If a, b, c, and d are in G. P., then $(a + b)^{2}$ , $(b + c)^{2}$ , and $(c + d)^{2}$ are also in G. P.

True

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False

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Solution

The correct option is A
True

Since a, b, c, and d are in G. P

∴ $\frac{b}{a}$ = $\frac{a}{b}$ = $\frac{d}{c}$

∴ $b^{2}$ = $a c$ , $c^{2}$ = $b d$ , $a d$ = $b c$ ...........(1)

Now, $(a + b)^{2} \times (c + d)^{2}$ = $(a c + b c + a d + b d)^{2}$

= $(b^{2} + c^{2} + 2 b c)^{2}$ ...[Using (1)]

= $((b + c)^{2})^{2}$

∴ $\frac{(c + d)^{2}}{(b + c)^{2}}$ = $\frac{(b + c)^{2}}{(a + b)^{2}}$

Thus, $(a + b)^{2}, (b + c)^{2}, (c + d)^{2}$ are in G. P

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