Question

The sum of 4th and 8th term of A.P. is 24 . The sum of 6 and 10 term is 44 . Find the first 3 terms of A.P.​

Answer 1

Answer:

-13, -8, -3

Step-by-step explanation:

T6 + T10 = 44

a + 5d + a + 9d = 44

2a + 14d = 44

a + 7d = 22  -- (1) => T8 = 22

T4 + T8 = 24

a + 3d + 22 = 24

a + 3d = 2  --    (2)

a + 7d = 22 -- (1)

So, (1) - (2) => 4d = 20  => d = 5

a + 15 = 2

a = -13

Therefore, first three terms of AP = a, a+d, a+2d

                                                        = -13, -8, -3

Answer 2

A4=a+3d

A8=a+7d

And

A6=a+5d

A10=a+9d

Then

a+3d+a+7d=24

a+5d+a+9d=44

By solving we get

(1)-- 2a+10d=24

(2)--2a+14d=44

By elimination we want to do that

We get d=5

Sub d value in (1) then

We get a=-13 then do

A1=a=-13

A2=a+d=-13+(5)=-8

A3=a+2d=-13+2(5)=-13+10=-3