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Question: If a, b, c and d are in G.P. show that \[{\rm{(}}{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2})({{\rm{b...

If a, b, c and d are in G.P. show that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2{\rm{(}}{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2})({{\rm{b}}^2} + {{\rm{c}}^2} + {{\rm{d}}^2}) = {({\rm{ab + bc + cd)}}^2}

Explanation

Solution

Here in this question, we will compare a, b, c, d with the basic G.P. equation and then substitute their value in the given equation. We will then expand both sides of equation one by one to find proof that the left side of the given equation is equal to the right side.

Complete step by step solution:
As we all know that the basic G.P. equation is a,ar,ar2,ar3,ar4,.........{\rm{a}},{\rm{ar}},{\rm{a}}{{\rm{r}}^2},{\rm{a}}{{\rm{r}}^3},{\rm{a}}{{\rm{r}}^4},.........
Where, a is the first term of the G.P. and r is the common ratio.
Here it is given that a, b, c, d are in G.P.
Therefore, a=a{\rm{a}} = {\rm{a}}
b=ar{\rm{b}} = {\rm{ar}}
c=ar2{\rm{c}} = {\rm{a}}{{\rm{r}}^2}
d=ar3{\rm{d}} = {\rm{a}}{{\rm{r}}^3}
Now we need to show that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2{\rm{(}}{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2})({{\rm{b}}^2} + {{\rm{c}}^2} + {{\rm{d}}^2}) = {({\rm{ab + bc + cd)}}^2}
First we will take LHS of the equation and put the values of a, b, c, d in it
After putting the values of a, b, c, d we get
(a2+b2+c2)(b2+c2+d2)=(a2+(ar)2+(ar2)2)((ar)2+(ar2)2+(ar3)2)=(a2+a2r2+a2r4)(a2r2+a2r4+a2r6)\begin{array}{l}{\rm{(}}{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2})({{\rm{b}}^2} + {{\rm{c}}^2} + {{\rm{d}}^2}) = {\rm{(}}{{\rm{a}}^2} + {({\rm{ar)}}^2} + {({\rm{a}}{{\rm{r}}^2}{\rm{)}}^2})({({\rm{ar}})^2} + {{\rm{(a}}{{\rm{r}}^2}{\rm{)}}^2} + {({\rm{a}}{{\rm{r}}^3})^2})\\\\{\rm{ = (}}{{\rm{a}}^2} + {{\rm{a}}^2}{{\rm{r}}^2} + {{\rm{a}}^2}{{\rm{r}}^4})({{\rm{a}}^2}{{\rm{r}}^2} + {{\rm{a}}^2}{{\rm{r}}^4} + {{\rm{a}}^2}{{\rm{r}}^6})\end{array}

Now by simplification, we get
(a2+b2+c2)(b2+c2+d2)=(a2(1+r2+r4))(a2r2(1+r2+r4)) =(a2)(a2r2)(1+r2+r4)(1+r2+r4) =a4r2(1+r2+r4)2\begin{array}{l}{\rm{(}}{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2})({{\rm{b}}^2} + {{\rm{c}}^2} + {{\rm{d}}^2}) = {\rm{(}}{{\rm{a}}^2}(1 + {{\rm{r}}^2} + {{\rm{r}}^4}))({{\rm{a}}^2}{{\rm{r}}^2}(1 + {{\rm{r}}^2} + {{\rm{r}}^4}))\\\ = {\rm{(}}{{\rm{a}}^2}{\rm{)(}}{{\rm{a}}^2}{{\rm{r}}^2})(1 + {{\rm{r}}^2} + {{\rm{r}}^4})(1 + {{\rm{r}}^2} + {{\rm{r}}^4})\\\ = {{\rm{a}}^4}{{\rm{r}}^2}{(1 + {{\rm{r}}^2} + {{\rm{r}}^4})^2}\end{array}

Now taking RHS of the equation and putting the values of a, b, c, d in it.
After putting the values of a, b, c, d , we get
(ab+bc+cd)2=(a×ar+ar×ar2+ar2×ar3)2=(a2r+a2r3+a2r5)2{({\rm{ab}} + {\rm{bc}} + {\rm{cd)}}^2} = {({\rm{a}} \times {\rm{ar}} + {\rm{ar}} \times {\rm{a}}{{\rm{r}}^2} + {\rm{a}}{{\rm{r}}^2} \times {\rm{a}}{{\rm{r}}^3}{\rm{)}}^2} = {({{\rm{a}}^2}{\rm{r}} + {{\rm{a}}^2}{{\rm{r}}^3} + {{\rm{a}}^2}{{\rm{r}}^5}{\rm{)}}^2}

Now by simplification, we get
(ab+bc+cd)2=(a2r(1+r2+r4))2=a4r2(1+r2+r4)2{({\rm{ab}} + {\rm{bc}} + {\rm{cd)}}^2} = {({{\rm{a}}^2}{\rm{r(}}1 + {{\rm{r}}^2} + {{\rm{r}}^4}{\rm{))}}^2} = {{\rm{a}}^4}{{\rm{r}}^2}{(1 + {{\rm{r}}^2} + {{\rm{r}}^4})^2}
We can clearly see that Left hand side of the equation is equal to the right hand side.

Hence we proved that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2{\rm{(}}{{\rm{a}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2})({{\rm{b}}^2} + {{\rm{c}}^2} + {{\rm{d}}^2}) = {({\rm{ab + bc + cd)}}^2}.

Note:
We always have to keep in mind the basic A.P. and G.P series.
A.P. series is a,a+d,a+2d,a+3d,............{\rm{a}},{\rm{a}} + {\rm{d}},{\rm{a}} + 2{\rm{d}},{\rm{a}} + 3{\rm{d,}}............
nthterm=a+(n−1)d{{\rm{n}}^{{\rm{th}}}}{\rm{term }} = {\rm{ a}} + \left( {{\rm{n}} - 1} \right){\rm{d}}
Where a is the first term and d is the common difference
G.P. series is a,ar,ar2,ar3,ar4,.........{\rm{a}},{\rm{ar}},{\rm{a}}{{\rm{r}}^2},{\rm{a}}{{\rm{r}}^3},{\rm{a}}{{\rm{r}}^4},.........
nthterm=arn−1{{\rm{n}}^{{\rm{th}}}}{\rm{term}} = {\rm{a}}{{\rm{r}}^{{\rm{n}} - 1}}
Where, a is the first term of the G.P. and r is the common ratio.