Question

The sum of four numbers which are in AP is 32 and the product of it's extreme is 55.find the number

Answer 1

Not fully sure. But it may help you.

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Assumption

First term be t

Common difference be n

Hence,

(t - 3n), (t - n), (t + n) and (t + 3n)

Situation ,

Sum of the four numbers = 32

Hence,

⇒ t - 3n + t - n + t + n + t + 3n = 32

⇒ 4t = 32

⇒ t = 8

Also it is given :-

$\textbf{\underline{Product\;of\;extremes = 55}}$

⇒ (t - 3n)(t + 3n) = 55

⇒ t² - 9n² = 55

⇒ 8² - 9n² = 55 {t = 8}

⇒ 9n² = 64 - 55

⇒ 9n² = 9

⇒ n² = 1

⇒ n = ± 1

When,

t = 8 and n = 1

$\textbf{\underline{Four\;numbers :- }}$

8 - 3(1) = 5

8 - 1 = 7

$\textbf{\underline{8 + 1 = 9 }}$

8 + 3(1) = 11

When t = 8 and n = - 1

$\textbf{\underline{Four\;numbers }}$

8 - 3(-1) = 11

8 - (- 1) = 9

$\textbf{\underline{8 + (- 1) = 7 }}$

8 + 3 (- 1) = 5