A) H.P.
B) G.P.
C) A.P.
D) None of these
Correct Answer: C
Solution :
\[{{a}^{2}},\ {{b}^{2}},\ {{c}^{2}}\] are in A.P. Therefore \[{{a}^{2}}+(ab+bc+ca)\], \[{{b}^{2}}+(ab+bc+ca)\], \[{{c}^{2}}+(ab+bc+ca)\] will be in A.P. \[\Rightarrow \]\[\{a(a+b)+c\,(a+b)\},\ \{b(b+a)+c\,(b+a)\},\ \]\[c(c+b)+a(b+c)\]will be in A.P. Þ\[(a+b)(a+c),\ (b+a)(b+c),\ (c+a)(c+b)\]will be in A.P. \[\Rightarrow \]\[\frac{1}{b+c},\ \frac{1}{c+a},\ \frac{1}{a+b}\] will be in A.P.{Dividing each term by \[f(n+3)+3f(n+1)=f(n)+3f(n+2)\]}
You need to login to perform this action.
You will be redirected in
3 sec
