Let the last term be n, then
Sum of the first and the last term = 66
a + ar^{n – 1} = 66
and ar. ar^{n – 2} = 128
a²r^{n – 1} = 128
From Eqs. (i) and (ii),
a (66 – a) = 128

Correct Option: B

Let the last term be n, then
Sum of the first and the last term = 66
a + ar^{n – 1} = 66
and ar. ar^{n – 2} = 128
a²r^{n – 1} = 128
From Eqs. (i) and (ii),
a (66 – a) = 128
⇒ a² – 66a + 128 = 0
⇒ a = 64 , 2

Given in question , x^th term = x⊃2 + 1
(x + 1)^th term = (x + 1)⊃2 + 1
According to question,
(x + 1)^th term – x^th term = (x + 1)² + 1 – (x² + 1)

Correct Option: C

Given in question , x^th term = x⊃2 + 1
(x + 1)^th term = (x + 1)⊃2 + 1
According to question,
(x + 1)^th term – x^th term = (x + 1)² + 1 – (x² + 1)
∴ Required answer = x² + 2x + 1 + 1 – x² – 1 = 2x + 1

According to question ,
Fourth number = sum of the squares of the other three numbers
By hit and trial or common sense, we have,
2 = (–1)² + (0)² + (1)²

Correct Option: C

According to question ,
Fourth number = sum of the squares of the other three numbers
By hit and trial or common sense, we have,
2 = (–1)² + (0)² + (1)²
Hence the numbers are –1 , 0 , 1 , 2 .

Here , first term = –14 and fifth term = 2
According to question,
T_n = a + (n – 1).d
T₅ = a + (5 – 1).d
2 = – 14 + 4d

Correct Option: B

Here , first term = –14 and fifth term = 2
According to question,
T_n = a + (n – 1).d
T₅ = a + (5 – 1).d
2 = – 14 + 4d

According to question,
a = –3, d = 0 - ( - 3 ) = 3 - 0 = 3
∴ T_n = a + (n – 1).d
T₁₀ = a + (10 – 1).d

Correct Option: C

According to question,
a = –3, d = 0 - ( - 3 ) = 3 - 0 = 3
∴ T_n = a + (n – 1).d
T₁₀ = a + (10 – 1).d
∴ T₁₀ = –3 + 9 × 3 = 24