Question

The sum of the three number in ap is 21 and their product is 231 find the numbers.

Answer 1

Hii There!!!


Let the AP be :-. (a-d), a, (a+d) 

=)(a-d)+a+(a+d) =21
=)3a=21
=)a=7

(a-d)×a×(a+d)=231
(a×a-d×d) ×a=231
(7×7-d×d)×7=231
(7×7-d×d)=231/7
(7×7-d×d)=33
d×d=49-33
d×d=16
d=4
So, the A. P. is 3, 7, 11.



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Answer 2

a = first term
d = common difference

Three terms of A.P. are

(a-b) +a +(a+d)

----------------------------

According to the question,

=> a -d +a +a-d = 21

=> a +a+a =21

=> 3a = 21

=> a = 21/3

=> a= 7

----------------------------

Product of three terms is 231,

=> (a-d)×a×(a+d) = 231

putting the value of a,

=> (7-d) ×7×(7+d) =231

=> (49-7d) ×(7+d) = 231

=> 343 +49d - 49d -7d² = 231

=> 343 -7d² =231

=> 7(49-d²) = 231

=> 49-d² = 231/7

=> 49-d² = 33

=> d² = -1(33-49)

=> d² = +16

=> d = ±√16

=> d = ±4


Numbers are,

Taking +4,
•(a-d) = 7-4=3
•a = 7
•(a+d) =7+4=11

Numbers are, 3,7,11

Taking -4,
• (a-d) = 7-(-4)=7+4=11
• a = 7
•(a+d) = 7+(-4)=7-4=3

Number are, 11,7,3




I hope this will help you


-by ABHAY