We can find the number of terms in a geometric progression by using the formula:
Number of terms = (logarithm of the last term over the first term) / (logarithm of the common ratio + 1)
In this case, the first term is 2, the last term is 128, and the common ratio is 4/2 = 2.
So,
Number of terms = (log 128 / log 2) / (log 2 + 1)
Number of terms = (7 / 1) / (1.301 + 1)
Number of terms = 7 / 2.301
Number of terms = 3.036
Therefore, there are approximately 3 terms in the geometric progression 2, 4, 8...,128.
How many terms are there in the geometric progression 2, 4, 8...,128?
3 answers
hmmm
What a lot of gobble-de-goop by the bot !!
we have a GS, where a = 2, r = 2 and term(n) = 128
term(n) = a r^(n-1)
128 = 2(2^(n-1))
2^(n-1) = 64 = 2^6
n - 1 = 6
n = 7
There are 7 terms in the GS
All you had to do it extend the series a few more terms, so easy !!
we have a GS, where a = 2, r = 2 and term(n) = 128
term(n) = a r^(n-1)
128 = 2(2^(n-1))
2^(n-1) = 64 = 2^6
n - 1 = 6
n = 7
There are 7 terms in the GS
All you had to do it extend the series a few more terms, so easy !!
2 answers