Question

there are five natural numbers not consecutive natural numbers if the average of first three is 10 the average of middle three is 14 and the average of last three is 35 and the middle number is 17 find the other number

Answer 1

Let the numbers be a , b , c , d & e

Given :

$\frac{a + b + c}{3} = 10 \\ a + b + c = 30 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: .......(eq1)$

$\frac{b + c + d}{3} = 14\\ b + c + d = 42 \: \: \: \: \: \: \: \: \: \: \: \: \: ........(eq2)$

$\frac{c + d + e}{3} = 35 \\ c + d + e = 105 \: \: \: \: \: \: \: \: \: \: \: \: \: \: .......(eq3)$

and Middle number c = 17

Eq1 ➡️ a + b + 17 = 30

Then a + b = 13 ...... (Eq1A)

Eq2➡️ b + 17 + d = 42

then b + d = 25 ...... (Eq2A)

Eq3➡️ 17 + d + e = 105

then d + e = 88 ....... (Eq 3A)

(Eq1A) - Eq2A ➡️ a - d = - 12

Then a + 12 = d

Eq3A➡️ a + 12 + e = 88 , then a + e = 76