\(\frac { 1 }{ 2 } ,\frac { 2 }{ 3 } ,\frac { 3 }{ 4 } \)
a₁ = \(\frac { 1 }{ 2 } \); a₂ = \(\frac { 2 }{ 3 } \); a₃ = \(\frac { 3 }{ 4 } \)
We see that the numerator of n^th term is n, and the denominator is one more than the numerator. Hence, \({ a }_{ n }=\frac { n }{ n+1 } \) n \(\in\) N

1 + 3 + 5 + ...40 terms = 40² = 1600

First term a = 3; common difference d = 6 - 3 = 3 ; last term l  = 111
We know that, n  = \(\left( \frac { l-a }{ d } \right) +1\)
n = \(\left( \frac { 111-3 }{ 3 } \right) +1\) = 37
Thus the A.P. contain 37 terms 

Starting from Tuesday we have to calculate the day after 45 days
The number for Tuesday is 2.
2 + 45 (mod 7) ≡ 47 (mod 7)
≡ 5 (mod 7)
Number 5 stands for Friday.
Uncle will be coming on Friday.

8x \(\equiv \) 1 (mod 11) can be written as 8x - 1 = 11k, for some integer k.
\(x=\frac { 11k+1 }{ 8 } \)
When we put k = 5, 13, 21,29 ,....then 11 k + 1 is divisible by 8
\(x=\frac { 11\times 5+1 }{ 8 } =7\)
\(x=\frac { 11\times 13+1 }{ 8 } =18\)
Therefore, the solutions are 7,18,29,40,.... 

Yes, the given number is a composite number, because
7 x 5 x 3 x 2 + 3 = 3 (7 x 5 x 2 + 1) = 3 x 71
Since the given number can be factorized in terms of two primes, it is a composite number.

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

We see that 6 boxes are required to pack 30 cakes with 4 cakes left over. This distribution of cakes can be understood as follows:

34	=	5	x	6	+	4
Total number of cakes	=	Number of cakes in each box	x	Number of boxes	+	Number of cakes left over
\(\downarrow\)		\(\downarrow\)		\(\downarrow\)		\(\downarrow\)
Dividend      \(\alpha\)	=	Divisor    \(\beta\)	x	Quotient     q	+	Remainder      r

9³ + 10³ +....21³ = (1³ + 2³ + 3³ + ...21³) - (1³ + 2³ + 3³ +...+ 8³)
= \(\left[ \frac { 21\times \left( 21+1 \right) }{ 2 } \right] ^{ 2 }=\left[ \frac { 8\times \left( 8+1 \right) }{ 2 } \right] ^{ 2 }\)= (231)² - (36)² = 52065

If S₁, S₂ and S₃ are sum of first n, 2n and 3n terms of an A.P respectively then 
\({ S }_{ 1 }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] ,{ S }_{ 2 }=\frac { 2n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] ,{ S }_{ 3 }=\frac { 3n }{ 2 } \left[ 2a+\left( 3n-1 \right) d \right]\)
Consider, S₂ - S₁ = \(\frac { 2n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] -\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \)
\(\frac { n }{ 2 } \left[ \left[ 4a+2\left( 2n-1 \right) d \right] -\left[ 2a+\left( n-1 \right) d \right] \right] \)
S₂ - S₁ = \(\frac { n }{ 2 } \left[ 2a+\left( 3n-1 \right) d \right] \)
3(S₁ - S₂) = \(\frac { n }{ 2 } \times \left[ 2a+\left( 3n-1 \right) d \right] \)
3(S₂ - S₁) = S₃

Let us take the four terms in the form (a - 3d), (a -d), (a + d) and (a + 3d).
Since sum of the four terms is 28,
a - 3d  + a - d + a + d + a + 3d = 28
4a = 28 gives a = 7
Similarly, since sum of their squares is 276,
(a - 3d)² + (a - d)² + (a + d)² + (a + 3d)² = 276
a² - 6ad + 9d² + a² - 2ad + d² + a² + 2ad  + d² + a² + 6ad  + 9d² = 276
4a² + 20d² = 276 \(\Rightarrow\) 4(7)² + 20d² = 276
d² = 4 gives d = \(\pm\)2
If d = 2 then the four numbers are 7 - 3(2), 7 - 2, 7 + 2, 7 + 3(2)
That is the four numbers are 1,5,9 and 13.
If a = 7, d = -2 then the four numbers are 13,9, 5 and 1
Therefore, the four consecutive terms of the  A.P are 1, 5, 9 and 13

Since the remainders are 4, 5 respectively the required number is the HCF of the number 445 - 4 = 441, 572 - 5 = 567.
567  = 441  x 1 + 126
441 = 126 x 3 + 63 
126 = 63 x 2 + 0
Therefore HCF of 441, 567 = 63 and so the required number is 63

(i) To check if h is one – one, we assume that h(b₁) = h(b₂)
Then we get 2.47 b₁ + 54.10 = 2.47 b₂ + 54.10
2.47b₁ = 2.47b₂ ⇒ b₁ = b₂
Thus, h(b₁) = h(b₂) ⇒ b₁ = b₂, So, the function h is one – one.
(ii) If the length of the thigh bone b = 50, then the height is
h(50) = (2.47 x 50) + 54.10 = 177.6 cms
(iii) If the height of a person is 147.96 cms, then h(b) = 147.96 and so the length of the thigh bone is given by 2.47b + 54.10 = 147.96.
b = \(\frac { 93.86 }{ 2.47 } \)
Therefore, the length of the thigh bone is 38 cms.