Extra amount he pays in installments
= Rs. 45750 - Rs. 40000
= 5750
He pays Rs. 5750 extra by installments

5² + 10² + 15² +...+ 105² = 52(1² + 2² + 3² +...+ 21²) 
= \(25\times \frac { 25\times \left( 21+1 \right) \left( 2\times 21+1 \right) }{ 6 } \)
= \(\frac { 25\times 21\times 22\times 43 }{ 6 } =82775\)

Let the amount he saved in the first year be x
Then x + (x + 100) + (x + 200) + ... 10 terms
= 16500
x+ x + 100 + x + 200 + ... 10 terms = 15500
(x + x + ... 10 terms) + (100 + 200 + ... 9 terms)
= 16500
\(10 x+\frac{9}{2}[2(100)+8(100)]=16500\)
10x = 16500 - 4500
10x = 12000
\(x=\frac{12000}{10}=1200\)
His 1^st year saving = Rs 1200

cow's milk = 175 litres
Buffalo's = 105 litres
The types of cans are of equal capacity
Capacity of a can = H.C.F. of 105, 175
By Euclid's division Algorithm
175 = 105 x 1 + 70
105 = 70 x 1 + 35
70.= 35 x 2 + 0
Remainder = 0
H.C.F. (105, 175) = 35
Capacity of a can = 35 litres
(ii) Number of cans of Cow's milk
\(=\frac{\text { Cow's Milk }}{\text { Capacity of a can }} \)
\(=\frac{175}{35}=5 \)
5 cans of cow's milk.
(iii) Number of cans of Buffalo's milk
\(=\frac{\text { Buffalo's milk }}{\text { Capacity of a can }} \)
\(=\frac{105}{35}=3 \)
3 cans buffalo's milk is there.

(i) (2³ - 1) + (4³ - 3³) + (6³ - 15³) + ...n terms
General term of the given series = (2n)3 - (2n - 1)3
= 8n³ - [(2n)³ - 3(2n)² (1) + 3(2n) (1) - 1³ ]
= 8n³ - [8n³ - 12n² + 6n - 1]
= 8n³ - 8n³ + 12n² - 6n + 1
= 12n² - 6n + 1
\(=12\left[\frac{n(n+1)(2 n+1)}{6}\right]-6\left[\frac{n(n+1)}{2}\right]+n\)
= 2n(n + 1)(2n + 1) - 3n(n + 1) + n
= 2n(2n² + n + 2n + l) - 3n² - 3n + n
= 4n³ + 6n² + 2n - 3n² - 3n + n
= 4n³ +3n²
Hence the sum of n terms = 4n³ + 3n²
(ii) (2³ - 1) + (4³ - 3³) + (6³ - 15³) + ....8 terms
Sum of n term = 4n³ + 3n²
Here n = 8
Sum = 4(8)³ + 3(8)²
= 2048 + 192 = 2240
Sum = 2240

Total amount saved in 6 years is S₆ = 7875
Since he saved half as much money as every year he saved in the previous year,
We have r = \(\frac { 1 }{ 2 } \) < 1
\(\frac { a\left( 1-{ r }^{ n } \right) }{ 1-r } =\frac { a\left( 1-\left( \frac { 1 }{ 2 } \right) ^{ 6 } \right) }{ 1-\frac { 1 }{ 2 } } =7875\)
\(\frac { a\left( 1-\frac { 1 }{ 64 } \right) }{ \frac { 1 }{ 2 } } \) = 7875 gives a x \(\frac { 63 }{ 32 } \) = 7875
a = \(a=\frac { 7875\times 32 }{ 63 } \) so, a = 4000
The amount saved in the first year is Rs.4000.

Given a, b, c are in A,P.
b - a =c - b
b + b = c + a
2b = c + a
Given x, y, z are three consecutive terms of a G.P.
x = a
y = ar
z = ar²
Now \(x^{b-c} \times y^{c-a} \times y^{a-b}=a^{b-c} \times(a r)^{c-a} \times\left(a r^{2}\right)^{a-b} \)
\(=a^{b-c} \times a^{c-a} \times r^{c-a} \times a^{a-b} r^{2(a-b)} \)
\(=a^{b-c+c-a+a-b} r^{c-a+2 a-2 b} \)
\(=a^{0} r^{(a+c)-2 b} \)
\(=1 . r^{2 b-2 b}[\text { from }(1) a+c=2 b]\)
= 1. ro
= 1 x 1 = 1
= RHS
\(x^{b-c} \times y^{c-a} \times z^{a-b}\) = 1
Hence proved.

Let the three terms be \(\frac{a}{r}\), a, ar
Given product = 27
\(\frac{a}{r}\) x a x ar = 27
a³ = 27
a³ = 3³
a = 3
Sum of the product taken two at a time = \(\frac { 57 }{ 2 } \)
\(\left(​​\frac{a}{r} \times a\right)+(a \times a r)+\left(a^{-1} \times a r\right)=\frac{57}{2} \)
\(\frac{a^{2}} {r }+a^{2} r+a^{2}=\frac{57}{2} \)
\(a^{2}\left(\frac{1}{r}+r+1\right)=\frac{57}{2} \)
\(3^{2}\left(\frac{1+r^{2}+r}{r}\right) =\frac{57}{2} \)
\(\frac{1+r^{2}+r}{r} =\frac{57}{2 \times 3 \times 3} \)
\(\frac{1+r^{2}+r}{r} =\frac{19}{6} \)
6 + 6r² + 6r = 19r
6r² + 6r - 19r + 6 = 0
6r² - 13r + 6 = 0
6r² - 9r - 4r + 6 = 0
3r(2r -3)- 2(2r - 3) = 0
(2r - 3) (3r - 2) = 0
2r - 3 = 0 (or) 3r - 2 = 0
2r = 3 (or) 3r = 2
\(r=\frac{3}{2}(\text { or })=\frac{2}{3}\)
\(\text { If } r=\frac{3}{2}, \text { the terms are } \frac{3}{3 / 2}, 3,3\left(\frac{3}{2}\right)=2,3, \frac{9}{2}\)
\(\text { If } r=\frac{2}{3} \text { the terms are } \frac{3}{2 / 3}, 3,3\left(\frac{2}{3}\right)=\frac{9}{2}, 3,2\)
 The three terms of G.P. are \(\frac{9}{2}, 3,2\)

Given sum of first n terms S_n = 2n² - 3n
sum of first term S₁ = 2 (1)² - 3 (1) = 2 - 3
= - 1 = t₁
Sum of first two terms S₂ = 2 (2)² - 3(2)
= 8 -  6 = 2
t₂ = S₂ - S₁
= 2 - (-1) = 2 + 1
t₂ = 3
Sum of first three terms S₃ = 2(3)² - 3(3)
= 18 - 9 = 9
t₃ = s₃ - s₂ = 9 - 2
t₃ = 7
Here t₂ - t₁ = 3 - (-1) = 3 + 1 = 4
t₃ - t₂ = 7 - 3 = 4
t₂ - t₁ = t₃ - t₃
The series is in A.P

1³ + 2³ + 3³ +...K³ = \(\left[\frac{k(k+1)}{2}\right]^{2}=44100=(210)^{2}\)
1 + 2 + 3 +...+ k = \(\frac{k(k+1)}{2}=210\)
1 + 2 + 3 +...+ k = 210