Question

find the three consecutive terms which are in AP whose sum is 24 and the product is 440​

Answer 1

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Answer 2

Let the terms be $\bf{( a - d ), a, ( a + d ).}$

Given that :

Sum of these terms = 24

a - d + a + a - d = 24

3a = 24

a = 24/3 = 8

Also,

Product of these terms = 440

( a - d ) ( a ) ( a + d ) = 440

Putting a = 8, we get

( 8 - d ) ( 8 ) ( 8 + d ) = 440

${\boxed{\bf{( a + b ) ( a - b ) = a^2 - b^2}}}$

( 8² - d² ) ( 8 ) = 440

( 64 - d² ) = 440/8 = 55

d² = 64 - 55

d² = 9

d = √9

d = ± 3

$\bf{ {\underline{Case \ I :}} \ When \ a = 8, d = 3}$

a - d = 8 - 3 = 5

a = 8

a + d = 8 + 3 = 11

$\bf{ {\underline{Case \ II :}} \ When \ a = 8, d = - 3}$

a - d = 8 - ( - 3 ) = 11

a = 8

a + d = 8 + ( - 3 ) = 5

Hence, $\bf\red{the \ terms \ are \ {\underline{11, 5 \ and \ 8}}.}$