Answer :
Answer:
Hey buddy Please mark me as brainlist
Explanation:
To determine if the given sequence is an arithmetic progression (A.P.), we need to check if the difference between consecutive terms is constant.
The given sequence is:
\[ 3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2} \]
First, let's find the differences between consecutive terms:
\[ (3 + \sqrt{2}) - 3 = \sqrt{2} \]
\[ (3 + 2\sqrt{2}) - (3 + \sqrt{2}) = \sqrt{2} \]
\[ (3 + 3\sqrt{2}) - (3 + 2\sqrt{2}) = \sqrt{2} \]
Since the common difference (\(d\)) is \(\sqrt{2}\), the sequence is an A.P. with the first term \(a = 3\) and common difference \(d = \sqrt{2}\).
**Finding the 11th term:**
The \(n\)-th term of an A.P. is given by:
\[ a_n = a + (n-1)d \]
For the 11th term (\(n = 11\)):
\[ a_{11} = 3 + (11-1)\sqrt{2} \]
\[ a_{11} = 3 + 10\sqrt{2} \]
So, the 11th term is \(3 + 10\sqrt{2}\).
**Finding the \(n\)-th term:**
The formula for the \(n\)-th term is:
\[ a_n = a + (n-1)d \]
Substituting \(a = 3\) and \(d = \sqrt{2}\):
\[ a_n = 3 + (n-1)\sqrt{2} \]
So, the \(n\)-th term of the sequence is:
\[ a_n = 3 + (n-1)\sqrt{2} \]
Answer:
Yes, it is in AP.
a¹¹ = 3+10√2
nth term = a +(n-1)d
Explanation:
d¹ = a²- a¹ = 3+√2-3 =√2
d² = a³-a² = 3+2√2 - ( 3+√2) = 3+2√2-3-√2 = √2
d³ = a⁴- a³ = 3+3√2 - ( 3+2√2) = 3+3√2-3-2√2 = √2
since, the common difference is √2 .
Hence , the sequence is in AP
a¹¹ = a + 10d = 3 +10×√2 = 3+10√2
nth term or tn = a +(n-1)d