Question

Question :-
⠀
☆ If a, b, c and d are in G.P. show that
⠀
$\bf\to{(a^2 + b^2 + c^2) (b^2 + c^2 + d^2) = (ab + bc + cd)^2}$
⠀
⚠️No spam⚠️
⠀
♦ Quality answer ✅
♦ Explanation ✅
♦ Copy paste ❌​

Answer 1

Topic :-

Geometric Progression

Given :-

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

To Prove :-

$(a^2+b^2+c^2)(b^2+c^2+d^2)=(ab+bc+cd)^2$

Solution :-

Assume 't' as first term and 'r' as common multiple of the GP.

nth term of GP is given by,

$a_n=ar^{n-1}$

where

$a_n=nth\:\:term$

a = First Term

r = Common multiple

So,

$a_1=tr^{1-1}=tr^0=t=a$

$a_2=tr^{2-1}=tr^1=tr=b$

$a_3=tr^{3-1}=tr^2=c$

$a_4=tr^{4-1}=tr^3=d$

Solving LHS,

$(a^2+b^2+c^2)(b^2+c^2+d^2)$

$(t^2+(tr)^2+(tr^2)^2)((tr)^2+(tr^2)^2+(tr^3)^2)$

$(t^2+t^2r^2+t^2r^4)(t^2r^2+t^2r^4+t^2r^6)$

$t^2(1+r^2+r^4).t^2r^2(1+r^2+r^4)$

$t^4r^2(1+r^2+r^4)^2$

Solving RHS,

$(ab+bc+cd)^2$

$(t.tr+tr.tr^2+tr^2.tr^3)^2$

$(t^2r+t^2r^3+t^2r^5)^2$

$(t^2r(1+r^2+r^4))^2$

$(t^2r)^2(1+r^2+r^4)^2$

$t^4r^2(1+r^2+r^4)^2$

We observe that LHS = RHS.

Hence, Proved !!

Answer 2

Given :-

• a, b, c, d are in G. P.

So,

Let

• first term of G.P. is 'a'.

and

• Common ratio is 'r'.

Let

$\bull \sf \: b \: = \: ar$

$\bull \sf \: c \: = \: a {r}^{2}$

$\bull \sf \: d \: = \: a {r}^{?}$

Consider,

$\rm :\implies\: \bf \: LHS$

$\rm :\implies\:( {a}^{2} + {b}^{2} + {c}^{2} )( {b}^{2} + {c}^{2} + {d}^{2} )$

On substituting the values of b, c, and d, we get

$\rm :\implies\: \bigg( {a}^{2} + {(ar)}^{2} + {(a {r}^{2} )}^{2} \bigg) \bigg( {(ar)}^{2} + {(a {r}^{2}) }^{2} + {a {r}^{3} )}^{2} \bigg)$

$\rm :\implies\:( {a}^{2} + {a}^{2} {r}^{2} + {a}^{2} {r}^{4} )( {a}^{2} {r}^{2} + {a}^{2} {r}^{4} + {a}^{2} {r}^{4} )$

$\rm :\implies\: {a}^{2} (1 + {r}^{2} + {r}^{4} ){a}^{2} {r}^{2} (1 + {r}^{2} + {r}^{4} )$

$\rm :\implies\: {a}^{4} {r}^{2} { \bigg(1 + {r}^{2} + {r}^{4} \bigg)}^{2}$

$\rm :\implies\: { \bigg(r {a}^{2} (1 + {r}^{2} + {r}^{4} )\bigg)}^{2}$

$\rm :\implies\: { \bigg( {a}^{2} {r}^{2} + {a}^{2} {r}^{3} + {a}^{2} {r}^{5} \bigg)}^{2}$

$\rm :\implies\: { \bigg(a.ar + ar. {ar}^{2} + {ar}^{2} .{ar}^{3} \bigg)}^{2}$

$\rm :\implies\: { \bigg(ab \: + \: bc \: + \: cd \bigg)}^{2}$

$\rm :\implies\: \bf \: RHS$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$