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

\(\frac{100}{19}\)

\(\frac{3}{2}\)

Here \(a=\frac { -3 }{ 4 } \), r =\(\frac { -1 }{ 4 } \)
\(\therefore\)  \({ S }_{ \infty }=\frac { a }{ 1-r } =\frac { \frac { -3 }{ 4 } }{ 1-\frac { -1 }{ 4 } } =\frac { \frac { -3 }{ 4 } }{ \frac { 5 }{ 4 } } =\frac { -3 }{ 5 } \)

a = \(\frac { -5 }{ 4 } \) and r = \(\frac { -1 }{ 4 } \)
S\(\infty\) = \(\frac { a }{ 1-r } =\frac { \frac { -5 }{ 4 } }{ 1-\left( \frac { -1 }{ 4 } \right) } =-1\)

Here a = 1, r = \(\frac { 2 }{ 3 } \)
S_\(\infty\)  = \(\frac { a }{ 1-r } =\frac { 1 }{ 1-\frac { 2 }{ 3 } } =3\)

\(\frac { n }{ 4 } (n+1)\left( { n }^{ 2 }+n+2 \right) \)

\(\frac { n }{ 2 } ({ n }^{ 3 }+4n^{ 2 }+4n-1)\)

2,6,18,54.....

Here a = 1, r = 4 and S_n = 5461
We know that S_n=\(\frac { a\left( { r }^{ n }-1 \right) }{ r-1 } \)
5461=\(\frac { 1.\left( { 4 }^{ n }-1 \right) }{ 4-1 } \)
=4ⁿ-1 =16383 \(\Rightarrow \)4ⁿ=16384 \(\Rightarrow \) 4ⁿ=4⁷\(\Rightarrow \)n=7.

2, 6, 10, 14

Here a_n = 2n² + 3
Putting n = 1, 2, 3, 4, 5, we have
a₁ = 2 \(\times\) (1)²+ 3 = 2 + 3 = 5
a₂ = 2 \(\times\) (2)²+ 3 = 8 + 3 = 11
a₃ = 2 \(\times\) (3)²+ 3 = 18 + 3 = 21
a₄ = 2 \(\times\) (4)²+ 3 = 32 + 3 = 35
a₅ = 2 \(\times\) (5)²+ 3 = 50 + 3 = 53
Thus first five terms of sequence are 5, 11, 21, 35, 53.

19 = a + ( 6 - 1 ) d and 41 = a + ( 17 - 1 ) d = 87

7900000

T_q = 9