Find the 20th term from the last term of the AP: 3, 8, 13, ..., 253.
20th term from the last term is 158.
Step 1: Identify the first term and common difference.
The AP is: 3, 8, 13, ... , 253.
First term \(a = 3\).
Common difference \(d = 8 - 3 = 5\).
The last term \(l = 253\).
Step 2: Find the total number of terms.
Use the nth term formula of an AP: \(a_n = a + (n - 1)d\).
Here, \(a_n = l = 253\), so:
\[253 = 3 + (n - 1) \cdot 5\]
Subtract 3 from both sides:
\[250 = (n - 1) \cdot 5\]
Divide by 5:
\[n - 1 = 50 \Rightarrow n = 51\]
So, the AP has 51 terms in total.
Step 3: Relate “20th from the last” to position from the beginning.
If an AP has \(n\) terms, then:
• 1st term from the last = \(n\)th term from the beginning.
• 2nd term from the last = \((n - 1)\)th term from the beginning.
• In general, \(k\)th term from the last = \((n - k + 1)\)th term from the beginning.
Here, \(n = 51\) and we need the 20th term from the last, so its position from the beginning is:
\[n - 20 + 1 = 51 - 20 + 1 = 32\]
So, we need the 32nd term from the beginning, \(a_{32}\).
Step 4: Find the 32nd term.
Use \(a_n = a + (n - 1)d\):
\[a_{32} = a + 31d = 3 + 31 \cdot 5\]
\[a_{32} = 3 + 155 = 158\]
Conclusion: The 20th term from the last term of the AP is 158.