Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73.
Step-by-Step Solution
Key Concept: Use the formula for the $n^{\text{th}}$ term of an arithmetic progression, $a_n = a + (n-1)d$, to set up two linear equations and solve for the first term $a$ and common difference $d$. Then substitute $n = 31$ to obtain the required term.
1. Let $a$ be the first term and $d$ the common difference of the AP.\
2. Using the $n^{\text{th}}$ term formula $a_n = a + (n-1)d$:
$$\begin{aligned}
a_{11} &= a + 10d = 38 \quad\text{(i)}\\
a_{16} &= a + 15d = 73 \quad\text{(ii)}\end{aligned}$$
3. Subtract equation (i) from equation (ii) to eliminate $a$:
$$\begin{aligned}(a + 15d) - (a + 10d) &= 73 - 38 \\
5d &= 35 \\
\Rightarrow \; d &= \frac{35}{5} = 7\end{aligned}$$
4. Substitute $d = 7$ back into equation (i) to find $a$:
$$\begin{aligned}a + 10(7) &= 38 \\
a + 70 &= 38 \\
\Rightarrow \; a &= 38 - 70 = -32\end{aligned}$$
5. Now find the 31st term $a_{31}$:
$$\begin{aligned}a_{31} &= a + (31-1)d \\
&= -32 + 30 \times 7 \\
&= -32 + 210 \\
&= 178\end{aligned}$$
6. Hence, the 31st term of the given AP is 178.
Correct Answer: 178