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
Feb 4, 2018

The sum of the arithmetic sequence is #120#

The sum of the geometric sequence is #88572# (assuming a common ratio #3#)

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 #a# is the initial term and #d# the common difference.

The sum of the first #N# terms of an arithmetic sequence is given by the formula:

#sum_(n=1)^N a + d(n-1) = Na + 1/2dN(N-1)#

This is essentially #N# times the value of the average of the first and last element.

In our example, we are told #a_5 = 11# and #d = 2#

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 #a# is the initial term and #r# the common ratio.

The sum of the first #N# terms of a geometric sequence is given by the formula:

#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 #(1-r)# we get:

#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 #3# - since that fits with the data given. We are given:

#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#