Question

the sum of 3 consecutive numbers in AP is 21 and product is 231.find the numbers

Answer 1

LET THE THREE CONSECTIVE NO. BE (A-D),(A),(A+D)
It is given that sum of three no. is 21
SO
(A-D)+(A)+(A+D)=21
A-D+A+A+D=21
3A=21
A=7
AND NOW IT IS GIVEN THAT PRODUCT IS 231
(A-D)(A)(A+D)=231
(7-D)(7+D)=231/7
7 SQ-D SQ=33
49-D SQ=11
49-33=D SQ
16=D sq
D=4
SO THE NO. ARE
A-D=7-4=3
A=7
A+D=7+4=11

Answer 2

from given conditions
(a+nd)+(a+(n+1)d)+(a+(n+2)d)=21
3a+d(n+(n+1)+(n+2))=21
3a+d(3n+3)=21
3a+3dn+3d=21
3a+3d(n+1)=21....(1)
also given product is 231 so
(a+nd)×(a+(n+1)d)×(a+(n+2)d)=231
now follow the figure