In a geometric series, the sum of the first 10 numbers of the series is 10, and the sum of the first 30 numbers of the series is 70. Find the sum of the first 40 numbers of the series.
+23252 The formula for the first n terms of a geometric series is: Sum = a · (1 - rn) / (1 - r)
where a is the first term and r is the common ratio.
We have: 10 = a · (1 - r10) / (1 - r) and 70 = a · (1 - r30) / (1 - r)
combining: [ a · (1 - r30) / (1 - r) ] / [ a · (1 - r10) / (1 - r) ] = 70 / 10
reducing: (1 - r30) / (1 - r10) = 7
rewriting: ( r30 - 1 ) / ( r10 - 1 ) = 7
If we divide ( r30 - 1 ) by ( r10 - 1 ) we get r20 + r10 + 1
so: r20 + r10 + 1 = 7
and: r20 + r10 - 6 = 0
Let x = r10 ---> x2 = r20
so: r20 + r10 - 6 = 0 ---> x2 + x - 6 = 0 ---> (x + 3)(x - 2) = 0
x can't be negative, so x = 2 ---> r10 = 2 ---> r = 21/10 ---> r = 1.071773463...
Since a · (1 - r10) / (1 - r) = 10 ---> a · (1 - 1.07177346310) / (1 - 1.071773463) = 10
Solving: a = 0.7177346254 [ or a = -10(1 - 21/10) ]
To find the sum of the first 40 numers: Sum = a · (1 - r40) / (1 - r) = 150
