Question

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

Answer 1

Answer:

t4+t8=24

a+3d+a+7d=24

2a+10d=24

a+5d=12 -(1)

t6+t10=44

a+5d+a+9d=44

2a+14d=44

a+7d=22 -(2)

subtract eq.1 from 2.

2d=10,d=5.

put the value of d in eq.1

a= -13.

A.P is -13,-8,-3,2,.........

Answer 2

Step-by-step explanation:

The nth term of the A.P. is given by

.Tn = a + ( n - 1 )d

T4 = a + 3d

T8 = a + 7d

T4 + T8 = 24

2a + 10d = 24

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

T6 = a + 5d

T10 = a + 9d

T6 + T10 = 44

2a + 14d = 44

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

Subtracting equation ( 1 ) from equation ( 2 ) we get

2d = 10

d = 5

Putting value of d in equation ( 1 )

a + 5 × 5 = 12

a = 12 - 25

a = - 13

So the required A.P. is

- 13 , - 8 , - 3 , 2 , ...

THANKS!!!