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7. If the 4th and 9th terms of à G.P. De anu l8. The 3rd term of a G.P. is the square of the first and 5th term is 64. Find the G.P0 0 and 2560 respectively |
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Answer» The third term of a GP = square of its firsttermar^2 = (a)^2r^2 = a^2÷ar^2 = a .......(1)now it is given that 5th term is 64 so ar^4 = 64a(r^2)^2 = 64from eq (1)a(a)^2 = 64a^3 = 64a = 4now from eq (1) r^2 = a r = a^(1/2)r. = 4^(1/2)r. = 2now we have a and r so we can calculate the first six terms of the GP.The first six terms if the GP are a, ar, ar^2, ar^3, ar^4, ar^5now putting the value if a and r we have first six terms of the GPSo the terms are 4, 8, 16, 32, 64, 128 |
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