Macam mana nak buat soalan add.maths ni?

The sum of the first and third term of a geometric progression is 2.5. Given that the first term is 0.5, find the possible values of the common ratio, r, of the progression. In the case where r>0, determine the last term of the progressions which is less than 1000.

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1

Jawapan

2014-01-14T22:04:35+08:00
First term = a = 0.5
Third term = a r^{2}

a + a r^{2} = 2.5
0.5 + 0.5 ( r^{2} ) = 2.5
0.5 ( r^{2} ) = 2.5 - 0.5
0.5 ( r^{2} ) = 2.0
( r^{2} ) = 4
r =  \sqrt{4}
r = 2 or -2
Since r > 0 , r = 2

Tn < 1000
ar^{n-1} < 1000

Masukkan a = 0.5 dan r = 2

(0.5) (2)^{n-1} < 1000
 (2)^{n-1}  < 2000
log (2)^{n-1} < 2000
(n-1) log 2 < log 2000
n-1 < log 2000
        --------------
            log 2
n -1 < 10.965
n <  11.965
Therefore, n = 11