4. Find three consecutive terms in A. P. whose sum is -3 and the prodive of their cubes is 3375 . (Take the value of \( d \) positive.)

Let the three consecutive terms in the arithmetic progression be (a – d), a, and (a + d), where d > 0. Step 1. The sum of the three terms is –3:   (a – d) + a + (a + d) = 3a = –3   ⇒ a = –1. Step 2. The product of their cubes is given as 3375. Notice that the product of the cubes is   [(a – d)³][a³][(a + d)³] = [(a – d)(a)(a + d)]³. So, we have   [(a – d)(a)(a + d)]³ = 3375 Taking cube roots on both sides:   (a – d)(a)(a + d) = 3375^(1/3) = 15. Step 3. Substitute a = –1 into the product:   (–1 – d)(–1)(–1 + d) = 15. We can simplify by multiplying the two factors (–1 – d) and (–1 + d) first:   (–1 – d)(–1 + d) = (–1)² – d² = 1 – d². Then,   (1 – d²)(–1) = d² – 1. So we have   d² – 1 = 15  ⇒  d² = 16  ⇒  d = 4  (since d is positive). Step 4. Now find the three terms:   First term: a – d = –1 – 4 = –5.   Second term: a = –1.   Third term: a + d = –1 + 4 = 3. Verification:   Sum: (–5) + (–1) + 3 = –3.   Product of cubes: (–5)³ · (–1)³ · 3³ = (–125) · (–1) · 27 = 3375. Thus, the three consecutive terms are –5, –1, and 3.