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How many terms of the series 54,51,48,..., be taken so that their sum is
513?Explain the double answer.
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Answers
Answered by
14
Answer:
Sn = n/2 ( 2a + (n-1)d)
513= n/2 (2* 54 +(n*-3) +3)
1026= 108n - 3n^2 +3n
3n^2 -111n +1026= 0
n^2 - 37n + 342 = 0
n= 18 or n = 19
Answered by
5
Answer:
hope it helps you
Step-by-step explanation:
a=54
d=51−54=−3
S
n
=513
S
n
=
2
n
[2a+(n−1)d]
⇒513=
2
n
[2(54)+(n−1)(−3)]
⇒1026=n[108−3n+3]
⇒1026=n[111−3n]
⇒3n
2
−111n+1026=0
⇒n
2
−37n+342=0
⇒n
2
−18n−19n+342=0
⇒(n−18)(n−19)=0
⇒n=18,19
T
18
=a+17d=54+17(−3)=3
T
19
=a+18d=54+18(−3)=0
Since the 19
th
term is 0, therefore no change will be seen in the sum of 18 and 19 terms. That is the reason why we are getting 2 answers.
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