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(i) The first term of a G.P. is 160 and the common ratio is . How many consecutive tould be taken to give a sum of 2110?2 |
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Answer» S = ar + ar^2 + ar^3 + ... + ar^nSr = ar^2 + ar^3 + ar^4 + ... + ar^(n + 1)Sr - S = ar^(n + 1) - arS * (r - 1) = ar * (r^n - 1)S = ar * (1 - r^n) / (1 - r) S is the sumar is the initial termr is the common ration is the number of terms S = 2110ar = 160r = 2/3n = ? Let's see if this will even ever sum to 2110 S = 160 * (1 - (2/3)^inf) / (1 - 2/3)S = 160 * 3 * (1 - 0)S = 480 This will never sum to 2110. Are you sure you copied this correctly |
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