First term of AP = a = -20 and last term = l = 28
Number of terms = n = 17
$$\eqalign{ & {\text{Sum of AP}} = \frac{{\text{n}}}{2}\left( {{\text{a}} + {\text{l}}} \right) \cr & = \frac{{17}}{2}\left( { - 20 + 28} \right) \cr & = 17 \times 4 \cr & = 68 \cr} $$

T₃ = a + 2d = 13 ...... (1)
T₆ = a + 5d = -5 ...... (2)
on solving (1) and (2) d = -6 & a = 25
T₁₁ = a + 10d
      = 25 + 10(-6)
      = -35

Side of first square is 4 cm.
side of second square = $$2\sqrt 2 {\kern 1pt} $$ cm.
Side of third square = 2 cm.
and so on, i.e. 4, 2, $$\sqrt 2 $$ , $$\sqrt 2 $$ , 1 ........
Thus, area of se square will be = 16, 8, 4, 2, 1, $$\frac{1}{2}$$ ..........
Hence, Sum of area of first, second, third square
$$\eqalign{ & = 16 + 8 + 4 + 2 + 1 + {\kern 1pt} \,...... \cr & = {\frac{{16}}{{ {1 - {\frac{1}{2}} } }}} \cr & = 32\,{\kern 1pt} c{m^2} \cr} $$

1^st Method:
8^th term = a + 7d = 39 ........... (i)
12^th term = a + 11d = 59 ........... (ii)
(i) - (ii);

Or, a + 7d - a - 11d = 39 - 59
Or, 4d = 20
Or, d = 5
Hence, a + 7 × 5 = 39
Thus, a = 39 - 35 = 4
2^nd Method (Thought Process):
8^th term = 39
And, 12^th term = 59
Here, we see that 20 is added to 8^th term 39 to get 12^th term 59 i.e. 4 times common difference is added to 39
So, CD = $$\frac{{20}}{4}$$  = 5
Hence, 7 times CD is added to 1^st term to get 39. That means 4 is 1^st term of AP

$$\eqalign{ & {\text{In}}\,{\text{an}}\,{\text{A}}{\text{.P}}{\text{.}} \cr & a = 2\,{\text{and}}\,d = 4,\,n = 40 \cr & \refore {S_n} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{{40}}{2}\left[ {2 \times 2 + \left( {40 - 1} \right) \times 4} \right] \cr & = 20\left[ {4 + 39 \times 4} \right] \cr & = 20 \times \left( {4 + 156} \right) \cr & = 20 \times 160 \cr & = 3200 \cr} $$

Sum of n terms of an A.P. = 3n² + 5n
Let a be first term and d be common difference
S_n = 3n² + 5n
S₁ = 3(1)² + 5 × 1 = 3 + 5 = 8
S₂ = 3(2)² + 5 × 2 = 12 + 10 = 22
∴ First term (a) = 8
a₂ = S₂ - S₁ = 22 - 8 = 14
d = a₂ - a₁ = 14 - 8 = 6
Now a_n = a + (n - 1)d
⇒ 164 = 8 + (n - 1) × 6
⇒ 6n - 6 = 164 - 8
⇒ 6n = 156 + 6
⇒ 6n = 162
⇒ n = $$\frac{{162}}{6}$$
⇒ n = 27
∴ 168 is 27^th term

7^th term (a₇) = a + 6d = 34
13^th term (a₁₃) = a + 12d = 64
Subtracting, 6d = 30 ⇒ d = 5
and a + 12 x 5 = 64
⇒ a + 60 = 64
⇒ a = 64 – 60 = 4
18^th term (a₁₈)
= a + 17d
= 4 + 17 x 5
= 4 + 85
= 89

$$\eqalign{ & {S_n} = \left[ {2a + \left( {n - 1} \right)d} \right] \times \frac{n}{2} \cr & {\text{Here,}} \cr & d = 1 \cr & a = 1\, \cr & {\text{and}}\,\,n - 1\,\,{\text{terms}} \cr} $$
$$ \Rightarrow {S_{n - 1}} = \left[ {2 + \left( {n - 1 - 1} \right)} \right]$$     $$ \times \frac{{\left( {n - 1} \right)}}{2}$$
$$ \Rightarrow {S_{n - 1}} = \left[ {2 + n - 2} \right]$$    $$ \times \frac{{\left( {n - 1} \right)}}{2}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{n\left( {n - 1} \right)}}{2}$$

First 20 odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, . . . . ., 39
Here a = 1, d = 2, n = 20
$$\eqalign{ & \refore {S_{20}} = \frac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right] \cr & = \frac{{20}}{2}\left[ {2 \times 1 + \left( {20 - 1} \right) \times 2} \right] \cr & = 10\left( {2 + 38} \right) \cr & = 10 \times 40 \cr & = 400 \cr} $$

First term (a₁) = a
Second term (a₂) = b
and last term (l) = 2a
∴ d = Second term - First term = b - a
∴ l = a_n = a + (n - 1)d
$$\eqalign{ & \Rightarrow 2a = a + \left( {n - 1} \right)\left( {b - a} \right) \cr & \Rightarrow \left( {n - 1} \right)\left( {b - a} \right) = a \cr & \Rightarrow n - 1 = \frac{a}{{b - a}} \cr & \Rightarrow n = \frac{a}{{b - a}} + 1 \cr & \Rightarrow n = \frac{{a + b - a}}{{b - a}} \cr & \Rightarrow n = \frac{b}{{b - a}} \cr & \refore {S_n} = \frac{n}{2}\left[ {a + l} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{b}{{2\left( {b - a} \right)}}\left[ {a + 2a} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{3ab}}{{2\left( {b - a} \right)}} \cr} $$