The sum of the \(4^{th}\) and \(8^{th}\) term of an AP is 24 and the sum of the \(6^{th}\) and \(10^{th}\) terms is 44. Find the first three terms of the AP.

The AP formula is \(T_n = a + (n − 1)d\)
\(4^{th}\) term, \(T_4 = a + 3d\)
\(8^{th}\) term, \(T_8 = a + 7d\)
Given,
\((a + 3d) + (a + 7d) = 24\)
⇒ \(2a + 10d = 24\)
⇒ \(a + 5d = 12\) --- (I)
Again,
\(6^{th}\) term, \(T_6 = a + 5d\)
\(10^{th}\) term, \(T_{10} = a + 9d\)
Given,
\((a + 5d) + (a + 9d) = 44\)
⇒ \(2a + 14d = 44\)
⇒ \(a + 7d = 22\) --- (II)

Subtracting (I) from (II), we get
\(2d = 10\)
⇒ \(d = 5\)
Substituting this into (I), we get
\(a + 25 = 12\)
⇒ \(a = −13\)

\(T_1 = a = −13\)
\(T_2 = a + d = −13 + 5 = −8\)
\(T_3 = a + 2d = −13 + 10 = −3\)