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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.

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Solution

## 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 $a+\left(n-1\right)d$.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,$\begin{array}{c}225=\left[a+\left(18-1\right)d\right]+\left[a+\left(19-1\right)d\right]+\left[a+\left(20-1\right)d\right]\\ 225=a+17d+a+18d+a+19d\\ 225=3a+54d\\ a=\frac{225}{3}-\frac{54}{3}d\\ a=75-18d..............\left(1\right)\end{array}$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,$\begin{array}{c}429=\left[a+\left(35-1\right)d\right]+\left[a+\left(36-1\right)d\right]+\left[a+\left(37-1\right)d\right]\\ 429=a+34d+a+35d+a+36d\\ 429=3a+105d\\ a=\frac{429}{3}-\frac{105}{3}d\\ a=143-35d..............\left(2\right)\end{array}$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:$\begin{array}{l}a,a+d,a+2d,a+3d...\\ ⇒\left(3\right),\left(3+4\right),\left(3+2×4\right).....\\ ⇒\left(3\right),\left(7\right),\left(3+8\right).....\\ ⇒3,7,11.......\end{array}$Hence, the A.P is $3,7,11.......$

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