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

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

\(a_4 + a_8 = 24\) ,and \(a_6 + a_{10} = 44\)

Using formula \(a_n = a + (n - 1)d\), to find nth term of arithmetic progression,

\(\Rightarrow \) a + (4 - 1) d + [a + (8 - 1) d] = 24 and, a + (6 - 1) d + [a + (10 - 1) d] = 44
\(\Rightarrow \) a + 3d + a + 7d = 24, and a + 5d + a + 9d = 44
\(\Rightarrow \) 2a + 10d = 24, and 2a + 14d = 44
\(\Rightarrow \) a + 5d = 12, and a + 7d =22

These are equations in two variables.

Using equation, a + 5d = 12, we can say that a = 12 - 5d…...(1)

Putting (1) in equation a + 7d = 22,
\(\Rightarrow \) 12 - 5d + 7d = 22
\(\Rightarrow \) 12 + 2d = 22
\(\Rightarrow \) 2d = 10
\(\Rightarrow \) d = 5

Putting value of d in equation
a = 12 - 5d,
\(\Rightarrow \) a = 12 – 5 (5)
= 12 – 25
= -13

Therefore, first term = a = -13
and, Common difference = d = 5

Therefore, AP is –13, –8, –3, 2…

Its first three terms are –13, –8 and –3.