Here, a = 1 and d = 1.
Sum to n terms of an AP is given by
S_n = n/2[ 2a + (n -1)d]
The sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Therefore
Sx-₁ = S₄₉ – Sx
⇒ (x -1/2)[2a + (x – 1 – 1)d]
⇒ 49/2[2a + (49 – 1)d] – x/2[2a + (x -1)d]
⇒ x -1/2[2(1) + (x -2)(1)]
⇒ 49/2[2(1) + 48(1)] – x/2[(1) + (x – 1)(1)]
⇒ (x – 1)[x] = 49[50] – x[x + 1]
⇒ x² – x = 2450 – x² – x
⇒ 2x² = 2450 x² = 1225 ⇒ x = 35
Hence, the value of x is 35.

The sum of n terms of an AP is given by
S_n = 4n – n²
Putting n = 1, we get
First term = a₁ = S₁ = 4(1) – (1) ² = 3
Putting n = 2, we get
sum of two terms = a₁ + a₂ = S₂ = 4(2) -(2)² = 4
⇒ a₁ + a₂ = 4
⇒ 3 + a₂ = 4 [∵ the first term a₁ = 3 ]
⇒ a₂ = 1
Hence, the second term is 1.
Common difference d = a₂ – a₁ = 1 – 3 = – 2
Therefore, the tenth term = a_10 = a + 9d = 3 + 9(-2) = – 16
Similarly, the nth term = a_n = a + (n -1)d = 3 + (n -1)(-2) = 5 – 2n

Let, the first term = a and common difference = d
the sum of the third and the seventh terms of the AP is 6, therefore
a₃ + a₇ = 6
⇒ a + 2d + a + 6d = 6
⇒ 2a + 8d = 6
⇒ a + 4d = 3
⇒ a = 3 – 4d …(1)
The product of the third and the seventh terms of the AP is 8, therefore
(a₃)(a₇) = 8
⇒ (a + 2d)(a + 6d) = 8
Putting the value of a from the equation(1), we get
(3 -2d)(3 +2d) = 8
⇒ 3² – 4d² = 8
⇒ 4d² = 1
⇒ d = ± 1/2
if, d = 1/2,
Putting the value of d in the equation (1), we get
a = 3 – 4(1/2) = 1
The sum of the 16 terms of this AP is given by
S₁₆ = 16/2[2a + 16 – 1)d] = 8[2(1) + 15 (1/2)] = 76
If, d = – 1/2,
Putting the value of d in the equation (1), we get
a = 3 – 4(-1/2) = 5
The sum of the 16 terms of this AP is given by
S₁₆ = 16/2[2a + 16 – 1)d] = 8 [2(5) + 15 (-1/2)] = 20
Hence, the sum of the 16 terms of AP is 20 or 76.