If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero?
Step-by-Step Solution
Key Concept: Use the nth term formula of an arithmetic progression, $T_n = a + (n-1)d$, to set up two linear equations from the given terms, solve for the first term $a$ and common difference $d$, and then find $n$ such that $T_n = 0$.
1. Let the first term be $a$ and the common difference be $d$.
2. Using the formula $T_n = a + (n-1)d$:
- For the 3rd term: $T_3 = a + 2d = 4$ … (i)
- For the 9th term: $T_9 = a + 8d = -8$ … (ii)
3. Subtract (i) from (ii):
$$ (a+8d) - (a+2d) = -8 - 4 \[4pt] 6d = -12 \[4pt] d = -2 $$
4. Substitute $d = -2$ into (i):
$$ a + 2(-2) = 4 \[4pt] a - 4 = 4 \[4pt] a = 8 $$
5. The AP is therefore $a = 8$, $d = -2$.
6. To find the term which is zero, set $T_n = 0$:
$$ a + (n-1)d = 0 \[4pt] 8 + (n-1)(-2) = 0 \[4pt] 8 - 2(n-1) = 0 \[4pt] 2(n-1) = 8 \[4pt] n-1 = 4 \[4pt] n = 5 $$
7. Hence the 5th term of the AP is zero.
Correct Answer: 5th term