If `a ,b ,c ,d`are in G.P. prove that:`(a^2+b^2),(b^2+c^2),(c^2+d^2)`are inG.P.`(a^2-b^2),(b^2-c^2),(c^2-d^2)`are inG.P.`1/(a^2+b^2),1/(b^2+c^2),1/(c^2+d^2)`are inG.P.`(a^2+b^2+c^2),(a b+b c+c d),(b^2+c^2+d^2)`

If `a ,b ,c ,d`are in G.P. prove that:`(a^2+b^2),(b^2+c^2),(c^2+d^2)`a

1.	If `a ,b ,c ,d`are in G.P. prove that:`(a^2+b^2),(b^2+c^2),(c^2+d^2)`are inG.P.`(a^2-b^2),(b^2-c^2),(c^2-d^2)`are inG.P.`1/(a^2+b^2),1/(b^2+c^2),1/(c^2+d^2)`are inG.P.`(a^2+b^2+c^2),(a b+b c+c d),(b^2+c^2+d^2)`
Answer» Let `a` is the first term and `r` is the common ratio. Then, `b = ar, c= ar^2,d = ar^3` `a^2+b^2 = a^2+a^2r^2 = a^2(1+r^2)` `b^2+c^2 = a^2r^2+a^2r^4 = a^2r^2(1+r^2)` `c^2+d^2 = a^2r^4+a^2r^6 = a^2r^4(1+r^2)` So, `a^2+b^2,b^2+c^2 and c^2+d^2` are in G.P. with first term `a^2(1+r^2)` and common ratio `r^2`. Now, `a^2-b^2 = a^2-a^2r^2 = a^2(1-r^2)` `b^2-c^2 = a^2r^2-a^2r^4 = a^2r^2(1-r^2)` `c^2-d^2 = a^2r^4-a^2r^6 = a^2r^4(1-r^2)` So, `a^2-b^2,b^2-c^2 and c^2-d^2` are in G.P. with first term `a^2(1-r^2)` and common ratio `r^2`. Now, `1/(a^2+b^2) = 1/(a^2+a^2r^2) = 1/(a^2(1+r^2))` `1/(b^2+c^2) = 1/(a^2r^2+a^2r^4) = 1/(a^2r^2(1+r^2))` `1/(c^2+d^2) = 1/(a^2r^4+a^2r^6) = 1/(a^2r^4(1+r^2))` So, `1/(a^2+b^2),1/(b^2+c^2) and 1/(c^2+d^2)` are in G.P. with first term `1/(a^2(1+r^2))` and common ratio `1/r^2`. Now, `a^2+b^2+c^2 = a^2+a^2r^2+a^2r^4 = a^2(1+r^2+r^4)` `ab+bc+cd = a(ar)+ar(ar^2)+ar^2(ar^3) = a^2r(1+r^2+r^4)` `b^2+c^2+d^2 = a^2r^2 + a^2r^4+a^2r^6 = a^2r^2(1+r^2+r^4)` So, `a^2+b^2+c^2,ab+bc+cd , b^2+c^2+d^2` are in G.P. with first term `a^2(1+r^2+r^4)` and common ratio `r`.