Question

Find four consecutive terms in an A.P. such that
the sum of the two middle terms is 24 and
product of the two end terms is 63.​

Answer 1

S O L U T I O N :

Let the four consecutive terms in an A.P. are;

• a - 3d
• a - d
• a + d
• a + 3d

$\underline{\underline{\tt{According\:to\:the\:question\::}}}$

$\underbrace{\bf{1^{st}\:case\::}}$

$\mapsto\tt{(a-d)+(a+d) = 24}$

$\mapsto\tt{a\cancel{-d}+a\cancel{+d} = 24}$

$\mapsto\tt{2a = 24}$

$\mapsto\tt{a= \cancel{24/2}}$

$\mapsto\bf{a = 12}$

$\underbrace{\bf{2^{nd}\:case\::}}$

$\mapsto\tt{(a-3d)\times (a+3d) = 63}$

$\mapsto\tt{a(a + 3d) - 3d(a + 3d) = 63}$

$\mapsto\tt{a^{2} \cancel{+ 3ad - 3ad} - 9d^{2} = 63}$

$\mapsto\tt{a^{2} - 9d^{2} = 63}$

$\mapsto\tt{(12)^{2} -9d^{2} = 63 \:\: [\therefore a= 12]}$

$\mapsto\tt{144 - 9d^{2} = 63}$

$\mapsto\tt{-9d^{2} = 63 - 144}$

$\mapsto\tt{-9d^{2} = -81}$

$\mapsto\tt{d^{2} =\cancel{-81/-9}}$

$\mapsto\tt{d^{2} = 9}$

$\mapsto\tt{d = \sqrt{9}}$

$\mapsto\bf{d = 3}$

Thus;

A R  I T H M E T I C  P R O G R E S S I O N :

$\bullet\sf{a-3d = [12 - 3(3) ] = [12 - 9] = \boxed{\bf{3}}}$

$\bullet\sf{a-d = [12 - 3 ] = \boxed{\bf{9}}}$

$\bullet\sf{a+d = [12 +3] = \boxed{\bf{15}}}$

$\bullet\sf{a+3d = [12 +3(3) ] = [12 + 9] = \boxed{\bf{21}}}$