Question

An A.P. consists of 37 terms, the sum of 3 middle terms is 225 and the sum of the last three terms is 429. Find the A.P.

Answer 1

Hint:The last term is the 37th term. Thus find the 3 middle terms and form the equation using their sum. Similarly, form an equation with the sum of the last three terms. Solve and find a and d. Thus form the series a, d and n.

Complete step-by-step answer:

It is said that the arithmetic series consist of 37 terms. We know that arithmetic progression is a sequence of numbers such that the difference of any two successive numbers is a constant.
As there are 37 terms in the series, the last term of the series will be the 37th term, i.e. n = 37, where n is the number of terms.
Thus the middle most term will be \[\left( \dfrac{n+1}{2} \right)\].
\[\therefore \]The middle term \[=\dfrac{37+1}{2}=\dfrac{38}{2}={{19}^{th}}\] term.
So, the 19th term is the middle most term.
From this we can say that the 3 middle most terms we can use the formula,
\[{{a}_{n}}=a+(n-1)d\], where a is the first term, d is the common difference.
18th term \[={{a}_{18}}\], where n = 18.
\[\therefore \]\[{{a}_{18}}=a+(18-1)d=a+17d\].
19th term \[={{a}_{19}}\], where n = 19.
\[\therefore \]\[{{a}_{19}}=a+(19-1)d=a+18d\]
20th term \[={{a}_{20}}\], where n = 20.
\[\therefore \]\[{{a}_{20}}=a+(20-1)d=a+19d\]
Thus we got,
\[\begin{align}
  & {{a}_{18}}=a+17d \\
 & {{a}_{19}}=a+18d \\
 & {{a}_{20}}=a+19d \\
\end{align}\]
It was said that the sum of 3 middle terms is 429.
Thus the sum of \[{{a}_{18}}+{{a}_{19}}+{{a}_{20}}=429\].
Substitute the values of \[{{a}_{18}},{{a}_{19}},{{a}_{20}}\].
\[\begin{align}
  & \left( a+17d \right)+\left( a+18d \right)+\left( a+19d \right)=429 \\
 & 3a+\left( 17+18+19 \right)d=429 \\
 & \Rightarrow 3a+54d=429 \\
\end{align}\]
Divide the entire equation by 3.
\[a+18d=75......(1)\]
It is also said that the sum of the last three terms = 429.
The last term is 37th term,
\[\therefore {{a}_{37}}=a+(37-1)d=a+36d\]
The last three terms of the A.P. will be \[{{a}_{n-2}},{{a}_{n-1}},{{a}_{n}}\].
Thus \[{{a}_{n}}={{a}_{37}}\], the other two terms are \[{{a}_{37-1}}={{a}_{36}}\] and \[{{a}_{37-2}}={{a}_{35}}\].
Thus let us find the term \[{{a}_{36}}\] and \[{{a}_{35}}\].
\[{{a}_{36}}=a+(36-1)d=a+35d\], which is the 36th term.
\[{{a}_{35}}=a+(35-1)d=a+34d\], which is the 35th term.
The sum of last 3 numbers = 429,
i.e. \[{{a}_{35}}+{{a}_{36}}+{{a}_{37}}=429\].
Substitute the values of \[{{a}_{35}},{{a}_{36}},{{a}_{37}}\].
\[\begin{align}
  & (a+34d)+(a+35d)+(a+36d)=429 \\
 & 3a+(34+35+36)d=429 \\
 & \therefore 3a+105d=429 \\
\end{align}\]
Divide the equation by 3.
\[a+35d=143.....(2)\].
Now we have 2 equations with variables a and d. Let us solve and find the values of a and b.
Subtract equation (2) from (1).
\[\begin{align}
  & (a+35d)-(a+18d)=143-75 \\
 & \Rightarrow a+35d-a-18d=68 \\
 & \therefore 17d=68 \\
 & d=\dfrac{68}{17}=4 \\
\end{align}\]
Thus we got the common difference, d = 4.
Let us substitute d = 4 in equation (1).
\[\begin{align}
  & a+18d=75 \\
 & \Rightarrow a+18\times 4=75 \\
 & a=75-72=93 \\
\end{align}\]
Thus first term, a =3.
Thus we have a = 3, and d = 4.
37th term, \[{{a}_{37}}=a+(n-1)d=3+(37-1)4=3+(36\times 4)=147.\]
1st term, \[{{a}_{1}}=a+(n-1)d=3+(1-1)4=3+0=3\]
2nd term, \[{{a}_{2}}=a+(n-1)d=3+(2-1)4=3+4=7\]
3rd term, \[{{a}_{3}}=3+(3-1)4=3+8=11\]
4th term, \[{{a}_{4}}=3+(4-1)4=3+\left( 3\times 4 \right)=3+12=15\]
Thus we can form the A.P. series as,
3, 7, 11, 15,…..,147.
Thus we got the required A.P.

Note:For finding the middle term there are 2 ways. If the number terms are even then we use \[{}^{n}/{}_{2}\]. If the number of terms are odd then we use \[{}^{n+1}/{}_{2}\] to get the middle term.