If 14th term of an A.P. is 4 and its 15th term is zero, then its first term is
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Notice that the common difference \(d\) is simply the difference between any two consecutive terms:
\[ d = a_{15} - a_{14} = 0 - 4 = -4 \]
Once you have \(d = -4\), you can quickly go backwards from the \(14^{\text{th}}\) term to the \(1^{\text{st}}\) term by adding \(13 \times 4\):
\[ a = a_{14} - 13d = 4 - 13(-4) = 4 + 52 = 56 \]
This reduces calculations to basic arithmetic.
Step 1: Write the nth term formula. For an A.P. with first term $a$ and common difference $d$: $a_n = a + (n-1)d$. Step 2: Set up equations for the 14th and 15th terms. $a_{14} = a + 13d = 4 \quad \cdots (1)$ $a_{15} = a + 14d = 0 \quad \cdots (2)$ Step 3: Subtract equation (1) from equation (2) to find d. \[ (a + 14d) - (a + 13d) = 0 - 4 \] \[ d = -4 \] Step 4: Substitute d back to find a. From equation (1): $a + 13(-4) = 4$ \[ a - 52 = 4 \implies a = 56 \] Step 5: Verify the answer. $a_{14} = 56 + 13(-4) = 56 - 52 = 4$ and $a_{15} = 56 + 14(-4) = 56 - 56 = 0$. Both match! Step 6: Conclusion. The first term of the A.P. is 56. \[ \boxed{a = 56} \]