Question

Find five numbers in an arithmetic progression whose sum is 30 and sum of whose squares is 190.

Answer 1

Answer:

The numbers are 4, 5, 6, 7, 8

or 8, 7, 6, 5, 4

Step-by-step explanation:

Let the numbers be

a-2d, a-d, a, a+d, a+2d.

⇒(a−2d)+(a−d)+a+(a+d)+(a+2d) =30

⇒5a =30

⇒a=6

[(a−2d)² +(a−d)² +a² +(a+d)² +(a+2d)² ] = 190

put a = 6

⇒(6−2d)²+(6–d)² + 6² +(6+d)² + (6+2d)² = 190

[(a+b)² +(a−b)² =2(a² +b² )]

⇒2(36+4d²) +2(36+d²) = 190 - 36

⇒2[(36+4d²) + (36+d²)] = 154

⇒72+5d² = 77

⇒5d² = 5

⇒d² = 1

⇒d = ±1

Numbers = 4,5,6,7,8 or 8,7,6,5,4

hope it helps you

thank you