Question

the sum of the 4th and 8 th 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

Answer 1

an = a + (n − 1) d
a4 = a + (4 − 1) d
a4 = a + 3d 
again
a8 = a + 7d
a6 = a + 5d
a10 = a + 9d
we know that a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 ..be equation. (i)
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ...be equation (ii)
substract (i) from (ii) we will 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
∴ the 1st 3 terms of this ap are −13, −8, and −3.

Answer 2

let the first term of an AP be 'a'
let the common difference of an AP be 'd'
according to the question,
a4+ a8 = 24
so , (a+3d) + (a+7d) =24
2a + 10d = 24
a + 5d = 12 ..................(1)

according to the question,
a6+a10 = 44
so , (a+5d) + (a+9d) = 44
7d-5d = 10
2d =10
d =5
therefore , common difference (d) =5
by substituting the value of 'd' in equation 1,
a + 5d =12
a +5(5) =12
a+ 25 = 12
a= -13
therefore the the first term (a) = -13

therefore the first three terms of an AP are -13 , -8 , -3..................