Question

If the sum of three no in an ap is 12 the sum of their cubes is 408 then the sum of their squares is

Answer 1

Answer:

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Answer 2

Answer:

Step-by-step explanation:

Let three terms (a - d) , a , (a + d ) are in AP.

A/C to question,

sum of three terms = 12

(a - d ) + a + (a + d ) = 12

3a = 12 ⇒ a = 4

again, question said ,

sum of their cubes = 408

(a - d)³ + a³ + (a + d)³ = 408

a³ - 3a²d + 3ad² - d³ + a³ + a³ + 3a²d + 3ad² + d³ = 408

3a³ + 6ad² = 408

now put the value of a = 4

3(4)³ + 6(4)d² = 408

3 × 64 + 24d² = 408

24d² = 408 - 192 = 216

d² = 9 ⇒ d = ±3

Hence, terms are (a - d) , a , (a + d) :

when d = +3

(a - d ) = (4 -3) = 1

a = 4

(a + d ) = 3 + 4 = 7

e.g., three terms are : 1, 4 ,7

when we take d = -3

(a - d) = 4 - (-3) = 7

a = 4

(a + d) = 4 + (-3) = 1

e.g., three terms are : 7, 4 , 1

Hope it helps you...