Answer:
20 475
Step-by-step explanation:
Sn = a1(1-r^n)/1-r
s12 = 5(1-2^12)/1-2
= 5(1-4096)/-1
= 5(-4095)/-1
= -20 475/-1
= 20 475
thank me later :>
230
Before we find the sum of the first 10 terms of the arithmetic sequence {5, 9, 12, 17, ...}, let's find first the common difference and the 10th term of the sequence.
d = a_{2}- a_{1}a
2
−a
1
= 9-5 = 49−5=4
Use the formula:
\begin{gathered}a_{n} =a_{1} (n-1)d\\\\a_{10} =5+ (10-1)4\\a_{10} =5 +(9)4\\a_{10} =5+ 36\\a_{10} =41\\\end{gathered}
a
n
=a
(n−1)d
10
=5+(10−1)4
=5+(9)4
=5+36
=41
I prefer this formula to find the su
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Answers & Comments
Answer:
20 475
Step-by-step explanation:
Sn = a1(1-r^n)/1-r
s12 = 5(1-2^12)/1-2
= 5(1-4096)/-1
= 5(-4095)/-1
= -20 475/-1
= 20 475
thank me later :>
Answer:
230
Step-by-step explanation:
Before we find the sum of the first 10 terms of the arithmetic sequence {5, 9, 12, 17, ...}, let's find first the common difference and the 10th term of the sequence.
d = a_{2}- a_{1}a
2
−a
1
= 9-5 = 49−5=4
Use the formula:
\begin{gathered}a_{n} =a_{1} (n-1)d\\\\a_{10} =5+ (10-1)4\\a_{10} =5 +(9)4\\a_{10} =5+ 36\\a_{10} =41\\\end{gathered}
a
n
=a
1
(n−1)d
a
10
=5+(10−1)4
a
10
=5+(9)4
a
10
=5+36
a
10
=41
I prefer this formula to find the su