1.

consider the Arithametic Sequence 135,141,147,.... can the sum of any 25consecutive terms of the sequence be 2016?​

Answer»

ANSWER:

(a) 6,10,14,....

SINCE t2−t1=t3−t2

Common DIFFERENCE, d=10−6=4

First TERM a=6

Sn=2n2a+(n−1)d=2n2×6+(n−1)4=2n12+4n−4=2n4n+8

=n(2n+4)

2n2+4n

(b) For the SUM to be 240 from the beginning

240=n2+4n

⇒240=n(2n+4)

⇒240=n(2n+4)

⇒120=n(n+2)

⇒n2+2n−120=0

⇒n2+


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