Question

An AP consists of $37$ terms. The sum of the three middlemost terms is $225$ and the sum of the last three is $429$. Find the AP.

Answer 1

Step 1: Calculate the middle terms.

Assume that the first term is $a$ and the common difference is $d$.

Know that here the total terms are $37$, which is an odd number.

So, the middle term is:

$\begin{array}{c}{\left(\frac{37+1}{2}\right)}^{th}\text{term}={\left(\frac{38}{2}\right)}^{th}\text{term}\\ ={19}^{th}\text{term}\end{array}$

Step 2: Calculate the sum of the three middle-most terms.

Know that the nth term is given as $a+(n-1)d$.

Understand that here the three middle-most terms are ${18}^{th}\text{term},{19}^{th}\text{term\hspace{0.17em}and\hspace{0.17em}}{20}^{th}\text{term}$.

Therefore,

$\begin{array}{c}225=[a+(18-1)d]+[a+(19-1)d]+[a+(20-1)d]\\ 225=a+17d+a+18d+a+19d\\ 225=3a+54d\\ a=\frac{225}{3}-\frac{54}{3}d\\ a=75-18d\mathrm{..............}\left(1\right)\end{array}$

Step 3: Calculate the sum of the last three terms.

Know that the last three terms are ${35}^{th}\text{term},{36}^{th}\text{term\hspace{0.17em}and}{37}^{th}\text{term}$.

Therefore,

$\begin{array}{c}429=[a+(35-1)d]+[a+(36-1)d]+[a+(37-1)d]\\ 429=a+34d+a+35d+a+36d\\ 429=3a+105d\\ a=\frac{429}{3}-\frac{105}{3}d\\ a=143-35d\mathrm{..............}\left(2\right)\end{array}$

Step 4: Calculate the common difference.

Equate equation $1\text{and}2$.

Therefore,

$\begin{array}{c}75-18d=143-35d\\ 35d-18d=143-75\\ 17d=68\\ d=4\end{array}$

Step 5: Form the A.P series.

Apply $d=4$in equation $2$.

Then,

$\begin{array}{c}a=143-35\left(4\right)\\ =143-140\\ =3\end{array}$

Therefore, the series:

$\begin{array}{l}a,a+d,a+2d,a+3d\mathrm{...}\\ \Rightarrow \left(3\right),(3+4),(3+2\times 4)\mathrm{.....}\\ \Rightarrow \left(3\right),\left(7\right),(3+8)\mathrm{.....}\\ \Rightarrow 3,7,\mathrm{11.......}\end{array}$

Hence, the A.P is $3,7,\mathrm{11.......}$