\) and \(b={ x }{ y }^{ 3 }=x\times y\times y\times y\)
HCF (a,b) = xy²

Let a be any positive integer.
We know that, any odd positive integer is of the form 2q + 1, where q is a whole number.
\(\therefore \)  a = 2q + 1
\(\Rightarrow \)  a² = (2q + 1)²           [squaring both sides]
\(\Rightarrow \)   a² = 4q(q + 1) + 1            ...(i)
Note that q(q + 1) is either '0' or even, for any whole number q.
So, let q(q + 1) = 2m where m is a whole number.
From Eq.(i), we get  a² = 4(2m) + 1 = 8m + 1

No, because by Euclid's division lemma, 
we have, a = 4q + r,  \(0\le r<4\)             ...(i)
a can be in the form 4q, 4q + 1, 4q + 2 or 4q + 3

Let the four consecutive number in AP are
a - 3d, a - d, a + d, a + 3d
Then, we have
(a - 3d) + (a - d) + (a + d) + (a + 3d)
⇒ 4a = 32
⇒ a = 8
Also, it is given that the ratio of the product of first and the last terms to the product of the two middle terms is 7:15, therefore we have
 \(\frac { (a-3d)(a+3d) }{ (a-d)(a+d) } =\frac { 7 }{ 15 } \ \)
\(\ \Rightarrow \ \frac { a^{ 2 }-9d^{ 2 } }{ a^{ 2 }-d^{ 2 } } =\frac { 7 }{ 15 } \Rightarrow 15a^{ 2 }-135d^{ 2 }=7a^{ 2 }-7d^{ 2 }\ \)
\(\ \Rightarrow 8a^{ 2 }-128d^{ 2 }\Rightarrow a^{ 2 }=16d^{ 2 }\)
\(\\ \Rightarrow \ 64=16d^{ 2 }\quad \quad \ d \left[ \because \quad a=8 \right] \)
\(\\ \Rightarrow \ d^{ 2 }=4\Rightarrow d=\pm 2\)
Hence, the number are 2, 6, 10, 14 or 14, 10, 6, 2.

8 n-3; 5,13,21,...

a₁₈ + a₁₉ + a₂₀ = 225
a₃₅ + a₃₆ + a₃₇ = 429
3, 7, 11, 15, ...

Here, first instalment, a = Rs 1000
and increases the instalment every month, d = Rs 100
Number of instalments, n  = 30
Then, list of numbers is 
 \(1000,(1000+100),(1000+2\times100),(1000+3\times100)...\) 
i.e. 1000, 1100, 1200, 1300,... which is an AP. 
Amount paid in 30th instalment,
 \({a}_{30}=1000+(30-1)100 \quad[\because {a}_{n}=a+(n-1)d]\) 
 \(= 1000 + 29 \times 100\) 
= 1000 + 2900 = Rs 3900
Amount paid in 30 instalments,
  \({S}_{30}=\frac{30}{2}[2 \times 1000+(30-1)100]\) 
   \([\because {S}_{n}=\frac {n}{2}=\{2a+(n-1)d\}]\) 
 \(= 15[2000+29 \times 100]=15[2000+2900]\) 
= 15[4900 ]= Rs 7500
Hence, amount of loan still, he has to pay
 = Rs 118000 - Rs 73500 = Rs 4500

(i) \((i)\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } \)
(ii)\((i)-\frac { 1 }{ 2 } ,2\)

45 km/h.

Since, Jaspal Singh repays his loan of Rs.118000, with first instalment of Rs 1000 and increases each instalment by Rs 100.
\(\therefore \) His instalments are Rs 1000, Rs 1100, Rs 1200, Rs 1300, ... which forms an A.P.
Here, first term is Rs 1000 and common difference is Rs 100.
\(\therefore \) 30th instalment = a₃₀ = a + 29d
 = Rs (1000 + 29 \(\times \) 100)
= Rs (1000 + 2900) = Rs 3900
Amount paid in 30 instalments = S₃₀
\(\Rightarrow \) S₃₀ =  Rs \(\quad \frac { 30 }{ 2 } (2\times 1000+29\times 100)\)
 \(=\ Rs\ 15 (2000+2900)\\ =\ Rs\ 15(4900)\ =\ Rs\ 73500\)
Amount of loan still have to pay
 =  Rs (118000 - 73500)
 =  Rs 44500 

A.T.Q., S₅ + S₇ = 167
\(\Rightarrow\) \({5\over2}[2a+14d]+{7\over2}[2a+6d]=167\)
\(\Rightarrow\) 5 ( a + 2d ) + 7 ( a + 3d ) = 167
\(\Rightarrow\)  12a + 31d = 167                   ...(i)
and  S₁₀ = 235
\(\Rightarrow\)  \({10\over2}[2a+9d]=235\)
\(\Rightarrow\)  2a + 9d = \({235\over4}=47\)                ...(ii)
Multiplying equation (ii) by 6 and then subtractigfrom (i), we have
12a + 31d = 167
12a + 54d = 282
-     -          -      
     -23d = -115
d= \({-115\over-23}=5\)
\(\therefore\)  From (ii), 2a + 9d = 47
\(\Rightarrow\)   2a + 9 x 5 = 47
\(\Rightarrow\)    2a = 47 - 45
\(\Rightarrow\)   2a = 2 \(\Rightarrow\) a = 1
Hence, S₂₀ = \({20\over2}[2a+19d]\)
= 10 [ 2 x 1 + 19x 5 ]
= 10 [ 2 + 95 ] = 10 x 97 = 970

Here     a = 1, d = 1 and n = 31
S_n = \({n\over2}[2a+(n-1)d]\)
\(={31\over2}[2\times1+(31-1)]\)
\(={31\over2}[2+30]={31\over2}\times 32\)
= 496
\(\therefore\)    Piggy bank amount =  Rs 496
Amount spent = Rs 204
Amount left = Rs 100
Total pocket money = Rs 800

\(\because\)         S₆ = 36 and S₁₆ = 256.
\(\Rightarrow\)   S₆ = \({6\over2}\) [ 2a + 5d ]
   [ \(\because\) S_n = \({n\over2} [ 2a + ( n - 1)d ]\)
\(\Rightarrow\)    S₆ = 3 ( 2a + 5d )
\(\Rightarrow\)    \({36\over3}\) = 2a + 5d 
\(\Rightarrow\)    12 = 2a + 5d                       ...(i)
and   S₁₆ = \({16\over2}[2a+15d]\)
\(\Rightarrow\)       \({256\over8}=2a+15d\)
\(\Rightarrow\)   32 = 2a + 15d                     ...(ii)
Subtracting (i) and (ii), 2n + 5d = 12
2a + 15d = 32
-    -           -    
     -10d = -20   \(\Rightarrow\)  d = 2
\(\therefore\)  From (i), 12 = 2a + 5(2)
12 - 10 = 2a  \(\Rightarrow\)  2a = 2 \(\Rightarrow\) a = 1
Hence   S₁₀ = \({10\over2}[2a+9d]\)
= 5 ( 2 x 1 +  9 x 2  )
= 5 ( 2 + 18 )
\(\Rightarrow\)   S₁₀ = 5 x 20 = 100.

Let present age of Asha be x years
and present age of Nisha be y years  
ATQ   x=y² + 2   
Difference in ages = (x years)
 Mother's age after (x — y) years is 
x+(x-y) = 10y-1
\(\Rightarrow \) 2x-y-10y+1=0
\(\Rightarrow \) 2(y²+2)-11y+1=0
\(\Rightarrow \) 2y²-+4-11y+5=0
\(\Rightarrow \) 2y²-11y+5=0
\(\Rightarrow \)2y²-10y-y+5=0
\(\Rightarrow \) (y-5) (2y-1) =0
\(\Rightarrow \) y=5 or y = \(\frac { 1 }{ 2 } \) [rejecting]
Nisha's present age = 5 years
Asha's present age = 5² + 2 = 27 years.