An AP consists of 37 terms. The sum of the three middle-most terms is 225 and the sum of the last three is 429. Find the AP.
First term \(a=3\), common difference \(d=4\) (AP: 3, 7, 11, …)
Step 1: Identify the middle terms.
The total number of terms is 37. The middle term is the \(19^{\text{th}}\) term. So, the three middle-most terms are the \(18^{\text{th}}, 19^{\text{th}}, 20^{\text{th}}\).
Step 2: Use the given sum of middle terms.
Sum of these three terms = 225.
Notice that the middle three terms are consecutive. Their sum is the same as \(3 \times a_{19}\) (because in an AP, the middle term is the average of three consecutive terms).
So, \(3a_{19} = 225 \Rightarrow a_{19} = 75\).
Step 3: Identify the last three terms.
The last three terms are the \(35^{\text{th}}, 36^{\text{th}}, 37^{\text{th}}\).
Their sum is given as 429.
Similarly, this sum is equal to \(3a_{36}\) (since \(a_{36}\) is the middle term of these three).
So, \(3a_{36} = 429 \Rightarrow a_{36} = 143\).
Step 4: Relating two terms of the AP.
We know:
Step 5: Eliminate \(a\) to find \(d\).
Subtract the two equations:
\((a + 35d) - (a + 18d) = 143 - 75\)
\(17d = 68 \Rightarrow d = 4\).
Step 6: Find the first term \(a\).
Substitute \(d = 4\) in \(a + 18d = 75\):
\(a + 18(4) = 75\)
\(a + 72 = 75 \Rightarrow a = 3\).
Final Answer:
First term \(a = 3\), common difference \(d = 4\). So the AP is: 3, 7, 11, 15, …