Question

find three consecutive in H.p whose sum is 24 and product is 440​

Answer 1

Answer:

Hey there !!

→ Let the required number be ( a- d ) , a and ( a + d ).

▶ Now,

A/Q,

=> ( a - d ) + a + ( a + d ) = 24.

=> a -d + a + a + d = 24.

=> 3a = 24.

=> a = 24/3.

=> a = 8.

▶ Now, again

A/Q,

=> ( a - d ) × a × ( a + d ) = 440.

=> a( a² - d² ) = 440.

=> 8 ( 8² - d² ) = 440.

=> 8 ( 64 - d² ) = 440.

=> 512 - 8d² = 440.

=> 8d² = 512 - 440.

=> 8d² = 72.

=> d² = 72/8.

=> d = √9.

=> d = ±3.

By putting the value of ‘a’ and ‘d’ ,

✔✔ Hence, the required numbers are ( 5,8,11 ) and ( 11,8,5 ) ✅✅.

Step-by-step explanation:

hope it helps:)

Answer 2

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

find three consecutive in AP whose sum is 24 and product is 440

${\large{\underline{\underline{\bf{Required\:Solution:}}}}}$

→ Let the required number be ( a- d ) , a and ( a + d ).

⇝ Now,

A/Q,

⇏( a - d ) + a + ( a + d ) = 24.

⇏ a -d + a + a + d = 24.

⇏3a = 24.

⇏ a = 24/3.

⇏a = 8.

⇝Now, again

A/Q,

⇏ ( a - d ) × a × ( a + d ) = 440.

⇏ a( a² - d² ) = 440.

⇏ 8 ( 8² - d² ) = 440.

⇏8 ( 64 - d² ) = 440.

⇏512 - 8d² = 440.

⇏ 8d² = 512 - 440.

⇏8d² = 72.

⇏ d² = 72/8.

⇏ d = √9.

⇏d = ±3.

By putting the value of ‘a’ and ‘d’ ,

Hence, the required numbers are

( 5,8,11 ) and ( 11,8,5 )

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