Question

Find for consecutive terms in an a.p. such that the some of themiddle two terms is 18 and the product of the two end terms is 45

Answer 1

Let the consecutive terms are ( a - 3d), (a - d), (a + d), (a + 3d)** [common difference is equal]

According to question

sum of two middle terms = 18

∴ a - d + a + d = 18

2a = 18

a = 9 --- ( i )

Also,

Product of two end terms (first and fourth) = 45

∴ ( a - 3d) (a + 3d) = 45

a² - (3d)² = 45 [ ∵ (a - b) (a + b) = a² - b² ]

a² - 9d² = 45

Putting value of a from ( i )

9² - 9d² = 45

81 - 9d² = 45

9d² = 81 - 45

9d² = 36

d = 2 or -2

Putting a = 9 and d = -2 or 2 we get four terms

3, 7, 11 and 15

**Remember, whenever you are asked about four consecutive numbers in AP, you consider four numbers as a - 3d, a - d, a + d, a + 3d. It will make your solution super easy