Question

if a,b,c,d are in geometric sequences then show.that (a-b+c)(b+c+d)=ab+bc+cd

Answer 1

If r is the ratio of the GP, then

a = a
b = ar
c = ar^2 and
d = ar^3

Therefore, substituting the value of b, c and d in (a-b+c)(b+c+d), we get

(a-ar+ar^2)(ar+ar^2+ar^3)

a is the common factor in the first bracket, while ar is the common factor in the second bracket. Taking the common factor out, we get

a(1-r+r^2)ar(1+r+r^2) or

a^2r(1-r+r^2)(1+r+r^2)

Expanding the brackets we get

a^2r(1+r+r^2-r-r^2-r^3+r^2+r^3+r^4)

a^2r(1+r^2+r^4)

This gives a^2r+a^2r^3+a^2r^5

Or a * ar + ar * ar^2+ ar^2*ar^3

but b = ar, c = ar^2 and d = ar^3. Substituting these in the last equation, we get

a*b+b*c+c*d or ab+bc+cd

Thus, if a, b, c and d are in GP, then (a-b+c)(b+c+d) = ab+bc+cd.

Thus, proved.

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