Question

G. 4) find the Som of natural numbers foom Ito 30,
by wth tum of an crithematic sequence is 5nta
find som of first 30 taims of the sequence.
c) Con the some
some of
some fans of this sequence be 2014. Justify
your answer.​

Answer 1

Answer:

(a) 6,10,14,....

Since t

2 −t 1 =t −t 2

Common difference, d=10−6=4

First term a=6S n = 2a 2a+(n−1)d= 2n 2×6+(n−1)4=2n 12+4n−4= 2n 4n+8=n(2n+4)2n 2 +4n

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

240=n 2+4n⇒240=n(2n+4)

⇒240=n(2n+4)

⇒120=n(n+2)

⇒n 2 +2n−120=0

⇒n 2 +12n−10n−120=0

⇒n(n+12)−10(n+12)=0

⇒(n+12)(n−10)=0

∴ n=−12,10

10 terms should be added to make the sum to be 240

(c) To make the sum to be 240, 'n' should be natural number.

250=n(2n+4)

⇒125=n(n+2)

⇒n 2 +2n−125=0

⇒n 2 +n−125=0n= 2−1 1−4×125

Since n is not a natural number, therefore the term 250 cannot exist in the series.

Step-by-step explanation:

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