Given that,

a = 17
l = 350
d = 9
Let there be n terms in the A.P.
l = a + (n − 1) d
350 = 17 + (n − 1)9
333 = (n − 1)9
(n − 1) = 37
n = 38
S_n = n/2 (a + l)
S₃₈ = 13/2 (17 + 350)
= 19 × 367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

d = 7
a₂₂ = 149
S₂₂ = ?
a_n = a + (n − 1)d
a₂₂ = a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2
S_n = n/2 (a + a_n)
= 22/2 (2 + 149)
= 11 × 151
= 1661

Given that,
a₂ = 14
a₃ = 18
d = a₃ − a₂ = 18 − 14 = 4
a₂ = a + d
14 = a + 4
a = 10
S_n = n/2 [2a + (n - 1)d]
S₅₁ = 51/2 [2 × 10 + (51 - 1) × 4]
= 51/2 [2 + (20) × 4]
= 51×220/2
= 51 × 110
= 5610

Given that,

S₇ = 49
S₁₇ = 289
S₇  = 7/2 [2a + (n - 1)d]
S₇ = 7/2 [2a + (7 - 1)d]
49 = 7/2 [2a + 16d]
7 = (a + 3d)
a + 3d = 7 ... (i)
Similarly,
S₁₇ = 17/2 [2a + (17 - 1)d]
289 = 17/2 (2a + 16d)
17 = (a + 8d)
a + 8d = 17 ... (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1
S_n = n/2 [2a + (n - 1)d]
= n/2 [2(1) + (n - 1) × 2]
= n/2 (2 + 2n - 2)
= n/2 (2n)
= n²

(i) a_n = 3 + 4n
a₁ = 3 + 4(1) = 7
a₂ = 3 + 4(2) = 3 + 8 = 11
a₃ = 3 + 4(3) = 3 + 12 = 15
a₄ = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a₂ − a₁ = 11 − 7 = 4
a₃ − a₂ = 15 − 11 = 4
a₄ − a₃ = 19 − 15 = 4
i.e., a_{k + 1} − a_k is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
S_n = n/2 [2a + (n - 1)d]
S₁₅ = 15/2 [2(7) + (15 - 1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35
= 525

(ii) a_n = 9 − 5n
a₁ = 9 − 5 × 1 = 9 − 5 = 4
a₂ = 9 − 5 × 2 = 9 − 10 = −1
a₃ = 9 − 5 × 3 = 9 − 15 = −6
a₄ = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a₂ − a₁ = − 1 − 4 = −5
a₃ − a₂ = − 6 − (−1) = −5
a₄ − a₃ = − 11 − (−6) = −5
i.e., a_{k + 1} − a_k is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
S_n = n/2 [2a + (n - 1)d]
S₁₅ = 15/2 [2(4) + (15 - 1) (-5)]
= 15/2 [8 + 14(-5)]
= 15/2 (8 - 70)
= 15/2 (-62)
= 15(-31)
= -465

Given that,
S_n = 4n − n²
First term, a = S₁ = 4(1) − (1)² = 4 − 1 = 3
Sum of first two terms = S₂
= 4(2) − (2)² = 8 − 4 = 4
Second term, a₂ = S₂ − S₁ = 4 − 3 = 1
d = a₂ − a = 1 − 3 = −2
a_n = a + (n − 1)d 
= 3 + (n − 1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore, a₃ = 5 − 2(3) = 5 − 6 = −1
a₁₀ = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3^rd, 10^th, and n^th terms are −1, −15, and 5 − 2n respectively.

The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S₄₀ = ?
S_n = n/2 [2a + (n - 1)d]
S₄₀ = 40/2 [2(6) + (40 - 1) 6]
= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920

The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S₁₅ = ?
S_n = n/2 [2a + (n - 1)d]
S₁₅ = 15/2 [2(8) + (15 - 1)8]
= 15/2[6 + (14) (8)]
= 15/2[16 + 112]
= 15(128)/2
= 15 × 64
= 960