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An A.P consists of $ 37 $ terms. The sum of three middle most of terms is $ 225 $ and the sum of the last three terms is $ 429 $ . Find the A.P.

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Last updated date: 13th Jun 2024
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Answer
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Hint: In this problem, we need to find the A.P, so we will assume the first term of the A.P as ‘ $ a $ ’ and the common difference of A.P as ‘ $ d $ ’. In the problem, they have mentioned the number of terms in the A.P and the sum of three middlemost of the terms. So, we will assume the three middlemost terms from the given data, then we will add them to equate with the given value. Here we will get an equation in terms of $ a $ and $ d $ . Again, we have the sum of the last three terms, from this also we will get an equation in terms of $ a $ and $ d $ , now we will solve both the equation and use them to write the A.P which is in the form of $ a,a+d,a+2d,... $ .

Complete step by step answer:
Given that,
An A.P consists of $ 37 $ terms.
Let the first term in A.P is $ a $ and the common difference of A.P is $ d $ .
Now the middle term of the A.P is $ \dfrac{37+1}{2}=19 $ , the middle most three terms are $ {{18}^{th}} $ term, $ {{19}^{th}} $ term, $ {{20}^{th}} $ term. Sum of the middle most three terms are
 $ S={{a}_{18}}+{{a}_{19}}+{{a}_{20}} $
We know that $ {{n}^{th}} $ term of the A.P is $ {{a}_{n}}=a+\left( n-1 \right)d $ , then
 $ \begin{align}
  & S=\left[ a+\left( 18-1 \right)d \right]+\left[ a+\left( 19-1 \right)d \right]+\left[ a+\left( 20-1 \right)d \right] \\
 & \Rightarrow S=a+17d+a+18d+a+19d \\
 & \Rightarrow S=3a+54d \\
\end{align} $
But in the problem, we have the sum of the middle most three terms as $ 225 $ , then we will get
 $ \begin{align}
  & \Rightarrow 3a+54d=225 \\
 & \Rightarrow 3\left( a+18d \right)=225 \\
 & \Rightarrow a+18d=75 \\
 & \Rightarrow a=75-18d.....\left( \text{i} \right) \\
\end{align} $
Now the last three terms of the A.P are $ {{37}^{th}} $ term, $ {{36}^{th}} $ term, $ {{35}^{th}} $ term. Sum of the last three terms are
 $ S={{a}_{37}}+{{a}_{36}}+{{a}_{35}} $
We know that $ {{n}^{th}} $ term of the A.P is $ {{a}_{n}}=a+\left( n-1 \right)d $ , then
 $ \begin{align}
  & S=\left[ a+\left( 37-1 \right)d \right]+\left[ a+\left( 36-1 \right)d \right]+\left[ a+\left( 35-1 \right)d \right] \\
 & \Rightarrow S=a+36d+a+35d+a+34d \\
 & \Rightarrow S=3a+105d \\
\end{align} $
But in the problem, we have the sum of the last three terms as $ 429 $ , then we will get
 $ \begin{align}
  & \Rightarrow 3a+105d=429 \\
 & \Rightarrow 3\left( a+35d \right)=429 \\
 & \Rightarrow a+35d=143 \\
\end{align} $
From equation $ \left( \text{i} \right) $ substituting the value of $ a $ in the above equation, then we will get
 $ \begin{align}
  & 75-18d+35d=143 \\
 & \Rightarrow 17d=143-75 \\
 & \Rightarrow 17d=68 \\
 & \Rightarrow d=4 \\
\end{align} $
From equation $ \left( \text{i} \right) $ the value of $ a $ is $ a=75-18\left( 4 \right)=3 $ .
Now the A.P is given by
 $ \begin{align}
  & 3,3+4,3+2\left( 4 \right),... \\
 & \Rightarrow 3,7,11,... \\
\end{align} $

Note:
 In this problem they have given the sum of the last three terms, if they have given the last term of the A.P, then we can use the formula $ {{a}_{37}}=a+36d $ and equate it to the given value, then we can easily get another equation to solve the problem.