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Geometric sequence problem: prove that if a1, a2, a3, a4,... is a geometric sequence? show work please?

  • then try to prove (a1)^2, (a2)^2, (a3)^2, (a4)^2 is also a geometric sequence. Show work please??? I'm not skipping all my homework, just two problems.

  • Answer:

    a2 = ra1 a3 = ra2 = r²a1 an = r^(n-1)a1 a2² = (ra1)² = r²a1² a3² = (r²a1)² = (r^4)a1² = (r²)²a1² an² = r^[2(n-1)]a² = (r²)^(n-1)a1² a1², a2², a3², ..., an² is a geometric sequence with the first term = a1² and common ratio = r²

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a_1, a_2, a_3, a_4 --------are in Geometric sequence ==> a_2 = a_1*r, a_3 = a_1*r^2 ....... (1) evaluate (a_2)^2 /(a_1)^2 and (a_3)^2 /(a_2)^2 using relation from equation (1) (a_2)^2 /(a_1)^2 = (a_1*r)^2 /(a_1)^2 = r^2 (a_3)^2 /(a_2)^2 = (a_1*r ^2)^2 /(a_1*r)^2 = r^2 since (a_2)^2 /(a_1)^2 = (a_3)^2 /(a_2)^2 (a_1)^2, (a_2)^2, (a_3)^2 ------ are also in geometric sequence

mohanrao d

if a1,a2,a3,a4.... is a geometric sequence , then a2/a1 = a3/a2 = a4/a3 ...= an/a(n-1) ...= R=radius (a2^2)/(a1^2) =(a2/a1)^2 =(a3^2)/(a2^2)= (a3/a2)^2= (a4^2)/(a3/a2)^2 = .....=R^2; SO a1^2,a2^2,a3^2......,an^2...is also a geometric sequence

Marcos

I'm not much at proofs, but this is pretty intuitive. 2, 4, 8 and 4, 16, 64 are both geometric sequences. Clearly, when you square all the terms in a geometric sequence with r=2, you also square the r value.

Infinity

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