It is given that the gap between two consecutive rungs is 25 cm and the top and bottom rungs are 2.5 metre i.e., 250 cm apart.
{tex}\therefore{/tex}  Number of rungs = {tex}\frac { 250 } { 25 } {/tex}+ 1 =  10 + 1 = 11.
It is given that the rungs are decreasing uniformly in length from 45 cm at the bottom to 25 cm at the top.
Therefore, lengths of the rungs form an A.P. with first term a = 45 cm and 11^th term l = 25 cm. n = 11
{tex}\therefore{/tex}  Length of the wood required for rungs = Sum of 11 terms of an A.P. with first term 45 cm and last term is 25 cm
{tex}= \frac { 11 } { 2 }{/tex} ( 45 + 25 ) cm {tex}\left[ \because S _ { n } = \frac { n } { 2 } ( a + l ) \right]{/tex}
{tex}= \frac { 11 } { 2 }{/tex}(70) cm
= 11 (35) cm
= 385 cm
Length of the wood required for rungs = {tex}\frac{385}{100}{/tex} = 3.85 metres ({tex}\because{/tex}100 cm = 1 m)
The length of the wood required for the rungs is 3.85 metres.

Let the number be (3q + r)
{tex}n = 3 q + r \quad 0 \leq r < 3{/tex}
{tex}\text { or } 3 q , 3 q + 1,3 q + 2{/tex}
{tex}\text { If } n = 3 q \text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}
{tex}3 q \text { is divisible by } 3{/tex}.
{tex}\text { If } n = 3 q + 1 \text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}
{tex}( 3 q + 3 ) \text { is divisible by } 3{/tex}.
{tex}\text { If } n = 3 q + 2 \text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}
{tex}( 3 q + 6 ) \text { is divisible by } 3{/tex}.
{tex}\therefore \text { out of } n , ( n + 2 ) \text { and } ( n + 4 ) \text { only one is divisible by } 3{/tex}.

According to the question, {tex}\triangle ABC {/tex} is an equilateral triangle.
In {tex}\triangle{/tex}ABD, using Pythagoras theorem,

{tex}\Rightarrow{/tex} AB² = AD² + BD²
{tex}\Rightarrow{/tex} BC² = AD² + BD², (as AB = BC = CA)
{tex}\Rightarrow{/tex} (2 BD)² = AD² + BD², (perpendicular is the median in an equilateral triangle)
{tex}\Rightarrow{/tex} 4BD² - BD² = AD²
{tex}\therefore{/tex} 3BD² = AD²

Let A(6, -1) and B(2,3) be the given point and P(x,y) be the required point.

Since point P is equidistant from points A and B,therefore,
PA = PB {tex}\Rightarrow{/tex} (PA)² = (PB)²
{tex}\Rightarrow{/tex}(x - 6)² + (y + 1)² = (2 - x)² + (3 - y)²
{tex}\Rightarrow{/tex} 36 + x² - 12x + y² + 1 +2y = 4 + x² - 4x + 9 + y² - 6y
{tex}\Rightarrow{/tex} -12x + 4x + 2y + 6y = 4 + 9 - 1 - 36
{tex}\Rightarrow{/tex} -8x + 8y = -24
-8 (x - y)=-24
{tex}\Rightarrow{/tex} x - y = 3
Hence, x - y = 3

Given:-In fig., {tex}\Delta{/tex}ABC D is the mid-point of BC and E is the mid-point of AD.

To prove:- BE : EX = 3 : 1
Construction: Through D, draw DF parallel to BX.
Proof:- In {tex}\Delta{/tex}AEX and {tex}\Delta{/tex}ADF
{tex}\angle EAX = \angle DAF{/tex} [Common]
{tex}\angle AXE = \angle AFD{/tex} [Corresponding angles]
Then, {tex}\Delta{/tex}AEX ~ {tex}\Delta{/tex}ADF [By AA similarity]
{tex}\therefore \frac{{EX}}{{DF}} = \frac{{AE}}{{AD}}{/tex} [Corresponding parts of similar {tex}Triangle{/tex}are proportional]
{tex}\Rightarrow \frac{{EX}}{{DF}} = \frac{{AE}}{{2AE}}{/tex} [AE = ED given]
{tex}\Rightarrow{/tex} DF = 2EX ....(i)
In {tex}\Delta{/tex}CDF ~ {tex}\Delta{/tex}CBX
{tex}\angle C = \angle C{/tex} [Common]
{tex}\angle CDF = \angle CBX{/tex} [Corresponding angles]
Then, {tex}\Delta{/tex}CDF ~ {tex}\Delta{/tex}CBX [By AA similarity] 

Therefore ,{tex} \frac{{CD}}{{CB}} = \frac{{DF}}{{BX}}{/tex} [Corresponding parts of similar {tex}\Delta{/tex}are proportional]
{tex}\Rightarrow \frac{1}{2} = \frac{{DF}}{{BE + EX}}{/tex} [BD = DC given]
{tex}\Rightarrow{/tex} BE + EX = 2DF
{tex}\Rightarrow{/tex} BE + EX = 4EX [By using (i)]
{tex}\Rightarrow{/tex} BE = 4EX - EX
{tex}\Rightarrow{/tex} BE = 3EX{tex}{/tex}
{tex}\Rightarrow \frac{{BE}}{{EX}} = \frac{3}{1}.{/tex}

Let n be an arbitrary positive integer.
On dividing n by 3, let q be the quotient and r be the remainder.
Then, by Euclid's division lemma, we have
n = 3q + r, where {tex}0 \leq r < 3{/tex}.
The possibilities of remainder = 0,1 or 2
n² = (3q + r)² [∵ (a + b) ² = a² + 2ab + b²]
{tex}\therefore{/tex} n² = 9q² + r² + 6qr ... ....(i), where {tex}0 \leq r < 3{/tex}.
Case I When r = 0.
Putting r = 0 in (i), we get
n² = 9q²
= 3(3q²)
n² =  3m, where m = 3q² is an integer.
Case II When r = 1.
Putting r = 1 in (i), we get
n² = (9q² + 1 + 6 q)
= 3(3q² + 2q) + 1
n²=  3 m + 1, where m = (3q² + 2q) is an integer.
Case lll When r = 2.
Putting r = 2 in (i), we get
n² = (9q² + 4 + 12q)
= 3(3q² + 4q + 1) + 1
n²= 3m + 1, where m = (3q² + 4q + 1) is an integer.
From all the above cases it is clear that the square of any positive integer is of the form 3m or (3m + 1) for some integer m.

Let A(-2, -3) and B(5, 6) be the given points.

1. Suppose x-axis divides AB in the ratio k:1 at point P
   Then, the coordinates of the point of division are
   {tex}P \left[ \frac { 5 k - 2 } { k + 1 } , \frac { 6 k - 3 } { k + 1 } \right]{/tex}
   Since P lies on x-axis, and y-coordinates of every point on x-axis is zero.
   {tex}\therefore \quad \frac { 6 k - 3 } { k + 1 } = 0{/tex}
   {tex}\Rightarrow{/tex} 6k - 3=0
   {tex}\Rightarrow{/tex} 6k = 3
   {tex}\Rightarrow{/tex} {tex}k = \frac { 3 } { 6 } \Rightarrow k = \frac { 1 } { 2 }{/tex}
   Hence, the required ratio is 1:2.
   Putting {tex}K = \frac { 1 } { 2 }{/tex}in the coordinates of P.
   We find that its coordinates are {tex}\left( \frac { 1 } { 3 } , 0 \right).{/tex}
2. Suppose y-axis divides AB in the ratio k:1 at point Q.
   Then, the coordinates of the point of division are
   {tex}Q \left[ \frac { 5 k - 2 } { k + 1 } , \frac { 6 k - 3 } { k + 1 } \right]{/tex}
   Since, Q lies on y-axis, and x-coordinates of every point on y-axis is zero.
   {tex}\therefore \quad \frac { 5 k - 2 } { k + 1 } = 0{/tex}
   {tex}\Rightarrow{/tex} 5k - 2 = 0​​​​​​​
   {tex}\Rightarrow \quad k = \frac { 2 } { 5 }{/tex}
   Hence, the required ratio is {tex}\frac { 2 } { 5 } : 1 = 2 : 5{/tex}
   Putting {tex}K = \frac { 2 } { 5 }{/tex}in the coordinates of Q.
   We find that the coordinates are {tex}\left( 0 , \frac { - 3 } { 7 } \right){/tex}​​​​​​​