the sum of 3 consecutive terms of an ap is 33 and their product is 792 the last of them is
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Answer:
a + d = 11 + 7 = 18
Step-by-step explanation:
Let ( a - d ) , a , ( a + d ) are three numbers
are in A. P
sum of the terms = 33
a - d + a + a + d = 33
3a = 33
a = 33/3
a = 11
product of the terms = 792
( a - d ) × a × ( a + d ) = 792
a ( a² - d² ) = 792
11 ( 11² - d² ) = 792
121 - d² = 792/11
121 - d² = 72
- d² = 72 - 121
- d² = -49
d² = 49
d = ± √49
d = ± 7
Therefore ,
Three numbers are ,
a - d = 11 - 7 = 4 ,
a = 11 ,
a + d = 11 + 7 = 18
Required smallest number = 4
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