If the 5th term of both arithmetic series and geometric serioes are 11 and 243 respectively. Let the common difference be 2. What is the sum of the first ten terms of both arithmetic series and geometric series?
1 Answer
The sum of the arithmetic sequence is
The sum of the geometric sequence is
Explanation:
Arithmetic sequence
An arithmetic sequence is a sequence of numbers with common difference between consecutive terms.
The general term of an arithmetic sequence can be written:
#a_n = a + d(n-1)#
where
The sum of the first
#sum_(n=1)^N a + d(n-1) = Na + 1/2dN(N-1)#
This is essentially
In our example, we are told
Hence:
#a = a_5 - 4d = 11 - 4(2) = 3#
and:
#sum_(n=1)^10 a_n = 10(3)+1/2(2)(10)(10-1) = 30+90 = 120#
Geometric sequence
A geometric sequence is a sequence of numbers with common ratio between consecutive terms.
The general term of a geometric sequence can be written:
#a_n = ar^(n-1)#
where
The sum of the first
#sum_(n=1)^N ar^(n-1) = (a(1-r^N))/(1-r)#
To see why consider:
#(1-r) sum_(n=1)^N ar^(n-1) = sum_(n=1)^N ar^(n-1) - r sum_(n=1)^N ar^(n-1)#
#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = sum_(n=1)^N ar^(n-1) - sum_(n=2)^(N+1) ar^(n-1)#
#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a+color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1)))) - ar^N#
#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a-ar^N#
#color(white)((1-r) sum_(n=1)^N ar^(n-1)) = a(1-r^N)#
Then dividing both ends by
#sum_(n=1)^N ar^(n-1) = (a(1-r^N))/(1-r)#
In our example, I will assume that the question should have specified that the common ratio is
#a_5 = 243" "# and#" "r = 3#
Hence:
#a = a_5/r^4 = 243/81 = 3#
Then:
#sum_(n=1)^10 a_n = (a(1-r^10))/(1-r) = (3(1-3^10))/(1-3) = (3(1-59049))/(-2) = 88572#