Respuesta :
Correct answer is 1276.
This an arithmetic series.
[tex]a_1=16,a_n=100,d=4,S_n=? \\ \\a_n=a_1+(n-1)d \\100=16+(n-1)4 \\4(n-1)=84 \\n-1=21 \\n=22 \\ \\S_n= \frac{(a_1+a_n)n}{2}= \frac{(16+100)22}{2}=116\times 11=1276[/tex]
This an arithmetic series.
[tex]a_1=16,a_n=100,d=4,S_n=? \\ \\a_n=a_1+(n-1)d \\100=16+(n-1)4 \\4(n-1)=84 \\n-1=21 \\n=22 \\ \\S_n= \frac{(a_1+a_n)n}{2}= \frac{(16+100)22}{2}=116\times 11=1276[/tex]
Answer: Sum of the multiples of 4 from 16 to 100 is 1276.
Step-by-step explanation:
Since we have given that
Multiples of 4 from 16 to 100:
[tex]16,20,24,.....100[/tex]
First we find the number of terms (n):
Here a = first term = 16
d = common difference = 20-16=4
So, we know the formula for "nth term":
[tex]a_n=a+(n-1)d\\\\100=16+(n-1)4\\\\100-16=(n-a)4\\\\\frac{84}{4}=n-1\\\\21=n-1\\\\n=21+1=22[/tex]
so, we need to calculate "Sum of 22 terms":
[tex]S_{22}=\frac{22}{2}(2a+(n-1)d)\\\\S_{22}=11(2\times 16+(22-1)4)\\\\s_{22}=11(32+21\times 4)\\\\S_{22}=1276[/tex]
Hence, Sum of the multiples of 4 from 16 to 100 is 1276.
