the sum of the three consecutive number in A.P. is 21 and their products is 231. Find the number.
falakKhan:
thanks u so much
Answers
Answered by
2
the three consecutive terms of an ap are= (x-d),x and (x+d)
ATQ:- x-d+x+x+d=21
=3x=21
x=7
and (x-d)(x)(x+1)=231
(7-d)7(7+d)=231
(7² - d²)=231/7
49-d²=33
d²=49-33
d²=16
d=4; d= -4
if d= 4
no.s are 7+4=11 , 7 , and 7 - 4 = 3..
if d= 4
no.s are 7-4=3 , 7, 7+4=11....
Hope this ans helps u.........
ATQ:- x-d+x+x+d=21
=3x=21
x=7
and (x-d)(x)(x+1)=231
(7-d)7(7+d)=231
(7² - d²)=231/7
49-d²=33
d²=49-33
d²=16
d=4; d= -4
if d= 4
no.s are 7+4=11 , 7 , and 7 - 4 = 3..
if d= 4
no.s are 7-4=3 , 7, 7+4=11....
Hope this ans helps u.........
Answered by
10
Here is your solution
let,
Three numbers be a,a-d,a+d
so a+a-d+a+d = 21
a = 7
A/q
(a)(a-d)(a+d) = 231
(7)(7-d)(7+d) = 231
49- d2 = 231/7
d2 = 16
therefore d = +4 or - 4
so no(s) r : 3, 7 and 10 or 10, 7 and 3
hope it helps you
let,
Three numbers be a,a-d,a+d
so a+a-d+a+d = 21
a = 7
A/q
(a)(a-d)(a+d) = 231
(7)(7-d)(7+d) = 231
49- d2 = 231/7
d2 = 16
therefore d = +4 or - 4
so no(s) r : 3, 7 and 10 or 10, 7 and 3
hope it helps you
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