Given, 1 = a₁ = a₂ and a_n = a_{n - 1} + a_{n - 2}, n > 2
On Putting n = 3, 4, 5 , 6 respectively, we get
For n =3, a₃ = a_{3 - 1} + a_{3 -2} = a₂ + a₁ = 1 + 1 = 2
For n =4, a₄ = a_{4 - 1} + a_{4 -2} = a₃ + a₂ = 2 + 1 = 3
For n =5, a₅ = a_{5 - 1} + a_{5 -2} = a₄ + a₃ = 3 + 2 = 5
For n =6, a₆ = a_{6 - 1} + a_{6 -2} = a₅ + a₄ = 5 + 3 = 8
Now, \(\frac { { a }_{ n+1 } }{ { a }_{ n } } \) ,  for n = 1, 2, 3, 4,5.
For n = 1, \(\frac { { a }_{ 2 } }{ { a }_{ 1 } } =\frac { 1 }{ 1 } =1\) :
For n =2,  \(\frac { { a }_{ 3 } }{ { a }_{ 2 } } =\frac { 2 }{ 1 } =2\) :
For n =3,  \(\frac { { a }_{ 4 } }{ { a }_{ 3 } } =\frac { 3 }{ 2 }\) :
For n =4,  \(\frac { { a }_{ 5 } }{ { a }_{ 4 } } =\frac { 5 }{ 3 }\) :
For n =5,  \(\frac { { a }_{ 6 } }{ { a }_{ 5 } } =\frac { 8 }{ 5 }\)
Hence, the required terms are 1, 2, \(\frac { 3 }{ 2 }\), \(\frac { 5 }{ 3 }\) and \(\frac { 8 }{ 5 }\)

Using \(S_\infty =\frac{a}{1-r}\)
\(S_1=\frac{1}{(1-\frac{1}{2})}=2\)
\(S_2=\frac{2}{(1-\frac{1}{3})}=3\)
\(S_3=\frac{3}{(1-\frac{1}{4})}=4\)
........................
........................
\(SP=\frac{P}{[1-\frac{1}{P-1}]}=P+1\)
\(\therefore\) S₁ + S₂ + S₃ + ..... S_p = 2 + 3 + 4 + .... + (P + 1)
 = \(\frac{P}{2}[2\times 2+(P-1)1]\)
= \(\frac{P}{2}(P+3)\)

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

Let a be the first term and r be the common ratio of given G.P. then
a₄ = 4 \(\Rightarrow\) ar³ = 4
Now product of first seven terms
= a₁· a₂· a₃· a₄· a₅· a₆· a₇
= (a)(ar)(ar²)(ar³)(ar⁴)(ar⁵)(ar⁶)
= a⁷r^2I = (ar³)7
= (4)⁷  [\(\therefore\)  ar³ = 4]
= 16384.

Here a = 2 and r = \(\frac { 8 }{ 2 } =4\)
Let a_n = 32768
\(\Rightarrow\) a_r^n-1 = 32768
\(\Rightarrow\) 2 x 4n-1 = 32768
\(\Rightarrow\) 4^n-1 = 16384
\(\Rightarrow\) 4^n-1 = 47
\(\Rightarrow\) n - 1 = 7
\(\Rightarrow\) n = 7 + 1
\(\Rightarrow\) n = 8 .

Here a = 4, d = 8 - 4 = 4 and an = 288
We know that a_n = a + (n - 1)d
\(\therefore\) 288 = 4 + (n - 1) x 4
\(\therefore\)  288 - 4 = (n - 1) x 4
\(\frac { 284 }{ 4 } =(n-1)\)
n = 71 + 1 = 72
Now a = 288 and d =- 4
\(\therefore\) 18^th term from end = 288 - (18 - 1) x 4
= 288 - 68 = 220

a_m+n = P and a_m-n = q  \(\Rightarrow\) ar^m+n-1 and ar^m-n-1 = q
\(\Rightarrow\) \(\frac { { ar }^{ m+n-1 } }{ { ar }^{ m-n-1 } } =\frac { p }{ q } \) \(\Rightarrow\) \({ r }^{ 2n }=\frac { p }{ q } \) \(\Rightarrow\) \(r=\left( \frac { p }{ q } \right) ^{ \frac { 1 }{ 2n } }\)
a_m = ar^m-1 = ar^m+n-1. \(\left( \frac { 1 }{ r } \right) ^{ n }\) = a_m+n \(\left( \frac { 1 }{ r } \right) ^{ n }\) = p \(\left( \frac { 1 }{ r } \right) ^{ n }\)  = p \(\left( \frac { p }{ q } \right) ^{ \frac { 1 }{ 2n } }\) = \(\sqrt { pq } \)
a_n = ar^n-1 = ar ^{m+n - 1}.\(\left( \frac { 1 }{ r } \right) ^{ m }\) = a_m+n \(\left( \frac { 1 }{ r } \right) ^{ m }\) = P \(p\left( \frac { p }{ q } \right) ^{ \frac { 3 }{ 2n } }\)

We have, a=3, l=57, and n=20
\(\because { S }_{ n }=\frac { n }{ 2 } [a+l]\)
\(\therefore { S }_{ 20 }=\frac { 20 }{ 2 } [3+57]=10\times 60=600\)

Given, a_n = 3n - 2
Replacing n by n + 1 in a_n, we get
a_n+1 = 3 (n+1) - 2 = 3n + 3 - 2 = 3n +1
Then, d = a_n+1 - a_n = (3n + 1) - (3n - 2)
= 3n + 1 - 3n + 2 = 3
It is independent of n, so given sequence is an AP and its 6th term, a₆ = 3 \(\times \) 6 - 2 = 16

We have, a_n = 2n² +1
On replacing n by n+1, we get
a_n+1 = 2 (n+1)² +1 = 2 (n² + 1 + 2n) +1
= 2n² + 2 + 4n + 1
= 2n² + 4n + 3
Here, a_n+1 -a_n = (2n² + 4n +3) - (2n² +1)
= 2n² + 4n +3 - 2n² - 1
= 4n + 2
It is not independent of n. So, given sequence is not an AP.

Here a₁ = a₂ = 2 and a_n = a_n-1 - 1.
Putting n = 3, 4 and 5, we have
a₃ = a_3-1-1=a_2-1=2-1 = 1 [\(\because\) a₂ = 2]
a₄=a_4-1-1=a_3-1=1-1=0[\(\because\) a₃ = 1]
a₅=a_5-1-1=a_4-1=0-1=-1[\(\because\) a₄= 0]
Thus, first five terms of the sequence are: 2, 2, 1, 0 and -1.
The corresponding series is 2 + 2 + 1+ 0 _ (-1) + ....

Here a₁ = 3, and a_n = 3a_n-1 + 2.
Putting n = 2, 3, 4 and 5, we have
a₂ = 3a_2-1 + 2=3a₁ + 2 = 3 \(\times\) 3 + 2
[\(\because\) a₁ = 3]
= 9 + 2 = 11
a₃ = 3a_3-1+2=3a₂ + 2 = 3 \(\times\) 11 + 2
= 33 + 2 = 35   [\(\because\) a₂ = 11]
a₄ = 3a_4-1+2=3a₃+2=3 \(\times\)35 + 2
[ \(\because\) a₃ = 35]
= 105 + 2 = 107
a₄ = 3a_5-1 + 2 = 3a₄+2 = 3 \(\times\) 107 + 2
[\(\because\) a₄ = 107]
= 321 + 2 = 323
Thus, first five terms of the sequences are 3, 11, 35, 107 and 323.
The corresponding series is 3 + 11 + 35 + 107 + 323+ ....

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

a₁₂ = 57, a₁₅ = 72

Here, \({ a }_{ n }=\frac { n }{ n+1 } \)
Putting n = 1, 2, 3, 4 and 5, we have
\({ a }_{ 1 }=\frac { 1 }{ 1+1 } =\frac { 1 }{ 2 } { a }_{ 2 }=\frac { 2 }{ 2+1 } =\frac { 2 }{ 3 } \)
\({ a }_{ 3 }=\frac { 3 }{ 3+1 } =\frac { 3 }{ 4 } { a }_{ 4 }=\frac { 4 }{ 4+1 } =\frac { 4 }{ 5 } \)
\({ a }_{ 5 }=\frac { 5 }{ 5+1 } =\frac { 5 }{ 6 } \)
Thus, the first five terms of sequence are  \(\frac { 1 }{ 2 }\),\(\frac { 2 }{ 3 }\),\(\frac { 3 }{ 4 }\), \(\frac { 4 }{ 5 }\) and \(\frac { 5 }{ 6 }\)