Question

If a, b, c, d are in G.P. then prove that
i) a+b, b+c, c+d are also in G.P.
ii) (b-c)² + (c-a)² + (d-b)² = (a-d)²

Answer 1

Answer:

Concept:

The general form of G.P is

a, ar, ar², ar³.............

a - first term of GP

r - common ratio

In any GP,$\frac{t_2}{t_1}=\frac{t_3}{t_2}$

1. since a, b, c, d are in G.P,

b=ar, c=ar², d=ar³

Now,

$\frac{b+c}{a+b}=\frac{ar+ar^2}{a+ar}=\frac{r(a+ar)}{a+ar}=r\\\\\frac{c+d}{b+c}=\frac{ar^2+ar^3}{ar+ar^2}=\frac{r(ar+ar^2)}{ar+ar^2}=r$

This implies,

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

By property of G.P,

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

2.

(b-c)² + (c-a)² + (d-b)²

=(ar-ar²)² + (ar²-a)² + (ar³-ar)²

=a²(r-r²)² + a²(r²-1)² + a²(r³-r)²

=a²[(r-r²)² + (r²-1)² + (r³-r)²}

$=a^2[r^2+r^4-2r^3+r^4+1-2r^2+r^6+r^2-2r^4] \\\\=a^2[r^6+1-2r^3] \\\\=a^2(1-r^3)^2\\\\ =[a(1-r^3)]^2\\\\ =[a-ar^3]^2 \\\\=[a-d]^2$