How many terms of the series 54,52,48,... must be taken so that their sum is 513?
Answers
Answered by
1
Answer:
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.
Answered by
4
Step-by-step explanation:
a=54
d=51−54=−3Sn
=513Sn
= 2n [2a+(n−1)d]
⇒513= 2n [2(54)+(n−1)(−3)]
⇒1026=n[108−3n+3]
⇒1026=n[111−3n]
⇒3n 2 −111n+1026=0
⇒n2 −37n+342=0
⇒n2 −18n−19n+342=0
⇒(n−18)(n−19)=0
⇒n=18,19
T 18 =a+17d=54+17(−3)=3
T19 =a+18d=54+18(−3)=0
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