Question

sum of three numbers in AP is 27 and product of the first and the last is 56find the numbers​

Answer 1

SOLUTION :

$\bigstar$ Firstly, let the three numbers in A.P. are;

• a-d
• a
• a+d

$\underbrace{\sf{According\:to\:the\:question\::}}}}$

$\longrightarrow\sf{a\cancel{-d}+a+a\cancel{+d}=27}\\\\\longrightarrow\sf{3a=27}\\\\\longrightarrow\sf{a=\cancel{27/3}}\\\\\longrightarrow\bf{a=9}$

&

$\longrightarrow\sf{(a-d)(a+d)=56}\\\\\longrightarrow\sf{a^{2} \cancel{+ad-ad}-d^{2} =56}\\\\\longrightarrow\sf{a^{2} -d^{2} =56}\\\\\longrightarrow\sf{(9)^{2} -d^{2} =56\:\:\:[\therefore a=9]}\\\\\longrightarrow\sf{81-d^{2} =56}\\\\\longrightarrow\sf{-d^{2} =56-81}\\\\\longrightarrow\sf{\cancel{-}d^{2}= \cancel{-}25}\\\\\longrightarrow\sf{d^{2}=25}\\\\\longrightarrow\sf{d=\sqrt{25} }\\\\\longrightarrow\bf{d=5}$

$\boxed{\bf{Three\:Numbers\:in\:A.P.}}}}$

$\bullet\:\sf{a-d=9-5=\boxed{\bf{4}}}}\\\\\bullet\:\sf{a=\boxed{\bf{9}}}}\\\\\bullet\:\sf{a+d=9+5=\boxed{\bf{14}}}}$

Answer 2

Answer

4,9,14...

Solution

Let three number in Ap =

a - d, a, a + d

Given

sum of three number in Ap = 27

=> a - d + a + a + d = 27

=> 3a = 27

=> a = 9

Products of first and last number is = 56

=> ( a - d) ( a + d) = 56

=> a^2 - d^2 = 56

=> 81 - d^2 = 56

=> - d^2 = - 25

=> d = 5

THE NUMBER ARE:-

=> a - d = 9 - 5 = 4

=> a = 9

=> a + d = 9 + 5 = 14

Hence, number are

=> 4,9,14.....