The smallest prime number, that is the fifth term of an increasing arithmetic sequence in which all the four preceding terms are also prime, is

### Concept & Logic For five consecutive terms in an increasing arithmetic progression (A.P.) to all be prime, the common difference ($d$) must be chosen carefully to avoid multiples of 2, 3, and 5. If a prime sequence has 5 terms and doesn't start with 5, one of the terms will inevitably be a multiple of 5 (and therefore composite). Thus, the sequence must start exactly with the prime number 5. Furthermore, to avoid multiples of 2 and 3, the common difference $d$ must be a multiple of $2 \times 3 = 6$. ### Step-by-Step Solution * **Determine the starting term:** As established, the sequence must begin with $a = 5$. * **Determine the common difference:** To keep the subsequent numbers prime, $d$ must be a multiple of 6. Let's test the smallest possible valid difference, $d = 6$. * **Generate the sequence:** Term 1: $5$ (Prime) Term 2: $5 + 6 = 11$ (Prime) Term 3: $11 + 6 = 17$ (Prime) Term 4: $17 + 6 = 23$ (Prime) Term 5: $23 + 6 = 29$ (Prime) * **Verify the condition:** All five terms are prime numbers. The fifth term is 29. ### Exam Strategy & Shortcut Instead of deriving the sequence from scratch, work backwards from the given options. Since the options represent the *fifth* term, subtract standard common differences ($d=2, d=4, d=6$) to see if you hit a composite number. For option (a) 17: Sequence ending in 17 with $d=2$ is 9, 11, 13, 15, 17 (9 and 15 are composite). For option (b) 29: Subtracting $d=6$ repeatedly gives 23, 17, 11, 5. All are primes. You have found the answer instantly. ### Common Pitfall Testing sequences with a common difference of 2 or 4. In any set of 3 or more consecutive odd numbers separated by 2 or 4, one will always be a multiple of 3 (e.g., 3, 5, 7, 9). You must use a common difference that is a multiple of 6 to step over the multiples of 3. ### Final Answer **Therefore, the correct answer is 29.**