Statement
A straight line drawn parallel to a side of triangle intersecting the other two sides, divides the sides in the same ratio.
Proof

In \(\Delta ABC\) ,D is a point on AB and E is a point on AC
To prove : \(\cfrac { AD }{ DB } =\cfrac { AE }{ EC } \) 
Construction: Draw a line DE || BC

No.	Statement	Reason
1.	\(\angle ABC=\angle ADE=\angle 1\)	Corresponding angles are equal because DE || BC
2.	\(\angle ACB=\angle AED=\angle 2\)	Corresponding angles are equal because DE || BC
3.	\(\\ \angle DAE=\angle BAC=\angle 3\)	Both triangles have a common angle
4.	\(\Delta ABC\sim \Delta ADE\)	By AAA similarity
	\(\frac { AB }{ AD } =\frac { AC }{ CE } \)	Corresponding sides are proportional
	\(\frac { AD+DB }{ AD } =\frac { AE+EC }{ AE } \)	Split AB and AC using the points D and E.
	\(1+\frac { DB }{ AD } =1+\frac { EC }{ AE } \)	On simplification
	\(\frac { DB }{ AD } =\frac { EC }{ AE } \)	Cancelling 1 on both sides
	\(\frac { AD }{ DB } =\frac { AE }{ EC } \)	Taking reciprocals
		Hence proved

The series is neither Arithmetic nor Geometric series. So it can be split into two series and then find the sum.
5 + 55 + 555 + ... + n terms  = 5 [1 + 11 + 111 + ...n terms]
= \(\frac { 5 }{ 9 } \) [9 + 99 + 999 +...+ n terms]
= \(\frac { 5 }{ 9 } \)[(10 - 1) + (100 - 1) + (1000 - 1) +...+ n terms]
= \(\frac { 5 }{ 9 } \)[10 + 100 + 1000 +...+ n terms)-n]
= \(\frac { 5 }{ 9 } \left[ \frac { 10\left( { 10 }^{ n }-1 \right) }{ \left( 10-1 \right) } -n \right] =\frac { 50\left( { 10 }^{ n }-1 \right) }{ 81 } =\frac { 5n }{ 9 } \)

Given 113400 = \({ p }_{ 1 }^{ { x }_{ 1 } }\times { p }_{ 2 }^{ { x }_{ 2 } }\times { p }_{ 3 }^{ { x }_{ 3 } }\times { p }_{ 4 }^{ { x }_{ 4 } }\)

The number 113400 can be factorized as 113400 = 2³ x 3⁴ x 5² x 7¹ where 2,3,5,7 are primes in ascending order.
Comparing 
\({ p }_{ 1 }^{ { x }_{ 1 } }\times { p }_{ 2 }^{ { x }_{ 2 } }\times { p }_{ 3 }^{ { x }_{ 3 } }\times { p }_{ 4 }^{ { x }_{ 4 } }\) = 2³ x 3⁴ x 5² x 7¹
P₁ = 2; P₂ = 3; P₃ = 5; P₄ = 7
x₁ = 3; x₂ = 4; x₃ = 2; x₄ = 1.

To find HCF of three given numbers, first we have to find HCF of the first two numbers.
To find HCF of 396 and 504
Using Euclid’s division algorithm we get 504 = 396  x  1 + 108
The remainder is 108 \(\neq \) 0
Again applying Euclid’s division algorithm 396  = 108  x 3 + 72 
The remainder is 72 \(\neq \) 0
Again applying Euclid’s division algorithm 108  = 72 x 1 + 36 
The remainder is 36 \(\neq \) 0
Again applying Euclid division algorithm 72 = 36 x 2 + 0
Here the remainder is zero. Therefore HCF of 396 , 504 = 36, To find the HCF of 636 and 36
Using Euclid’s division algorithm we get 636 = 36 x 17 + 24
The remainder is 24 \(\neq \) 0
Again applying Euclid's division algorithm 36 = 24 x 1 + 12
The remainder is 12 \(\neq \) 0
Again applying Euclid's division algorithm 24 = 12 x 2 + 0
Here the remainder is zero. Therefore HCF of 636,36 = 12
Therefore Highest Common Factor of 396, 504 and 636 is 12. 

n^th term of G.P = ar^{n - 1}
Here a = 729
\(r =\frac{t_{2}}{t_{1}}=\frac{243}{729} \)
\(r =\frac{1}{3} \)
\(t_{7} =a r^{7-1}=a r^{6}=729\left(\frac{1}{3}\right)^{6} \)
\(=729 \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3} \)
t₇ = 1

Given the A.P. - 11, - 15, - 19,....
Here First term a = - 11
Common difference d = t₂ - t₁ = - 15 - ( - 11)
= -15 + 11
d = - 4
n^th term of an A.P. is t_n = a + (n - 1)d
19^th term (t₁₉) = -11 + (19 - 1) (-4)
= -11 + 18 (- 4)
= -11 + (- 72) = -83 
19th term of -11,-15, -19,...is - 83.

In \(\triangle\)ABC, AD is the bisector of \(\angle\)A
Therefore by Angle Bisector of \(\angle\)A
\(\frac { AB }{ AC } =\frac { BD }{ DC } \)
\(\frac{4}{3}=\frac{6}{A C}\) gives 4AC = 18. Hence, AC \(=\frac{9}{2}=4.5 \mathrm{~cm}\)

Since the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides, we have
\(\frac { Area(\Delta ABC) }{ Area(\Delta DEF) } =\frac { { BC }^{ 2 } }{ { EF }^{ 2 } } \) gives \(\frac { 54 }{ Area(\Delta DEF) } =\frac { { 3 }^{ 2 } }{ { 4 }^{ 2 } } \)
\(Area(\Delta DEF)=\frac { 16\times 54 }{ 9 } =96{ cm }^{ 2 }\)

The number 13824 can be factorized as
2^a x 3^b = 13824 = 2⁹ x 3³
a = 9 and b = 3.

The number 800 can be factorized as
800 = 2 x 2 x 2 x 2 x 2 x 5  x 5 = 2⁵ x 5²
Hence a^b x b^a = 2⁵ x 5²
This implies that a = 2 and b = 5  or a = 5 and b = 2

Given number of flower pots = 532
By Euclid's Division Lemma
a = bq + r where 0 < r < lbl
532 = 21 x 25 + 7
where 0 < r < l21l
Number of completed rows = 25
Number of flower pots left over = 7