Question

The sumof 4 condective term of ap is 32 and the ratio of the product of the first &the last term to the product of 2 middle term is 7:5 .find the number

Answer 1

Correct Question :

The sum of 4 consecutive terms of an A.P. is 32 & the ratio of the product of the first & last term of the product of two middle term is 7:15. Find the number.

Solution :

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

• (a-3d) , (a-d) , (a+d) ,(a+3d)

A/q

$\longrightarrow\sf{(a-3d)+(a-d)+(a+d)(a+3d) = 32}\\\\\longrightarrow\sf{a-3d + a-d+a+d+a+3d = 32}\\\\\longrightarrow\sf{4a \cancel{-3d+3d - d + d }= 32}\\\\\longrightarrow\sf{4a=32}\\\\\longrightarrow\sf{a=\cancel{32/4}}\\\\\longrightarrow\bf{a=8}$

Now;

$\longrightarrow\sf{(a-3d) (a+3d) : (a-d)(a+d) = 7:15}\\\\\longrightarrow\sf{\dfrac{(a-3d)(a+3d)}{(a-d)(a+d)} =\dfrac{7}{15} }\\\\\\\longrightarrow\sf{\dfrac{a^{2} \cancel{+ 3ad -3ad} -9d^{2}}{a^{2} \cancel{+ad - ad }- d^{2}} =\dfrac{7}{15} }\\\\\\\longrightarrow\sf{\dfrac{a^{2} -9d^{2} }{a^{2} -d^{2} } =\dfrac{7}{15} }\\\\\longrightarrow\sf{15(a^{2}-9d^{2})=7(a^{2} -d^{2}) }\\\\\longrightarrow\sf{15a^{2} -135d^{2}=7a^{2} -7d^{2}}\\\\\longrightarrow\sf{15a^{2} -7a^{2} =-7d^{2} + 135d^{2}}$

$\longrightarrow\sf{8a^{2} =128d^{2} }\\\\\longrightarrow\sf{8(8)^{2} =128d^{2}\:\:[\therefore a =8] }\\\\\longrightarrow\sf{8\times 64 = 128d^{2} } \\\\\longrightarrow\sf{512 = 128d^{2} }\\\\\longrightarrow\sf{d^{2} = \cancel{512/128}}\\\\\longrightarrow\sf{d^{2} = 4}\\\\\longrightarrow\sf{d=\sqrt{4} }\\\\\longrightarrow\bf{d=\pm 2}$

$\boxed{\bf{Arithmetic\:progression\:Numbers\::}}}}$

For using +ve common difference :

$\bullet\:\sf{(a-3d) = [8-3(2) ]=[8-6]=\boxed{\bf{2}}}\\\\\bullet\sf{(a-d) = [8-2 ]=\boxed{\bf{6}}}\\\\\bullet\sf{(a+d) = [8+2 ]=\boxed{\bf{10}}}\\\\\bullet\sf{(a+3d) = [8+3(2) ]=[8+6]=\boxed{\bf{14}}}$

For using -ve common difference :

$\bullet\:\sf{(a-3d) = [8-3(-2) ]=[8-(-6)]=[8+6]=\boxed{\bf{14}}}\\\\\bullet\sf{(a-d) = [8-(-2) ]=[8+2]=\boxed{\bf{10}}}\\\\\bullet\sf{(a+d) = [8+(-2 )]=[8-2]=\boxed{\bf{6}}}\\\\\bullet\sf{(a+3d) = [8+3(-2) ]=[8-6]=\boxed{\bf{2}}}$