\(19,\frac{38}{3},\frac{76}{9},....\)

Given series is (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ......
\(\Rightarrow \frac{1}{(x-y)}[(x-y)(x+y)+(x-y)(x^2+xy+y^2)+(x-y)(x^3+x^2y+xy^2+y^3)+...]\)
\(\Rightarrow \frac{1}{(x-y)}[(x^2-y^2)+(x^3+y^3)+(x^4-y^4)+....\infty]\)
\(\Rightarrow \frac{1}{(x-y)}[(x^2+x^3+x^4+...\infty)-(y^2+y^3+y^4+....\infty)]\)
\(\Rightarrow \frac{1}{(x-y)}[\frac{x^2}{1-x}-\frac{y^2}{1-y}]\)
\(\Rightarrow \frac{1}{(x-y)}[\frac{x^2-x^2y-y^2+xy^2}{(1-x)(1-y)}]\)
\(\Rightarrow \frac{1}{(x-y)}[\frac{(x^2-y^2)-xy(x-y)}{(1-x)(1-y)}]\)
\(\Rightarrow\frac{x+y-xy}{(1-x)(1-y)}\)

Here a = 6 \(r=\frac { 1.2 }{ 6 } \) = \(\frac { 1 }{ 5 } \)
S\(\infty\) = \(\frac { a }{ 1-r } =\frac { 6 }{ 1-\frac { 1 }{ 5 } } =7.5\)

n = 8

It is given that a = \(\frac { b+c }{ 2 } \Rightarrow \)b+c =2a
AlsoG₁, G₂ are two G.M. between b and c. So b, G₁, G₂,c is aG.P. with common ratio n= \(\left( \frac { c }{ b } \right) ^{ \frac { 1 }{ 3 } }\)
\(\therefore \) G₁= br = b x\(\left( \frac { c }{ b } \right) ^{ \frac { 1 }{ 3 } }\)=\({ b }^{ \frac { 2 }{ 3 } }{ c }^{ \frac { 1 }{ 3 } }\)
G₂= br² = b x\(\left( \frac { c }{ b } \right) ^{ \frac { 1 }{ 3 } }\)=\({ b }^{ \frac { 1 }{ 3 } }{ c }^{ \frac { 2 }{ 3 } }\)
\(\therefore \) G³₁+G₂³=\(\left( { b }^{ \frac { 2 }{ 3 } }{ c }^{ \frac { 2 }{ 3 } } \right) ^{ 3 }\)+\(\left( { b }^{ \frac { 1 }{ 3 } }{ c }^{ \frac { 1 }{ 3 } } \right) ^{ 3 }\)=b²c+bc²
= bc(b + c) = bc x 2a = 2abc.

Side of the First Square =10 cm
Side of the Second Square = \(\sqrt { { 5 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 50 } =5\sqrt { 2 } cm\)
Side of the third square = \(\sqrt { \left( \frac { 5\sqrt { 2 } }{ 2 } \right) ^{ 2 }+\left( \frac { 5\sqrt { 2 } }{ 2 } \right) ^{ 2 } } =\sqrt { \frac { 50 }{ 4 } +\frac { 50 }{ 4 } } =5cm\)
Sum of areas of squares =  [10² + (5\(\sqrt { 2 } \))²]+5²+......]
= [10 + 50 + 25 + ...] = \(\frac { 100 }{ 1-\frac { 1 }{ 2 } } \) = 200 sq.cm

Given, depreciation in value of machine = 20%
After each year the value of the machine is 80 %
(100-20)% of its value of the previous year, so at the end of 5 yrs, the machine will depreciate as many times as 5.
Hence, we have to find the 6th terms of the G P whise first term a₁ is 1250 and common ratio r is 8.
Hence, value at the end 5 y6rs.
=t₆=a_rr⁵
=1250(0.8)⁵=409.6

\(\frac { { n }^{ 2 }-n+2 }{ 2 } \)

T_n = (nth term of 2, 4, 6....) x (nth term of 4, 6, 8,...)
= [ 2 + ( n - 1 ) 2 ] [4 + ( n - 1 ) 2 ]    [\(\because \) T_n = a + ( n - 1 ) d ]
= 2n ( 2n + 2 )
Ans. T₂₀ = 1680

In the given AP, let the first term = a and the common difference = d.
According to the question, a_m = \(\frac { 1 }{ n } \)  and a_n = \(\frac { 1 }{ m } \) 
\(\therefore \)  a + (m - 1)d = \({ 1 }/{ n }\) ......(i)
and   a + (n -1)d = \({ 1 }/{ m }\) .....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
(m - n)d = \(\left( \frac { 1 }{ n } -\frac { 1 }{ m } \right)\) \(\Longrightarrow \) d = \(\frac { 1 }{ mn } \) 
On putting d = \(\frac { 1 }{ mn } \) in Eq. (i), we get
\(a+\frac { (m-1) }{ mn } \) = \(\frac { 1 }{ n } \)  \(\Longrightarrow \) a = \(\left\{ \frac { 1 }{ n } -\frac { (m-1) }{ mn } \right\} \) = \(\frac { 1 }{ mn } \),
Now, mnth term = a + (mn - 1)d
= \(\left\{ \frac { 1 }{ n } -\frac { (mn-1) }{ mn } \right\} \) = \(\frac { mn }{ mn } \) = 1
Hence, the mnth term of the given P is 1

740

(i) b₇ = -48
b₁₃ = -168
(ii) \(a_{7}=\frac{49}{128}\)

Let the GP be A,AR,AR²,AR³....
Given , first term , 
A = a ....(i)
and nth term , AR^n-1 = b ....(ii)
Now , p = Product of n terms
p = A\(\times\)AR¹\(\times\)AR²\(\times\)AR³\(\times\).....\(\times\)n terms
p = A^{1+1+1+1+.....+n} terms R^{1+2+3+....(n-1)}
p = AⁿR \(\frac{n(n-1)}{2}\)
p2 = Aⁿ Aⁿ R^n(n-1) = Aⁿ (AR^n-1)n
p2 = aⁿbⁿ  [using Eqs.(i)  and (ii)]

173

Ref. fig.
(i) As we have studied in previous classes that the line joining the mid points of any two sides of a triangle is half the third side.
\(\therefore\) Sum ofthe perimeters of all the triangle is given by
S_p = 64 + 27 + 13.5 + ... \(\infty\)
= \(\frac{54}{1-\frac{1}{2}}=54\times 2 = 108\) cm.
(ii) We know that the area of the triangle formed by joining the mid points of the sides one-forth of the given triangle.
\(\therefore\) area of equilateral \(\Delta\)ABC
= \(\frac{\sqrt{3}}{4}\times (18)^2=81\sqrt{3}\) cm.
Area of \(\Delta\)DEF = \(\frac{1}{4}\times \) area of \(\Delta\)ABC
 = \(\frac{1}{4}\times 81\sqrt{3}=\frac{81}{4}\sqrt{3}\) cm.
and the area of \(\Delta\)GH1 = \(\frac{1}{4}\times \) area of \(\Delta\)DEF
= \(\frac{1}{4}\times \frac{81}{4}\sqrt{3}=\frac{81}{16}\sqrt{3}\) cm.
\(\therefore\) Sum of the areas of all the triangle is
\(S_a=81\sqrt{3}+\frac{81}{4}\sqrt{3}+\frac{81}{16}\sqrt{3}+....\infty\)
Clearly, it is Geometric series whose
\(a=81\sqrt{3}\)  and \(r=\frac{1}{4}\)
\(\therefore S_a=\frac{a}{1-r}=\frac{81\sqrt{3}}{1-\frac{1}{4}}=81\sqrt{3}\times \frac{4}{3}\)
= \(108\sqrt{3}\) cm².

After one year, value of machine
=[15625-15625 x \(\frac { 20 }{ 100 } \)]
=15625-3125=Rs 12500
After two years, value of machine
=[12500-12500 x \(\frac { 20 }{ 100 } \)]
=12500-2500 = Rs 10000
After three years, value of machine
=[10000-10000 x \(\frac { 20 }{ 100 } \)]
=10000-2000 = Rs 8000
The sequences ofvalues of machine is 12500, 10000, 8000, ... is a G.P
Here a=12500  r=\(\frac { 10000 }{ 12500 } =\frac { 4 }{ 5 } \)
∴ a₅=ar⁴=12500 x \(\left( \frac { 4 }{ 5 } \right) ^{ 4 }\)
=12500 x \(\frac { 256 }{ 625 } \) = Rs 5120
Thus the value of machine at the end of 5 years =Rs 5120

Since a, b, c are in AP.
b - a = c - b \(\Rightarrow\) 2b = a + c ..(i)
Now \({ A }_{ 1 }=\frac { 1 }{ 2 } \left( a+b \right) \)
and \(A_{ 2 }=\frac { 1 }{ 2 } (b+c)\)
\(\therefore\)  A.M of A₁ and A₂ = \(\frac { 1 }{ 2 }\) (A₁ + A₂)
= \(\frac { 1 }{ 2 } \left[ \frac { 1 }{ 2 } \left( a+b \right) +\frac { 1 }{ 2 } \left( b+c \right) \right] \)
= \(\frac { 1 }{ 2 } \times \frac { 1 }{ 2 } \left[ a+2b+c \right] \)
= \(\frac { 1 }{ 4 } [2b+2b]\left[ \because \quad a+c=2b \right] \)
= \(\frac { 1 }{ 4 } \times 4b=b\)

Here a = 25, d = 22 - 25 = -3.
Let sum of n terms be 116 then S_n = 116
We know that S_n = \(\frac { n }{ 2 } \)[2a + (n-1)d]
\(\therefore 116=\frac { n }{ 2 } [2\times 25+(n-1)\times -3]\)
\(\Rightarrow\)  232 = n[50 - 3n + 3]
\(\Rightarrow\) 232 = 53n-3n² \(\Rightarrow\) 3n² - 53n + 232 = 0
\(\therefore n=\frac { -(-53)\pm \sqrt { ({ -53) }^{ 2 }-4\times 3\times 232 } }{ 2\times 3 } \)
\(=\frac { 53\pm \sqrt { 2809-2784 } }{ 6 } \)
\(=\frac { 53\pm \sqrt { 25 } }{ 6 } =\frac { 53\pm 5 }{ 6 } \)
\(\therefore n=\frac { 53\pm 5 }{ 6 } \ or \ n=\frac { 53-5 }{ 6 } \)
\(n=\frac { 58 }{ 6 } \quad or\quad n=\frac { 48 }{ 6 } =8\)
\(n=\frac { 58 }{ 6 } \) is not possible. Thus n = 8.
Now a_n = a + (n-1)d
\(\therefore\) a₈ = 25 + (8-1) \(\times\) -3 = 25 - 21 = 4
Thus last term of the given A.P. is 4.