consider the arithmetic sequence 135,141,147, can the sum of 25 consecutive terms of the sequence be 2016 ? justify your answer
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Answer:
(a) 6,10,14,....
Since t
2
−t
1
=t
3
−t
2
Common difference, d=10−6=4
First term a=6
S
n
=
2
n
2a+(n−1)d=
2
n
2×6+(n−1)4=
2
n
12+4n−4=
2
n
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=0
n=
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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