Question

find the four numbers in ap whose sum is 6 and product of whose extreme is 10 times the product of the two intermediate terms

Answer 1

$\text{Let the four numbers in A.P be}$

$a-3d,a-d,a+d,a+3d$

$\textbf{Given:}$

$\text{Sum of the four numbers is 6 and}$

$\text{Product of whose extreme is 10 times the product of the two intermediate terms}$

$\textbf{To find:}$

$\text{The four numbers}$

$\textbf{Solution:}$

$\text{Sum of the four numbers is 6}$

$\implies\,a-3d+a-d+a+d+a+3d=6$

$\implies\,4a=6$

$\implies\bf\,a=\dfrac{3}{2}$

$\text{Also,}$

$(a-3d)(a+3d)=10(a-d)(a+d)$

$a^2-9d^2=10(a^2-d^2)$

$a^2-9d^2=10\,a^2-10\,d^2)$

$10\,d^2-9d^2=10\,a^2-a^2$

$d^2=9(\dfrac{9}{4})$

$d^2=\dfrac{81}{4}$

$\implies\bf\,d=\pm\dfrac{9}{2}$

$\text{when}\;d=\dfrac{9}{2}$

$\text{The four numbers are}$

$\dfrac{3}{2}-\dfrac{27}{2},\,\dfrac{3}{2}-\dfrac{9}{2},\,\dfrac{3}{2}+\dfrac{9}{2},\,\dfrac{3}{2}+\dfrac{27}{2}$

$\bf\,-12,-3,6,15$

$\text{when}\;d=\dfrac{-9}{2}$

$\text{The four numbers are}$

$\dfrac{3}{2}+\dfrac{27}{2},\,\dfrac{3}{2}+\dfrac{9}{2},\,\dfrac{3}{2}-\dfrac{9}{2},\,\dfrac{3}{2}-\dfrac{27}{2}$

$\bf\,15,6,-3,-12$

$\textbf{Find more:}$

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3.AP given that the first term (a) = 54, the common difference

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4.If 8 times the 8th term of an AP is equal to 15 times its 15th term then find the 23rd term.

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