Tolong dijawab ☺

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dari deret aritmatika, diketahui U11=53 merupakan suku tengah. Jumlah n suku pertama deret itu adalah..

Tolong dijawab ☺

U11=a+10b=53
U11=U10+U12/2
2xU11=U10+U12
2×53=U10+U12
106=U10+U12
U10=106-(U12)
=106-(U11+b)
=106-(53+b)
=106-53-b
=53-b
U12=106-(U10)
=106-(U11-b)
=106-(53-b)
=106-53+b
=53+b
U10,U11,U12
53-b,53,53+b
b=(53+b)-(53-b)/k+1
=b+b/1+1
=2b/2
2b=2b
b=1
52,53,54
a+10b=53
a+10(1)=53
a+10=53
a=53-10
a=43
U10=a+9b
=43+9(1)
=43+9
=52
U12=a+11b
=43+11(1)
=43+11
=54
Un=a+(n-1)b
54=43+(n-1)1
=43+n-1
=42+n
54-42=n
12=n

Sn=n/2(a+Un)
S12=12/2(43+U12)
=6(43+54)
=6(97)
=582
Demikian semoga membantu.