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

Since the mosaic is in the shape of an equilateral triangle of 12 ft, and the tile is in the shape of an equilateral triangle of 12 inch (1 ft), there will be 12 rows in the mosaic.

From the figure, it is clear that number of white tiles in each row are 1, 2, 3, 4, …, 12 which clearly forms an Arithmetic Progression.
Similarly the number of blue tiles in each row are 0, 1, 2, 3, …, 11 which is also an Arithmetic Progression.
Number of white tiles = 1 + 2 + 3 +....+12 = \(\frac { 12 }{ 2 } \left[ 1+12 \right] \) = 78
Number of blue tiles = 0 + 1+ 2 + 3 +...+\(\frac { 12 }{ 2 } \left[ 0+11 \right] \) = 66
The total number of tiles in the mosaic  = 78 + 66 = 144

The natural numbers between 300 and 600 which are divisible by 7 are 301, 308, 315, …, 595.
The sum of all natural numbers between 300 and 600 is 301 + 308 + 315 +...+ 595
The terms of the above series are in A.P.
First term a = 301; common difference d = 7; Last term l = 595.
\(n=\left( \frac { l-a }{ d } \right) +1=\left( \frac { 595-301 }{ 7 } \right) +1=43\)
Since, \({ S }_{ n }=\frac { n }{ 2 } \left[ a+l \right] \), we have \({ s }_{43 }=\frac { 43 }{ 2 } \left[ 301+595 \right] \) = 19264 

Given the 13^th term = 3 so, t₁₃ = a + 12d = 3..... (1)
Sum of first 13 terms = 234 gives \(\frac { 13 }{ 2 } \) [2a + 12d] = 234
2a + 12 = 36...(2)
Solving (1) and (2) we get , a = 33, d = \(\frac { -5 }{ 2 } \)
Therefore, common difference is \(\frac { -5 }{ 2 } \).
Sum of first 21 terms S₂₁ = \(\frac { 21 }{ 2 } \left[ 2\times 33+\left( 21-1 \right) \times \left( -\frac { 5 }{ 2 } \right) \right] =\frac { 21 }{ 2 }\)[66 - 50] = 168.

Here we have to find the value of m, such that S_n = 190.
Sum of first n terms of an A.P
S_n = \(\frac { n }{ 2 } \)[2a + (n - 1)d] = 190
\(\frac { n }{ 2 } \)[2 x 1 + (n - 1) x 4 ] = 190
n [4n - 2] = 380
2n² -n - 190 = 0
(n - 10) (2n + 19) = 0
But n = 10 as n = \(\frac { 19 }{ 2 } \) is impossible , therefore, n = 10

Here the value of n is not given. But the last term is given. From this, we can find the value of n.
Given a = 0.40 and l = 1, we find d = 0.43 - 0.40 = 0.03
Therefore, n = \(\left( \frac { l-a }{ d } \right) +1\)
= \(\left( \frac { 1-0.40 }{ 0.03 } \right) +1=21\)
Sum of first n terms of an A.P S_n = \(\frac { n }{ 2 } \left[ a+l \right] \)
Here, n = 21. Therefore, S₂₁ = \(\frac { 21 }{ 2 } \left[ 0.40+1 \right] =14.7\)
So, the sum of 21 term of the given series is 14,7.

Let the first month earning be
t1 = Rs. 15,000
Increase per year d = Rs. 1,500
Expenses for first year =  Rs. 13,000
Expenses increase per year = Rs. 900
Saving for every year will be
(15000 - 13000), (16500 = 13900), (18000 - 14800),...
i.e. 2000, 2600, 3200, ...
Since t₂ - t₁ - t₃ - t₂, this sequence form an A.P
a = 2000, d = 2600 - 2000 = 600
Let the n^th years saving be Rs 20,000 then
a + (n - 1)d = t_n
2000 + (n - 1) (600) = 20,000
(n - 1) 600 = 20,000 - 2000
(n - 1) 600 = 18000
\(n-1=\frac{18000}{600}=30\)
n = 30 + 1 = 31
To save Rs. 20000, it takes 31 years.

Given the temperatures are in A.P.
Let the temperatures of Monday, Tuesday, Wednesday, Thursday and Friday be t₁, t₂, t₃, t₄ and t₅ respectively
Given t₁ + t₂ + t₃ = 0^oC
a + a + d + a + 2d = 0^oC
3a + 3d = 0
And t₃ + t₄ + t₅ = 18^oC
a + 2d + a + 3d + a + 4d = 18^oC
3a + 9d = 18^oC            .........(2)
Subtracting (1) from (2)
6d = 18^oC
d = 3^oC
Put d = 3^oC in (1)
3a + 3 x 3 = 0
3a = -9
\(a=\frac{-9}{3}=-3^{\circ} \mathrm{C}\)
t₁ = -3^oC, t₂ = (-3) + 3 = 0^oC,
t₃ = -3 + 2(3) = 3^oC,
t₄ = -3 + 3 (3) = 6^oC , t₅ = - 3 + 4(3) = 9^oC.
Five days temperatures are -3^oC, 0^oC, 3^oC, 6^oC, 9^oC

Given in an A.P. 6^th term : 8^th term = 7 : 9
a+ (6 - 1)d : a + (8 - 1)d = 7 : 9
a + 5d : a + 7d = 7 : 9
Product of the extremes = product of the means.
9(a + 5d) = 7(a + 7d)
9a + 45d = 7a + 49d.
9a - 7a =.49 d - 45d
2a = 4d
a = 2d
To find the ratio of 9^th term : 13^th term
a + (9 - 1) d : a + (13 - 1) d = a + 8d  : a + 12 d
= 2d + 8d : 2d + 17 d
= 10 d : 14 d
= 5 : 7 
The ratio of 9^th term to 13^th term is 5 : 7.

Let the three consecutive terms be a - d, a, a + d
Given their sum is 27
(a - d) + a + (a + d) = 27
a - d + a + a + d = 27
3a = 27
\(a=\frac{27}{3}=9\)
Product = 288
(a - d) a (a + d) = 288
a(a² - d²) = 288
9(9² - d²) = 288
\(81-d^{2}=\frac{288}{9}\)
81 - d² = 32
d² = 81 - 32 = 49
d x d = 7 x 7
d = \(\pm\)7
(i) a = 9, d = 7, The three terms are,
= 9 -7, 9, 9 + 7
= 2,9, 16
(ii) a = 9, d = -7, The three terms are
9-(-7), 9, 9-7 \(\Rightarrow\) 16,9,2
The required three consecutive terms of the A.P are 2,9,16.

Let the amount received by the three children be in the form of A.P. is given by
a - d, a, a + d, Since, Sum of the amount is Rs. 207, we have
(a - d) + a + (a + d) = 207
3a = 207 gives a = 69
It is given that product of the two least amounts is 4623.
(a - d)a = 4623 
(69 - d)69 = 4623
d = 2
Therefore, amount given by the mother to her three children are
Rs. (69 - 2), Rs. 69, Rs. (69 + 2). That is Rs. 67, Rs. 69 and Rs. 71.   

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

Starting time from Chennai = 23.30 hrs
Travelling time = 11 hrs
Here we use modulo 24.
Reaching time = 23.30 + 11 (mod 24)
= 34.30 (mod 24)
= 10.30 (mod 24)
Since 11 = 0 x 24 + 11
It reaches London on Monday at 10.30 a.m
Chennai time = 4.30 hrs + London time
London time = Chennai time - 4.30 a.m
= 10.30 - 4.30 = 6 a.m
The flight will land at London Airport on Monday at 5 a.m.

First take 2⁵ ≡ 15 (mod 17)
(25)² ≡ 15² (mod 17)
≡ 4 (mod 17)
2¹⁰ ≡ 4 (mod 17)
(2¹⁰)⁴ ≡ 4⁴ (mod 17)  
2⁴⁰ ≡ 1 mod 17
(2⁴⁰)² ≡ 1² mod 17
2.2⁸⁰ ≡ 2 x 1 (mod 17)
2⁸¹ ≡ 2 mod 17
Remainder when 2⁸¹ is divided by 17 is 2.

7 is divisible by 7
7 ≡ 0 (mod 7)
7 + 2 ≡ 0 + 2 (mod 7)
[Adding the constant 2]
9 ≡ 2 (mod 7)
9ⁿ ≡ 2ⁿ (mod 7)
6 x 9ⁿ ≡ 6 x 2ⁿ (mod 7)
[multiplying by the constant 6]
2ⁿ + 6 x 9ⁿ ≡ 2ⁿ + 6 x 2ⁿ (mod 7)
≡ 2ⁿ (1 + 6) (mod 7)
≡ 2ⁿ x 7 (mod 7)
≡ 2ⁿ x 0 (mod 7)
≡ 0 (mod 7)
When 2ⁿ + 6 x 9ⁿ is divided by 7 we get remainder 0,
Hence 2ⁿ + 6 x 9ⁿ is always divisible by 7 for any positive integer n.

Since 70000 is divisible by 7
70000 \(\equiv \) 0 (mod 7)
70000 + 4 \(\equiv \) 0 + 4 (mod 7)
70004 \(\equiv \)  4 (mod 7)
Therefore, the remainder when 70004 is divided 7 is 4 
Since 777 is divisible by 7
777 \(\equiv \) 0 (mod 7)
777 + 1 \(\equiv \) 0 + 1 (mod 7)
778 \(\equiv \) 1 (mod 7)
Therefore, the remainder when 778 is divided by 7 is 1.

The required number is the L.C.M. of first ten natural numbers
i.e., L.C.M. of (1,2,3,4,5,6,7,8,9,10)

L.C.M. is 5 x 2 x 3 x 2x7 x 2 x 3 = 2520
The least number divisible by first ten natural numbers is 2520.

The required number = L.C.M. of  (35, 56, 91) + 7
LCM of 35, 56, 91 = 7 x 5 x 8 x 3
= 3640
Required number = 3647 + 7
= 3647

L.C.M.= 3 x 2 x 2 x 2 x 5 x 3 = 360
Greatest number of 6 digit is 999999
L.C.M. of 24,15 and 36 = 360

On dividing 999999 by 360 remainder obtained is 279.
Greatest number of 6 digit, divisible by 24,15 and 36 = 999999 - 279 = 999720
Hence the required number is = 999720

By fundamental. theorem, every composite number can be expressed as a product of primes.

Factorizing 408 and 170 we get
408 = 2³ x 3¹ x 17¹
170 = 2¹ x 5¹ x 17¹
H.C.F. of 408 and 170 = 2¹ x 17¹ = 34
Also we know that H.C.F. x L.C.M.
= product of two numbers 
34 x L.C.M. = 408 x 170
L.C.M = \(\frac{408 \times 170}{34}=2040\)
H.C.F. (408, 170) = 34; L.C.M. (408, 170) = 2040