Question

Consider the arithmetic sequence 135, 141, 147.... can the sum of any 25 consecutive terms of the sequence be2020? Justify

Answer 1

Answer:

Just check if the the sum of first 25 terms of given A.P is 2020 or not

Step-by-step explanation:

here a=135 , d= 141-135=6 , n=25

Sum of first 25 terms is given by

S=n/2[2a+(n-1)d]

= 25/2[2(135)+(25-1)6]

=25/2[270+24(6)]

=25/2[270+144]

=25/2[414]

=25(207)

=5175

≠2020 , therefore , if the sum of first 25 terms isn't 2020 , it isn't possible for any consecutive 25 terms in this A.P

So the answer is no.