An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Step-by-Step Solution
Key Concept: Use the nth term formula of an arithmetic progression, $a_n = a + (n-1)d$, to determine the first term $a$ and common difference $d$ from the given terms, then compute the required term.
1. Let the first term be $a$ and the common difference be $d$.
2. Given:
- 3rd term $a_3 = 12$ \[ $a_3 = a + 2d = 12$ \]
- Last term (50th term) $a_{50} = 106$ \[ $a_{50} = a + 49d = 106$ \]
3. Express $a$ from the first condition:
$$a = 12 - 2d$$
4. Substitute $a$ into the second condition:
$$ (12 - 2d) + 49d = 106$$
$$ 12 + 47d = 106$$
$$ 47d = 94$$
$$ d = \frac{94}{47} = 2$$
5. Find the first term $a$:
$$ a = 12 - 2d = 12 - 2\times 2 = 12 - 4 = 8$$
6. Find the 29th term using $a_{29} = a + 28d$:
$$ a_{29} = 8 + 28\times 2 = 8 + 56 = 64$$
7. Result: The 29th term of the AP is $64$.
Correct Answer: 64