$$\eqalign{ & {T_3} = a + 2d = - 1.....\,\left( 1 \right) \cr & {T_8} = a + 7d = 19\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 4\,\& \,a = - 9 \cr & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr} $$
$${S_{11}} = \frac{{11}}{2}$$ $$\left[ {2\left( { - 9} \right) + \left( {11 - 1} \right)\left( 4 \right)} \right]$$
$$\eqalign{ & {S_{11}} = \frac{{11}}{2}\left[ {\left( { - 18} \right) + \left( {40} \right)} \right] \cr & {S_{11}} = \frac{{11}}{2}\left[ {22} \right] \cr & {S_{11}} = 121 \cr} $$

T₇ = a + 6d = -15 ....... (1)
T₁₂ = a + 11d = 5 ...... (2)
on solving (1) and (2)
d = 4 & a = -39
T₁₆ = a + 15d
      = -39 + 15(4)
      = -39 + 60
      = 21

$$\eqalign{ & {1^{st}}{\kern 1pt} {\text{Method}}: \cr & {1^{st}}{\kern 1pt} {\text{term}} = 5; \cr & {3^{rd}}{\kern 1pt} {\text{term}} = 15; \cr & {\text{Then}},{\kern 1pt} \,d = 5; \cr & {16^{th}}{\kern 1pt} {\text{term}} = a + 15d \cr & = 5 + 15 \times 5 = 80 \cr & {\text{Sum}} = {n \times \frac{{\left( {a + l} \right)}}{2}} \cr} $$
$$ = {{\text{no}}{\text{.}}{\kern 1pt} {\text{of}}{\kern 1pt} {\text{terms}} \times \frac{{ {{\text{first}}{\kern 1pt} {\text{term + last}}{\kern 1pt} {\text{term}}} }}{2}} $$
$$\eqalign{ & = {16 \times \frac{{\left( {5 + 80} \right)}}{2}} \cr & = 16 \times \frac{{85}}{2} \cr & = 8 \times 85 \cr & = 680 \cr} $$

2^nd Method(Thought Process):
Sum = number of terms × average of that AP
$$\eqalign{ & {\text{Sum}} = 16 \times {\frac{{\left( {5 + 80} \right)}}{2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 16 \times \frac{{85}}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 8 \times 85 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 680 \cr} $$

$$\eqalign{ & {T_3} = a + 2d = - 13\,.....\,\left( 1 \right) \cr & {T_6} = a + 5d = - 4\,.....\,\left( 2 \right) \cr & {\text{on solving}}\left( 1 \right)\,{\text{and}}\,\left( 2 \right) \cr & d = 3\& a = - 19 \cr & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr} $$
$${S_{12}} = \frac{{12}}{2}$$ $$\left[ {2\left( { - 19} \right) + \left( {12 - 1} \right)\left( 3 \right)} \right]$$
$$\eqalign{ & {S_{12}} = \left( 6 \right)\left[ { - 38 + 33} \right] \cr & {S_{12}} = - 30 \cr} $$

First term is -64. The common difference is -2. The last term is -100.
Sum of first n terms of an AP =
$$\frac{n}{2}\left[ {2{a_1} + \left( {n - 1} \right)d} \right]$$
To compute sum, we know first term a1 = -64 and common difference d = -2.
We do not know number of terms n. Let us first compute number of terms and n find sum of terms.
a_n = a₁ + (n - 1)d
-100 = -64 + (n - 1)(-2)
Therefore, n = 19
Sum =
$${S_n} = \frac{{19}}{2}$$ $$\left[ {2\left( { - 64} \right) + \left( {19 - 1} \right)\left( { - 2} \right)} \right]$$
$$\eqalign{ & {S_n} = \frac{{19}}{2}\left[ { - 128 - 36} \right] \cr & {S_n} = 19 \times \left( { - 82} \right) \cr & {S_n} = - 1558 \cr} $$

S_n is sum of first n terms a is first term and d is common difference
$$\eqalign{ & {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & \frac{{{S_x}}}{{{S_{kx}}}} = \frac{{\frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}{{\frac{{kx}}{2}\left[ {2a + \left( {kx - 1} \right)d} \right]}} \cr & \because \frac{{{S_x}}}{{{S_{kx}}}}\,{\text{is}}\,{\text{independent of}}\,x \cr} $$
$$\refore \frac{{\frac{n}{2}\left[ {2a + \left( {x - 1} \right)d} \right]}}{{\frac{{kx}}{2}\left[ {2a + \left( {kx - 1} \right)d} \right]}}\,$$    is independent of x
$$\refore \frac{{\frac{n}{2}\left[ {2a + xd - d} \right]}}{{\frac{{kx}}{2}\left[ {2a + kdx - d} \right]}}$$    is independent of x
$$ \Rightarrow \frac{{2a - d}}{{k\left( {2a - d} \right)}}$$   is in dependent of x if 2a - d $$ \ne $$ 0
If 2a - d =0, n d = 2a

(2K - 1) - K = (2K + 1) - (2K - 1)
2K - 1 - K = 2
⇒ K = 3

Let a be first term and d be common difference of an A.P. and
$$\eqalign{ & {S_n} = 2{n^2} + 5n \cr & \refore {S_1} = 2{\left( 1 \right)^2} + 5 \times 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 2 + 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 7 \cr & \refore {S_2} = 2{\left( 2 \right)^2} + 5 \times 2 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 8 + 10 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18 \cr & \refore {\text{First}}\,{\text{term}}\,\left( {{a_1}} \right) = 7\,{\text{and}} \cr & {\text{Second}}\,{\text{term}}\,{a_2} = {S_2} - {S_1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 18 - 7 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 11 \cr & \refore d = {a_2} - {d_1} \cr & \,\,\,\,\,\,\,\,\,\,\, = 11 - 7 \cr & \,\,\,\,\,\,\,\,\,\,\, = 4 \cr & Now\,{a_n} = a + \left( {n - 1} \right)d \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 + \left( {n - 1} \right)4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 7 + 4n - 4 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 4n + 3 \cr} $$

a_n = a + (n - 1) × d
⇒ 4 = a + (7 - 1) × -4
⇒ 4 = a - 24
⇒ a = 4 + 24
       = 28

The third term t₃ = a + 2d
The ninth term t₉ = a + 8d
t₃ + t₉ = 2a + 10d = 8
Sum of first 11 terms of an AP is given by
$$\eqalign{ & \Rightarrow {S_{11}} = \frac{{11}}{2}\left[ {2a + 10d} \right] \cr & \Rightarrow {S_{11}} = \frac{{11}}{2} \times 8 \cr & \Rightarrow {S_{11}} = 44 \cr} $$