Let us assume the three consecutive natural numbers be P, P + 1 and P + 2.
∴  According to the given question,
P² + (P + 1)² + (P + 2)² = 2030

Correct Option: B

Let us assume the three consecutive natural numbers be P, P + 1 and P + 2.
∴  According to the given question,
P² + (P + 1)² + (P + 2)² = 2030
   P² + P² + 2P + 1 + P² + 4P + 4 = 2030
⇒    3P² + 6P + 5 = 2030
⇒    3P² + 6P − 2025 = 0
⇒    P² + 2P – 675 = 0
⇒    P² + 27P – 25P – 675 = 0
⇒    P (P + 27) – 25 (P+ 27) = 0
⇒    (P – 25) (P + 27) = 0
∴  P = 25 and – 27
∴  Required number = P + 1 = 25 + 1 = 26

Series of all natural numbers from 75 to 97 is in A.P. whose first term, a = 75, last term, l = 97
If number of terms be n, then
a_n = a + ( n – 1 )d
⇒  97 = 75 + ( n – 1 )
⇒  n = 97 – 74 = 23

Correct Option: D

Series of all natural numbers from 75 to 97 is in A.P. whose first term, a = 75, last term, l = 97
If number of terms be n, then
a_n = a + ( n – 1 )d
⇒  97 = 75 + ( n – 1 )
⇒  n = 97 – 74 = 23

Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2. Sum of n terms in arithmetic progression is given by.

Correct Option: C

Series of first 20 odd natural numbers is an arithmetic progression with 1 as the first term and the common difference 2. Sum of n terms in arithmetic progression is given by.

Let us assume the 3 consecutive odd natural numbers are P, P + 2, P + 4;
According to question;
∴  P + P + 2 + P + 4 = 147
⇒  3P + 6 = 147
⇒  3P = 147 – 6 = 141

Correct Option: C

Let us assume the 3 consecutive odd natural numbers are P, P + 2, P + 4;
According to question;
∴  P + P + 2 + P + 4 = 147
⇒  3P + 6 = 147
⇒  3P = 147 – 6 = 141

Unit’s digit in 3 × 38 × 537 × 1256
= Unit’s digit in 3 × 8 × 7 × 6

Correct Option: D

Unit’s digit in 3 × 38 × 537 × 1256
= Unit’s digit in 3 × 8 × 7 × 6
= 4 × 2 = 8