The sum of the first 30 terms of an A.P. is 1920. If the fourth term is 18, find its $11^{th}$ term.

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Step 1: $\frac{30}{2}[2a + 29d] = 1920 \Rightarrow 2a + 29d = 128$ ... (i)
Step 2: Also, $a_4 = 18 \Rightarrow a + 3d = 18$ ... (ii)
Step 3: From equation (i) & (ii): $a = 6$, $d = 4$
Step 4: $	herefore a_{11} = a + 10d = 6 + 40 = 46$

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