find the 10th term of the sequnce below

Question

40,__,__,__,__,10

Answer 1

To find the 10th term of the sequence \(40, _, _, _, _, 10\), we first must identify a pattern or rule. Given the first term (40) and the last term (10), we can assume that this sequence decreases consistently.

1. Identifying the common difference:

   • The first term is 40, and the fifth term is 10.
   • The number of terms between them, from 1 to 5, is 4 (counting 5 terms: 1 to 5).
   • Therefore, the total decrease from 40 to 10 occurs over 4 intervals (the gaps between terms).
   • The total decrease is: \[ 40 - 10 = 30 \]
   • The common difference (d) is calculated as: \[ d = \frac{-30}{4} = -7.5 \]

2. Finding intermediate terms:

   • Now we can find the other terms in the sequence using this common difference.
   • The second term: \[ 40 - 7.5 = 32.5 \]
   • The third term: \[ 32.5 - 7.5 = 25 \]
   • The fourth term: \[ 25 - 7.5 = 17.5 \]
   • The fifth term (already known): \[ 17.5 - 7.5 = 10 \]

Putting it all together, the sequence so far is: \[ 40, 32.5, 25, 17.5, 10 \]

Now, to find the 10th term, we need to determine the value of \(a_n\) where \(n = 10\). We can use the formula for the nth term of an arithmetic sequence: \[ a_n = a_1 + (n-1) \cdot d \] where \(a_1 = 40\), \(d = -7.5\) and \(n = 10\): \[ a_{10} = 40 + (10-1) \cdot (-7.5) \] \[ a_{10} = 40 + 9 \cdot (-7.5) \] \[ a_{10} = 40 - 67.5 \] \[ a_{10} = -27.5 \]

Thus, the 10th term of the sequence is \(\boxed{-27.5}\).

Answer 2

you missed the 5th term the 10 is the 6th term

Answer 3

I apologize for the oversight. Thank you for your correction. Since 10 is the 6th term of the sequence \(40, _, _, _, _, 10\), we need to find the 10th term based only on these six terms.

1. Identify the terms:

   • The first term \(a_1 = 40\)
   • The 6th term \(a_6 = 10\)

2. Calculate the common difference:

   • The difference between the first term and the 6th term is: \[ 40 - 10 = 30 \]
   • Since there are 5 gaps between the first term and the 6th term (from \(a_1\) to \(a_6\)), we can find the common difference \(d\) by dividing the total change by the number of intervals (5): \[ d = \frac{-30}{5} = -6 \]

3. Find the intermediate terms:

   • Now we can find the values of the intermediate terms:
     • 2nd term: \[ a_2 = 40 + (-6) = 34 \]
     • 3rd term: \[ a_3 = 34 + (-6) = 28 \]
     • 4th term: \[ a_4 = 28 + (-6) = 22 \]
     • 5th term: \[ a_5 = 22 + (-6) = 16 \]
     • 6th term (already known): \[ a_6 = 16 + (-6) = 10 \]

So the sequence now is: \[ 40, 34, 28, 22, 16, 10 \]

4. Find the 7th, 8th, 9th, and 10th terms:
   • 7th term: \[ a_7 = 10 + (-6) = 4 \]
   • 8th term: \[ a_8 = 4 + (-6) = -2 \]
   • 9th term: \[ a_9 = -2 + (-6) = -8 \]
   • 10th term: \[ a_{10} = -8 + (-6) = -14 \]

Thus, the 10th term of the sequence is \(\boxed{-14}\).