Question

the sum of the 4th and 8th terms
of an AP is 24and the sum of the
6th and 10th terms is 44. find
the First three terms of the AP.​

Answer 1

a4+a8 = 24

a+(4-1)d + a+(8-1)d = 24

a+3d + a+7d = 24

2a+10d = 24

a + 5d = 12............(1)

a6+a10 = 44

a+(6-1)d+a+(10-1)d = 44

a+5d+a+9d = 44

2a + 14d = 44

a + 7d = 22.........(2)

subtract (1) from (2)

we get,

7d-5d = 22-12

2d = 10

d = 5

put d = 5 in equation (1)

we ger,

a+5d = 12

a+5×5 = 12

a +25 = 12

a = 12-25 = - 13

a = -13 , d = 5

first three term of AP are

a , a+d , a+2d

a = -13

a+d = -13+5 = -8

a + 2d = -13+10 = -3

AP is :-

-13, -8 , - 3

Answer 2

Step-by-step explanation:

let first term be a and common difference d.

using nth term= a+(n-1)d

4th term= a+3d

8th term=a+7d

given,

a+3d+a+7d=24

2a+10d=24

a+5d=12 --(1)

6th term=a+5d

10th term=a+9d

given,

a+5d+a+9d=44

2a+14d=44

a+7d=22 ---(2)

on subtracting eq.1 from 2

2d=10

:d=5

therefore,

on substituting the value of d in eq. 1

a+5d=12

a=-13

:: hence,

first 3 terms are :

a+d,a+2d,a+3d

-13,-8,-3 (ans.).