the fifth term of an arithmetic sequence is 38 and the nineth term is 66 what is its 25th term
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Answer:
The 5th term of an arithmetic sequence is 38 and 9th term is 66.What is its 25th term.
Given that,
a5 = 38
a9 = 66
We can write these terms as...
a5 → a + 4d = 38... (1)
a9 → a + 8d = 66... (2)
Now, subtract the equations (1) & (2), we get
➡ - 4d = - 28
➡ 4d = 28
➡ d = 7
∴d=7
Now, substitute the value of d in (1)
➡ a + 4(7) = 38
➡ a + 28 = 38
➡ a = 38 - 28
➡ a = 10
∴a=10
To find the value of 25th term,
↪ an = a + (n - 1)d
a = 10
d = 7
n = 25
➡ a25 = 10 + (25 - 1)(7)
➡ a25 = 10 + 24(7)
➡ a25 = 10 + 168
➡ a25 = 178
Answer:
The 5th term of an arithmetic sequence is 38 and 9th term is 66.What is its 25th term.
Given that,
a5 = 38
a9 = 66
We can write these terms as...
a5 → a + 4d = 38... (1)
a9 → a + 8d = 66... (2)
Now, subtract the equations (1) & (2), we get
➡ - 4d = - 28
➡ 4d = 28
➡ d = 7
∴d=7
Now, substitute the value of d in (1)
➡ a + 4(7) = 38
➡ a + 28 = 38
➡ a = 38 - 28
➡ a = 10
∴a=10
To find the value of 25th term,
↪ an = a + (n - 1)d
a = 10
d = 7
n = 25
➡ a25 = 10 + (25 - 1)(7)
➡ a25 = 10 + 24(7)
➡ a25 = 10 + 168
➡ a25 = 178
Step-by-step explanation:
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