Consider the following two series:

Series I: $8, 9, 15, 25, 42, 68, 105$

Series II: $60, 180, 450, 900, 1350, 1380, 675$

Identify the wrong term in Series II and determine which of the following statements about $\left(\frac{N}{30} + 1\right)$ is true, where $N$ is the wrong term:

(i) It is a prime number.

(ii) The sum of its digits is less than 9.

(iii) It is the nearest multiple of 5 and leaves a remainder of 4 when divided by 5.

Question

Consider the following two series:

Series I: $8, 9, 15, 25, 42, 68, 105$

Series II: $60, 180, 450, 900, 1350, 1380, 675$

Identify the wrong term in Series II and determine which of the following statements about $\left(\frac{N}{30} + 1\right)$ is true, where $N$ is the wrong term:

(i) It is a prime number.

(ii) The sum of its digits is less than 9.

(iii) It is the nearest multiple of 5 and leaves a remainder of 4 when divided by 5.

Accepted Answer

**Solution:**

1. First, identify the pattern in Series II:
   - $60 	imes 3 = 180$
   - $180 	imes 2.5 = 450$
   - $450 	imes 2 = 900$
   - $900 	imes 1.5 = 1350$
   - $1350 	imes 1 = 1350$
   - $1350 	imes 0.5 = 675$

2. The term $1380$ does not fit this pattern because after $1350$, the next term should be $1350$ again (multiplied by 1), not $1380$.

3. Therefore, the wrong term $N$ is $1380$.

4. Calculate $\frac{N}{30} + 1$:
   $$\frac{1380}{30} + 1 = 46 + 1 = 47$$

5. Check the statements:
   - (i) Is 47 a prime number? Yes, 47 is prime.
   - (ii) Sum of digits of 47 is $4 + 7 = 11$, which is not less than 9, so (ii) is false.
   - (iii) Check if 47 is a nearest multiple of 5 and leaves remainder 4:
     - Nearest multiples of 5 around 47 are 45 and 50.
     - $47 \div 5$ leaves remainder 2, not 4, so (iii) is false.

**Answer:** Only (i) follows.

Answer

Answer

Answer