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If a,b,c are in A.P., then prove that the following are also in A.P (i) a^(2)(b+c),b^(2)(c+a),c^(2)(a+b) ltbr gt(ii) 1/(sqrtb+sqrtc),1/(sqrtc+sqrta),1/(sqrta+sqrtb) (iii) a(1/b+1/c),b(1/c+1/a),c(1/a+1/b) |
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Answer» SOLUTION :Let `a^(2)(b+C),b^(2)(c+a),c^(2)(a+b)` are in A.P. Then, `b^(2)(c+a)-a^(2)(b+c)=c^(2)(a+b)-b^(2)(c+a)` or `c(b^(2)-a^(2))+ab(b-a)=a(c^(2)-b^(2))+BC(c-b)` or `(b-a)(ab+bc+ca)=(c-b)(ab+bc+ca)` or b-a=c-b or 2b=a+c Thus, a,b,c are in A.P., which is given. (ii) Let `1/(sqrtb+sqrtc),1/(sqrtc+sqrta),1/(sqrta+sqrtb)` are in A.P.then, `1/(sqrtc+sqrta)-1/(sqrtb+sqrtc)=1/(sqrta+sqrtb)-1/(sqrtc+sqrta)` or `(sqrtb-sqrta)/((sqrtc+sqrta)(sqrtb+sqrtc))=((sqrtc-sqrtb))/((sqrta+sqrtb)(sqrtc+sqrta))` or `(sqrtb-sqrta)/((sqrtc+sqrta)(sqrtb+sqrtc))=((sqrtc-sqrtb))/((sqrta+sqrtb)(sqrtc+sqrta))` or `(sqrtb-sqrta)/(sqrtb+sqrtc)=(sqrtc-sqrtb)/(sqrta+sqrtb)` or b-a=c-b or 2b=a+c Thus, a,b,c are in A.P., which is given. Hence, `1/(sqrtb+sqrtc),1/(sqrtc+sqrta),1/(sqrta+sqrtb)` are in A.P. (iii) a,b,c are in A.P. Then, `a/(abc),b/(abc),c/(abc)` are in A.P. [On DIVIDING each term by abc] `rArr1/(bc),1/(ca),1/(ab)` are in A.P. `rArr(ab+bc+ca)/(bc),(ab+bc+ca)/(ca),(ab+bc+ca)/(ab)` are in A.P. [On multiplying each termby ab+bc+ca] `rArr(ab+bc+ca)/(bc)-1,(ab+bc+ca)/(ca)-1,(ab+bc+ca)/(ab)-1` are in A.P. [On adding -1 to each term] `rArr(ab+ac)/(bc),(ab+bc)/(ca),(bc+ca)/(ab)` are in A.P. `rArra(1/b+1/c),b(1/c+1/a),c(1/a+1/b)` are in A.P. |
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