Question

There are 37 terms in an A.P., the sum of three terms placed exactly at the middle is 225 and the sum of last three terms is 429. Write the A.P.

Answer 1

Let the first term be a and the common difference be d.
Since, the A.P. contains 37 terms, the middle term is \[\left( \frac{37 + 1}{2} \right)^{th} = {19}^{th}\] term.

According to the question,

\[a_{18} + a_{19} + a_{20} = 225\] 

\[ \Rightarrow \left( a + \left( 18 - 1 \right)d \right) + \left( a + \left( 19 - 1 \right)d \right) + \left( a + \left( 20 - 1 \right)d \right) = 225\]

\[ \Rightarrow 3a + 17d + 18d + 19d = 225\]

\[ \Rightarrow 3a + 54d = 225\]

\[ \Rightarrow 3\left( a + 18d \right) = 225\]

\[ \Rightarrow a + 18d = \frac{225}{3}\]

\[ \Rightarrow a + 18d = 75\]

\[ \Rightarrow a = 75 - 18d . . . \left( 1 \right)\]

Also,

\[a_{35} + a_{36} + a_{37} = 429\]

\[ \Rightarrow \left( a + \left( 35 - 1 \right)d \right) + \left( a + \left( 36 - 1 \right)d \right) + \left( a + \left( 37 - 1 \right)d \right) = 429\]

\[ \Rightarrow 3a + 34d + 35d + 36d = 429\]

\[ \Rightarrow 3a + 105d = 429\]

\[ \Rightarrow 3\left( a + 35d \right) = 429\]

\[ \Rightarrow a + 35d = \frac{429}{3}\]

\[ \Rightarrow 75 - 18d + 35d = 143 \left( \text { from } \left( 1 \right) \right)\]

\[ \Rightarrow 17d = 143 - 75\]

\[ \Rightarrow 17d = 68\]

\[ \Rightarrow d = 4\]

\[ \Rightarrow a = 75 - 18\left( 4 \right) \left(\text { from } \left( 1 \right) \right)\]

\[ \Rightarrow a = 75 - 72\]

\[ \Rightarrow a = 3\]

Hence, the resulting A.P is 3, 7, 11, ....