Question:
prove that in an infinite G.P. whose common ratio r is numerically less than 1, the ratio of any terms to the sum of all succeeding terms is 1-r
Answer:
Let the GP is:
a, ar, ar2 , ar3 , ................
where a is the first term and r is the common ratio of the GP
Now, tn /(S∞ - Sn ) = (a*rn-1 )/{a/(1-r) - a(1-rn )/(1 - r)}
=> tn /(S∞ - Sn ) = (a*rn-1 )/[a/(1-r)* {1 - (1 - rn )}]
=> tn /(S∞ - Sn ) = (a*rn-1 )/{(a*rn )/(1 - r)}
=> tn /(S∞ - Sn ) = (1 - r)/r
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