If $a, b, c, d$ are in a G.P., then prove that $a + b, b + c, c + d$ , are also in G.P.

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Solution

If $a, b, c, d$ are in G.P then $b = a r$ , $c = a r^{2}$ and $d = a r^{3}$ where $a$ is the first term of G.P and $r$ is the common ratio.

Consider

$(b + c)^{2} = (a r + a r^{2})^{2} = (a r)^{2} + (a r^{2})^{2} + 2 (a r) (a r^{2}) = a^{2} r^{2} + a^{2} r^{4} + 2 a^{2} r^{3} = a^{2} r^{2} (1 + r^{2} + 2 r)$
$= a^{2} r^{2} (1 + r)^{2} . . . . (1)$
$(u s i n g (x + y)^{2} = x^{2} + y^{2} + 2 x y)$

Now consider,

$(a + b) (c + d) = (a + a r) (a r^{2} + a r^{3}) = (a \times a r^{2}) + (a \times a r^{3}) + (a r \times a r^{2}) + (a r \times a r^{3}) = a^{2} r^{2} + a^{2} r^{3} + a^{2} r^{3} + a^{2} r^{4}$
$= a^{2} r^{2} + 2 a^{2} r^{3} + a^{2} r^{4} = a^{2} r^{2} (1 + 2 r + r^{2}) = a^{2} r^{2} (1 + r)^{2} . . . . . (2)$
$(u s i n g (x + y)^{2} = x^{2} + y^{2} + 2 x y)$

From equation 1 and 2, we get

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

Hence, $a + b, b + c, c + d$ are in G.P.

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