The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44.

Solution:

$\mathrm{t}_{4}+\mathrm{t}_{8}=24 ; \mathrm{t}_{6}+\mathrm{t}_{10}=44$

$\Rightarrow(a+3 d)+(a+7 d)=24$

$(a+5 d)+(a+9 d)=44$

$\Rightarrow 2 \mathrm{a}+10 \mathrm{~d}=24 ; 2 \mathrm{a}+14 \mathrm{~d}=44$

We have $a+5 d=12$ ...(1)

and $\mathrm{a}+7 \mathrm{~d}=22$ ...(2)

Subtracting (1) from (2), we get

$2 d=10 \Rightarrow d=5$

From (i) $a+5 \times 5=12, a=-13$

$\mathrm{t}_{1}=-13, \mathrm{t}_{2}=-8, \mathrm{t}_{3}=-3$