Question

If $a,b,c,d$ are in a geometric sequence, then show that $(a-b+c)(b+c+d)=ab+bc+cd$

Answer 1

We are given $a,b,c,d$ are in G.P.

$\Rightarrow b^2=ac$ ....(i)

and $c^2=bd$ .....(ii)

Now given statement

$(a-b+c)(b+c+d)=ab+ac+ad-b^2-bc-bd+cb+c^2+cd$

Using (i) and (ii)

$(a-b+c)(b+c+d)=ab+ad+cd$ .....(iii)

Multiply (i) and (ii)

$\Rightarrow bc=ad$

Using this in $bc=ad$ in equation (iii)

$(a-b+c)(b+c+d)=ab+bc+cd$

Hence, shown.