18. The sum of 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the A.P.

We know that, the nth term of the AP is; an = a+(n−1)d a4 = a+(4−1)d a4 = a+3d In the same way, we can write, a8 = a+7d a6 = a+5d a10 = a+9d Given that, a4+a8 = 24 a+3d+a+7d = 24 2a+10d = 24 a+5d = 12 …………………………………………………… (i) a6+a10 = 44 a +5d+a+9d = 44 2a+14d = 44 a+7d = 22 …………………………………….. (ii) On subtracting equation (i) from (ii), we get, 2d = 22 − 12 2d = 10 d = 5 From equation (i), we get, a+5d = 12 a+5(5) = 12 a+25 = 12 a = −13 a2 = a+d = − 13+5 = −8 a3 = a2+d = − 8+5 = −3 Therefore, the first three terms of this A.P. are −13, −8, and −3.