\({ z }_{ n }=\begin{cases} n \\ { a }_{ n-1 }+{ a }_{ n-2 }+{ a }_{ n-3 } \end{cases}\)
a₁ = 1, a₂ = 2, a₃ = 3
a₄ = a₃ + a₂ + a₁ = 3 + 2 + 1 = 6
a₅ = a₄ + a₃ + a₂ = 6 + 3 + 2 = 11
a₆ = a₅ +a₄ +a₃ = 11 + 6 + 3 = 20
the first 6 terms are 1, 2, 3, 6, 11, 20

\({ a }_{ n }=\begin{cases} n+1\quad if\quad n\quad is\quad odd \\ n\quad \quad if\quad n\quad is\quad even \end{cases}\)
a₁ = 1 + 1 = 2, a2 = 2, a₃ = 3 + 1 = 4
a₄ = 4, a₅ = 5 +1 = 6, a₆ = 6
hence the first 6 terms are 4, 2, 2, 4, 6, 6...

Let \(a_n={3n-2\over 3^{n-1}}\)
\(a_1={1\over 30}=1\)
\(a_2={3(2)-2\over 3^1}={4\over 3}\)
\(a_3={3(3)-2\over 3^2}={7\over9}\)
\(a_4={3(4)-2\over 3^3}={10\over 27}\)
∴ The sequence \({1\over1},{4\over 3},{7\over 9},{10\over 27},...\)
\(=1,4\left(1\over3\right),7\left(1\over3\right)^2+10\left(1\over3\right)^3+...\)
∴ 1, 4, 7, 10 is an A.P and \(\left( \frac { 1 }{ 3 } \right) ^{ 0 },\left( \frac { 1 }{ 3 } \right) ^{ 1 },\left( \frac { 1 }{ 3 } \right) ^{ 2 }\) g.P
Hence the given sequence is an arithmetic - geometric progression.

2018
Let a_n = 2018
then the first 6 terms are 2018, 2018, 2018, 2018, 2018, 2018
It is not an AP, GP, AGP and HP.

Let a_n = \(\frac { 2n+3 }{ 3n+4 } \)
\({ a }_{ 1 }=\frac { 2+3 }{ 3+4 } =\frac { 5 }{ 9 } \)
\({ a }_{ 2 }=\frac { 4+3 }{ 6+4 } =\frac { 7 }{ 10 } \)
\({ a }_{ 3 }=\frac { 6+3 }{ 9+4 } =\frac { 9 }{ 13 } \)
\({ a }4=\frac { 8+3 }{ 12+4 } =\frac { 11 }{ 16 } \)
\({ a }_{ 5 }=\frac { 10+3 }{ 15+4 } =\frac { 13 }{ 19 } \)
\({ a }_{ 6 }=\frac { 12+3 }{ 18+4 } =\frac { 15 }{ 22 } \)
\(\frac { 5 }{ 9 } ,\frac { 7 }{ 10 } ,\frac { 9 }{ 13 } ,\frac { 11 }{ 16 } ,\frac { 13 }{ 19 } ,\frac { 15 }{ 22 } ...\)
this is neither A.P, G.P nor AGP

Let \(a_n={(-1)^n\over 1}\)
\(a_1={(-1)^1\over 1}=-1, a_2={(-1)^2\over 2}={1\over 2}, a_3={(-1)^3\over 3}={-1\over 3}\)
\(a_4 ={(-1)^4\over 4}={1\over 4},a_5 ={(-1)^5\over 5}=-{1\over 5},a_6 ={(-1)^6\over 6}={1\over 6}\)
∴ The sequence is  \(-1,{1\over 2},-{1\over 3},{12\over4},-{1\over 5}, {1\over 6},...\)
That is \(-1,{1\over 2},-{1\over 3},{12\over4},-{1\over 5}, {1\over 6},...\)
Consider 1, 2, 3, 4, .... which is an A.P.
Since d = 2 - 1 = 3 - 2 = 1
and -1, 1, -1,1, ... is a G.P. where
\(r=\frac { 1 }{ -1 } =\frac { -1 }{ 1 } =-1\)
Hence this is an arithmetico-geometric progression.

Let a_n = 4\(\left( \frac { 1 }{ 2 } \right) ^{ n }\)
\({ a }_{ 1 }= 4\left( \frac { 1 }{ 2 } \right) ^{ 1 }=\frac { 4 }{ 2 } =2\)
\({ a }_{ 21 }= 4\left( \frac { 1 }{ 2 } \right) ^{ 2 }=\frac { 4 }{ 4 } =1\)
\({ a }_{ 3 }= 4\left( \frac { 1 }{ 2 } \right) ^{ 3 }=\frac { 4 }{ 8 } =\frac { 1 }{ 2 } \)
\({ a }_{ 5 }= 4\left( \frac { 1 }{ 2 } \right) ^{ 4 }=\frac { 4 }{ 32 } =\frac { 1 }{ 8 } \)
\({ a }_{ 6 }= 4\left( \frac { 1 }{ 2 } \right) ^{ 5 }=\frac { 4 }{ 64 } =\frac { 1 }{ 16 } \)
the sequence is \(2,1,\frac { 1 }{ 2 } ,\frac { 1 }{ 4 } ,\frac { 1 }{ 8 } ,\frac { 1 }{ 16 } \)
a = 2; r = \(\frac{1}{2}\)
It is of the form. a, ar, ar²
It is geometric progression.

\({ a }_{ 1 }=\frac { \left( 1+1 \right) \left( 1+2 \right) }{ \left( 1+3 \right) (1+4) } =\frac { 2(3) }{ 4(5) } =\frac { 6 }{ 20 } =\frac { 3 }{ 10 } \)
\({ a }_{ 2 }=\frac { \left( 2+1 \right) \left( 2+2 \right) }{ \left( 2+3 \right) (2+4) } =\frac { 3(4) }{ 5(6) } =\frac { 12 }{ 30 } =\frac { 2 }{ 5 } \)
\({ a }_{ 3 }=\frac { \left( 3+1 \right) \left( 3+2 \right) }{ \left( 3+3 \right) (3+4) } =\frac { 4(5) }{ 6(7) } =\frac { 10 }{ 21 } \)
\({ a }_{ 4 }=\frac { 5(6) }{ 7(8) } =\frac { 15 }{ 28 } \)
\({ a }_{ 5 }=\frac { 6(7) }{ 8(9) } =\frac { 7 }{ 12 } \)
\({ a }_{ 6 }=\frac { 7(8) }{ 9(10) } =\frac { 28 }{ 45 } \)
The sequence is \(\frac { 3 }{ 10 } ,\frac { 2 }{ 5 } ,\frac { 10 }{ 21 } ,\frac { 15 }{ 28 } ,\frac { 7 }{ 12 } ,\frac { 28 }{ 45 } \)
None of A.P, G.P or H.P

\(\frac { 1 }{ 2^{ n+1 } } \)
\({ a }_{ 1 }\frac { 1 }{ 2^{ n+1 } } =\frac { 1 }{ { 2 }^{ 2 } } ,{ a }_{ 2 }=\frac { 1 }{ { 2 }^{ 2+1 } } \frac { 1 }{ { 2 }^{ 3 } } \)
\({ a }_{ 2 }=\frac { 1 }{ 2^{ 3+1 } } =\frac { 1 }{ { 2 }^{ 4 } } \)
\({ a }_{ 3 }=\frac { 1 }{ 2^{ 4+1 } } =\frac { 1 }{ { 2 }^{ 5 } } ,{ a }_{ 5 }=\frac { 1 }{ 2^{ 5+1 } } =\frac { 1 }{ { 2 }^{ 6 } } ,{ a }_{ 6 }=\frac { 1 }{ { 2 }^{ 6+1 } } =\frac { 1 }{ { 2 }^{ 7 } } \)
ஃ  the first 6 terms of the sequence are \(\frac { 1 }{ { 2 }^{ 2 } } ,\frac { 1 }{ { 2 }^{ 3 } } ,\frac { 1 }{ { 2 }^{ 4 } } ,\frac { 1 }{ { 2 }^{ 5 } } ,\frac { 1 }{ { 2 }^{ 6 } } and\frac { 1 }{ { 2 }^{ 7 } } \)
Since \({ a }_{ 1 }=\frac { 1 }{ { 2 }^{ 2 } } \& r=\frac { 1 }{ { 2 }^{ 3 } } \div \frac { 1 }{ { 2 }^{ 2 } } =\frac { 1 }{ { 2 }^{ 3 } } \times { 2 }^{ 2 }=\frac { 1 }{ 2 } \)
\(r=\frac { 1 }{ { 2 }^{ 4 } } \div \frac { 1 }{ { 2 }^{ 3 } } =\frac { 1 }{ 2 } \times { 2 }^{ 3 }=\frac { 1 }{ 2 } \)
the given sequence is a geometric progression

T_n = a + (n - 1)d
T_m+n = a + (m + n - 1)d
& T_m-n = a + (m - n - 1)d
T_m+n + T_m-n = a + (m + n - 1)d + a + (m - n - 1)d
= 2a + d(m + n - 1 + m - n - 1)
= 2a + d(2m - 2)
= 2[a + (m - 1)d]
T_m+n + T_m-n = 2. T_m