If a-d2- a2-a-d2 are in GP- then d2-a2 equals to -a - 0- d - 0-

The correct option is B 2 If x,y,z are in GP xz = y^2 (a d)^2 , a^2 , (a + d)^2 are in GP ⇒ (a^2)^2 = (a d)^2 (a + d)^2 (a^2)^2 = (a^2 d^2)^2 ⇒ a^2 = + ...

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Byju's Answer

Standard XI

Mathematics

Sum of n Terms in GP

If a-d2, a2,a...

Question

If $(a - d)^{2}$ , $a^{2}$ , $(a + d)^{2}$ are in GP, then $\frac{d^{2}}{a^{2}}$ equals to [a ≠ 0,d ≠ 0]

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Solution

The correct option is B
2

If x,y,z are in GP

xz = $y^{2}$

$(a - d)^{2}$ , $a^{2}$ , $(a + d)^{2}$ are in GP

$\Rightarrow$ $(a^{2})^{2}$ = $(a - d)^{2}$ $(a + d)^{2}$

$(a^{2})^{2}$ = $(a^{2} - d^{2})^{2}$

$\Rightarrow$ $a^{2}$ = $¯ ¯¯ ¯ +$ $(a^{2} - d^{2})$

$a^{2}$ = $a^{2} - d^{2}$ Or $a^{2}$ = - $a^{2} + d^{2}$

$\Rightarrow$ d = 0 [d ≠ 0] or 2 $a^{2}$ = $d^{2}$

$\Rightarrow$ $\frac{d^{2}}{a^{2}}$ = 2

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If (a−d)2 , a2 , (a+d)2 are in GP, then d2a2 equals to [a ≠ 0,d ≠ 0]

The correct option is B 2 If x,y,z are in GP xz = y2 (a−d)2 , a2 , (a+d)2 are in GP ⇒ (a2)2 = (a−d)2 (a+d)2 (a2)2 = (a2−d2)2 ⇒ a2 = ¯¯¯¯+ (a2−d2) a2 = a2−d2 Or a2 = -a2+d2 ⇒ d = 0 [d ≠ 0] or 2a2 = d2 ⇒ d2a2 = 2

If $(a - d)^{2}$ , $a^{2}$ , $(a + d)^{2}$ are in GP, then $\frac{d^{2}}{a^{2}}$ equals to [a ≠ 0,d ≠ 0]