Question

EXERCISE 2.3
de un
tree opnutie mumbers in AP
18 and the sum of their
Find four numbers in AP who sumi 20 and wam of whom were 130
2000
F) Pour numbers are in AP. Their sum is 16 and the sum of their an Tiad the
number
The sum of three consecutive ter of an AP in 21 and their peluet m und the
numbers,
The sum of three consecutive sumber in AP 27 and their product is said the
numbers
vide 32 into four parts which are in AP. wuch that the product of the extremes into the produto
eans as 7:15
MD.U. 2015​

Answer 1

Answer:

2  ,  6   , 10  , 14

Step-by-step explanation:

Divide 32 into four parts such that they are in AP

& product of the extremes is in ratio of 7:15

four numbers  are

a - 3d , a - d , a + d , a + 3d

first term = a - 3d

common difference = 2d

Adding all   4a = 32

=> a = 8

so numbers are

8 - 3d , 8 - d , 8 + d , 8 + 3d

(8 - 3d) * (8 + 3d) : (8 - d) * (8 + d) :: 7 :15

=> 15 (8 - 3d) * (8 + 3d) = 7 *  (8 - d) * (8 + d)

=> 15 ( 64 - 9d²) = 7(64 - d²)

=> 8 * 64 = 128d²

=> d² = 4

=> d = ± 2

Hence 4 numbers are

2  ,  6   , 10  , 14

or

14 , 10  , 6,  2

all other Questions have missing information