find-four-consecutive-multiples-of-5-such-that-twice-the-sum-of-the-two-greatest-integer-exceed-five-times-the-least-by-5-

Question Number 114863 by mathdave last updated on 21/Sep/20

f i n d f o u r c o n s e c u t i v e m u l t i p l e s o f 5

s u c h t h a t t w i c e t h e s u m o f t h e t w o

g r e a t e s t i n t e g e r e x c e e d f i v e t i m e s t h e

l e a s t b y 5

Answered by PRITHWISH SEN 2 last updated on 21/Sep/20

let the integrs are

(5 m - 10), (5 m - 5), 5 m, (5 m + 5)

by the problem

2 {5 m + (5 m + 5)} = 5 (5 m - 10) + 5

\Rightarrow m = 11

∴ the numbers are

45, 50, 55, 60