The eighth term of an AP is half its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15th term.
3
Step 1: Write the general formula of an AP
The (n^{th}) term of an AP is: \(a_n = a + (n-1)d\), where \(a\) = first term, \(d\) = common difference.
Step 2: Translate the first condition
The 8th term is half of the 2nd term.
8th term = \(a + 7d\)
2nd term = \(a + d\)
So, \(a + 7d = \tfrac{1}{2}(a + d)\).
Step 3: Simplify this equation
Multiply both sides by 2: \(2a + 14d = a + d\)
Bring terms together: \(2a - a + 14d - d = 0\)
\(a + 13d = 0\)
So, \(a = -13d\). … (Equation 1)
Step 4: Translate the second condition
The 11th term is 1 more than one-third of the 4th term.
11th term = \(a + 10d\)
4th term = \(a + 3d\)
Condition: \(a + 10d = \tfrac{1}{3}(a + 3d) + 1\)
Step 5: Simplify this equation
Multiply everything by 3 to remove the fraction: \(3a + 30d = a + 3d + 3\)
Bring terms together: \(3a - a + 30d - 3d = 3\)
\(2a + 27d = 3\) … (Equation 2)
Step 6: Substitute value of a
From (Equation 1): \(a = -13d\)
Substitute in (Equation 2): \(2(-13d) + 27d = 3\)
\(-26d + 27d = 3\)
\(d = 3\)
Now, \(a = -13d = -13 × 3 = -39\)
Step 7: Find the 15th term
15th term = \(a + 14d\)
= \(-39 + 14 × 3\)
= \(-39 + 42\)
= 3
Final Answer: The 15th term is 3.