write the four first terms of the geometric sequence satisfying the following conditions a2=11and a3=-121
The common ratio r of the above GP i= a3/a2 = -121/11 = -11 a1 = a2/r = 11/-11 = -1 a4 = -121(-11) = 1331
so the terms are -1, 11, -121 and 1331.
thank you have helped me so much i will be asking further questions
Hint: A geometric series has a constant which is found by dividing any of its two consecutive terms.
So the common ratio of the above GP is given by a3/a2 which is -11.
so a2/a1 = -11
11/a1 = -11
11 = -11a
a = - 1.
consequently, a4/a3 = -11
a4/-121 = -11
a4 = -11 x -121
a4 = 1331
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