Question

if the sum of three numbers which are in A.P is 27 and the product of first and last is 77, then the numbers are

Answer 1

$let \: a \: b \: \: and \: c \: are \: in \: ap$
then
$a \times c= 77$
let common difference is d then
b-d+b+b+d=27
imply that b=9
now (b-d)(b+d)=77
imply that :-
$b {}^{2} - d ^{2} = 77$
putting b =9, we get
$81 - d ^{2} = 77 \\ d ^{2} = 4 \\ d = + 2or - 2$
Now required AP is b-d,b,&b+d
put d=+2&-2
we get,AP as 7,9&11 or 11,9&7.