Q84 of 84 Page 5

The sum of four consecutive numbers in an AP is 32, and the ratio of the product of the first and the last term to the product of two middle terms is 7: 15. Find the numbers. (CBSE 2018)

Let the numbers be : (a – 3d), (a – d), (a + d), (a + 3d)

(Note the common difference is “2d”)


According to Question,

Sum of four consecutive terms in A.P = 32

• (a – 3d) + (a – d) + (a + d) + (a + 3d) = 32

⇒ 4a = 32
⇒ a = 32/4
⇒ a = 8


So, the numbers are: (8 – 3d), (8 – d), (8 + d), (8 + 3d)


Also,

The ratio of the product of the first and the last term to the product of two middle terms is 7: 15. Therefore,


⇒


(Using (a + b)(a – b) = a2 – b2)


960 - 135d2 = 448 - 7d2

⇒ 128d2 = 512


⇒ d2 = 4


⇒ d = ±2


So, putting d = 2, numbers are: 2, 6, 10, 14


(Note: putting d = – 2 gives the same numbers but in reverse order)


Numbers are 2, 6, 10, 14

More from this chapter

All 84 →
80

If demotes the sum of first n terms of an A.P., prove that S12 = 3 (s8 – S4). (CBSE 2015)

 81

The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X. (CBSE 2016)

 82

A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in the first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief? (CBSE 2016)

 83

If the mth term of an A. P. is 1/n and nth term is 1/m then show that its (mn)th term is 1. (CBSE 2017)