Question

Consider the arithmetic sequence 135,141,147,....can the sum of any 25 consecutive terms of the sequence be 2016

Answer 1

no
a1=135
d=141-135=6
Sn=n/2[2a1+(n-1)d]
S25=25/2[2(135)+(24)6]
S25=25/2[270+144]
S25=25/2[414]
S25=5175
which is ≠2016

Answer 2

Answer: No

Step-by-step explanation:

The sum of n terms is given by :-

$S_n=\frac{n}{2}(2a+(n-1)d)$

From the given arithmetic sequence 135,141,147,....

First term = a=135

d=$141-135=6$

Then, $S_{25}=\frac{25}{2}(2(135)+(25-1)6)\\\Rightarrow\ S_{25}=\frac{25}{2}(270+144)\\\Rightarrow\ S_{25}=\frac{25}{2}(414)\\\Rightarrow\ S_{25}=25\times207\\\Rightarrow\ S_{25}=5175\neq2016$