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Question
Find three consecutive terms which are in AP . Whose sum is 24 and product is 440
Find three consecutive terms which are in AP . Whose sum is 24 and product is 440
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Solution
→ Let the required number be ( a- d ) , a and ( a + d ).....................(i)
▶ 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’ in equation (i) we get the required Number.
Hence, the required numbers are ( 5,8,11 ) and ( 11,8,5 ) .
▶ 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’ in equation (i) we get the required Number.
Hence, the required numbers are ( 5,8,11 ) and ( 11,8,5 ) .
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