Question

Find three terms of an AP such that their sum is 3 and product is -8

Answer 1

$\large\bf\underline{Given:-}$

• Product of three terms = -8
• sum of three terms = 3

$\large\bf\underline {To \: find:-}$

• Three terms

$\huge\bf\underline{Solution:-}$

Let the three terms are :-(a -d) , a and ( a + d)

▶️ According to Question :-

Sum of three terms = 3

↪ (a - d) + a + ( a + d) = 3

↪ a - d + a + a + d = 3

↪ 3a = 3

↪ a = 3/3

↪ a = 1 .....(1)

Product of three terms = -8

↪ (a - d) (a+ d) (a) = -8

↪ (a² - d²)a = -8

↪ a³ - ad² = -8

Substituting value of a from (1)

↪1³ - 1 × d² = -8

↪ 1 - d² = -8

↪ -d² = -8 -1

↪ -d² = -9

↪ d² = 9

↪ d = √9

↪ d = 3

So, Three terms are :-

↣ a - d = 1 - 3 = -2

↣a = 1

↣ a + d = 1 + 3 = 4

✝️Hence,

Three terms of an AP such that their sum is 3 and product is -8 is -2 , 1 and 4

$\rule{200}3$