\(\Delta A O F \sim \Delta B O D \Rightarrow \frac{O A}{O B}=\frac{F A}{D B}\) .......(i)
\(\Delta F A C \sim \Delta E B C \Rightarrow \frac{F A}{E B}=\frac{A C}{B C} \Rightarrow \frac{F A}{D B}=\frac{A C}{B C} \)  ......(ii) [∵ EB = DB]
From Eqs. (i) and (ii), we get
\(\frac{O A}{O B}=\frac{A C}{B C} \Rightarrow \frac{O A}{O B}=\frac{O C-O A}{O B-O C}\)
\(\Rightarrow (O B+O A) \cdot O C=2 O A \cdot O B\)
Divide both sides by OA. OB. OC.

Given ∠AEF = ∠AFE and E is the mid-point of CA.

To prove \(\frac{B D}{C D}=\frac{B F}{C E}\)
Construction Draw a line CG parallel to DF.
Proof Given, ∠AEF = ∠AFE and E is mid-point of CA.
\(\therefore C E=A E=\frac{A C}{2}\) ........(i)
In ΔBDF, CG II DF 
By basic proportionality theorem,
\(\frac{B D}{C D}=\frac{B F}{G F}\) .......(ii)
In ΔAFE, ∠AEF = ∠AFE [given]
⇒ AF = AE [since, sides opposite to equal angles are equal]
⇒ AF = AE = CE [from Eq. (i)] ... (iii)
In ΔACG, E is the mid-point of AC and EF II CG.
⇒ FG = AF [∵ AE = CEl ... (iv)
From Eqs. (ii), (iii) and (iv),
\(\frac{B D}{C D}=\frac{B F}{C E} [\because G F=A F=C E]\)

Cost of one pen = Rs.10, cost of one pencil box = Rs.15

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

p=-1, q=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=\(\frac {-3}{2}\)

By solving both equations. Find the values of x and y and then put these values in \(y=\lambda x+5\) to get required values of \(\lambda .\)
x=340, y=-165, \(\lambda =-\frac { 1 }{ 2 } \)

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 = \(\frac{-3}{1}=\frac{Constant \quad term}{Coefficient \quad of \quad x^{2}}\)

We have,
\(ax^{ 2 }+bx+c=(px^{ 3 }+qx^{ 2 }+rx+5).0+ax^{ 2 }+bx+c\)
By division algorithm, we get
Quotient=0 and remainder =ax²+bx+c

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

True, because the product of any two consecutive numbers, say n(n + 1) will always be even as one out of n or (n+1) must be even.

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,...

28

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 

Here, in L.H.S. of the given equation, we have
a = - 4 and d = - 1 - ( - 4 ) = - 1 + 4 = 3 and l = x
\(\therefore\)    - 4 + ( - 1 ) +  2 + ... + x = 437
\(\Rightarrow\)    \({n\over2}(-4+x)=437\)   [ \(\because\) S_n = \({n\over2}(a+l)\) ]
\(\Rightarrow\)     n ( - 4 + x ) = 874                  ...(i)
Also,  n ( - 4 + x ) = 874                  ...(ii)
   [ \(\because\) a_n = a + ( n - 1 )d ]
From (i) and (ii), we have
n ( - 4 - 4 + ( n - 1)d) = 874
\(\Rightarrow\)   - 8n + n ( n - 1 )3 = 874
\(\Rightarrow\)  - 8n + 3n² - 3n - 874 = 0  
\(\Rightarrow\)   3n² - 11n - 874 = 0
\(n={{{11\pm\sqrt{(-11)^{2}-4\times3\times(-874)}}}\over{2\times3}}\)
\(={{11\pm\sqrt{121+10488}}\over{5}}\)
\(={{11\pm103}\over{6}}={{11+103}\over{6}},{{11-103}\over{6}}\)
\(=19,{-{92}\over{6}}\) ( Rejecting )
n = 19
From(ii), we obtain
x = - 4 + ( 19 - 1 )3
x = - 4 + 54
x = 50

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.

n = 27
Middle most term = \(\frac { n+1 }{ 2 } =\frac { 28 }{ 2 } =14\)
t₁₃ + t₁₅ = 2 + 2 = 4
t₁₂ + t₁₆ = 4 + 4 = 8
t₁₁ + t₁₇ = 6 + 6 = 12
t₁₀ + t₁₈ = 8 + 8 = 16
t₁ + t₂₇ = 26 + 26 = 52
Hence AP becomes 4,8,12,16,......,52
a = 4, d = 4, a_n = 52, n = 13
\({ S }_{ 13 }=\frac { 13 }{ 2 } \left[ 4+52 \right] =364m\)
and distance covered to collect the books = 364 m
Total distance covered = 364 + 364 = 728 m
Maximum distance she travelled carrying a flag is 26 m

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.