Question #e4948

2 Answers
Dec 18, 2016

Answer:

#970#

Explanation:

For the standard #color(blue)"arithmetic sequence"#

#a,a+d,a+2d,a+3d,......,a+(n-1)d#

where a#=a_1# is the first term, d the common difference and n the number of terms.

and #d=a_2-a_1=a_3-a_2= ....=a_n-a_(n-1)#

#color(blue)"The sum to n terms" = color(red)(bar(ul(|color(white)(2/2)color(black)(S_n=n/2[2a+(n-1)d])color(white)(2/2)|)))#

A series is the sum of the terms in the sequence.

Here #a=1, d=6-1=11-6=5" and " n=20#

#rArrS_20=20/2[(2xx1)+(19xx5)]#

#=10(2+95)=970#

Dec 18, 2016

Answer:

#970#

Explanation:

We have: #1+6+11+...+96#

This is an arithmetic sequence with a common difference of #7#.

First, let's determine the number of terms in the sequence:

#=> T_(n) = T_(1) + (n - 1) d#

#=> 96 = 1 + (n - 1) (6 - 1)#

#=> 95 = 5 (n - 1)#

#=> n - 1 = 19#

#therefore n = 20#

Then, let's evaluate the sum of the #20# terms of this arithmetic sequence:

#=> S_(20) = (20) / (2) (1 + 96)#

#=> S_(20) = 10 cdot 97#

#therefore S_(20) = 970#