Question

find three terms in AP such that their sum is12 ans the sum of their product is 108​

Answer 1

EXPLANATION.

→ Three terms of an Ap their sum is = 12

→ The sum of their products is = 108

→ Let the three terms of an Ap =

( a - d) , a, ( a + d)

→ a - d + a + a + d = 12.

→ 3a = 12

→ a = 4

→ ( a - d) ( a)(a + d) = 108

→ (a) ( a² - d² ) = 108

→ Put the value of a = 4

→ 4 ( 4² - d² ) = 108

→ 4 ( 16 - d² ) = 108

→ 64 - 4d² = 108

→ - 4d² = 108 - 64

→ - 4d² = 44

→ -d² = 11

→ d² = -11

→ d = -√11

→ Three number are

( a - d) = ( 4 - (-√11)) = ( 4 + √11 )

( a) = 4

( a + d) = 4 - √11

→ series are → ( 4 + √11), 4, ( 4 - √11 ) .