High School

The sum of the 8th and 4th terms of an arithmetic progression (AP) is 24, and the sum of the 6th and 10th terms is 44. Find the AP.

Answer :

Given:

Sum of 8th & 4th terms of an AP = 24

Sum of 6th & 10th terms = 44

Find:

A.P.

Solution:

We know that,

nth term of an AP = a + (n - 1)d

Hence,

⟹ a + (8 - 1)d + a + (4 - 1)d = 24

⟹ a + 7d + a + 3d = 24

⟹ 2a + 10d = 24 -- equation (1)

Similarly,

⟹ a + 5d + a + 9d = 44

⟹ 2a + 14d = 44 -- equation (2)

Subtract equation (1) from (2).

⟹ 2a + 14d - (2a + 10d) = 44 - 24

⟹ 2a + 14d - 2a - 10d = 20

⟹ 4d = 20

⟹ d = 20/4

⟹ d = 5

Substitute the value of d in equation (1).

⟹ 2a + 10(5) = 24

⟹ 2a = 24 - 50

⟹ 2a = 24 - 50

⟹ 2a = - 26

⟹ a = - 26/2

⟹ a = - 13

We know,

General form of an AP is a , a + d , a + 2d...

Hence,

⟹ required AP = - 13 , - 13 + 5 , - 13 + 2(5)...

⟹ required AP = - 13 , - 8 , - 3....

I hope it will help you.

Regards.

required AP = - 13 , - 13 + 5 , - 13 + 2(5)...

required AP = - 13 , - 8 , - 3....

Given:

Sum of 8th & 4th terms of an AP = 24

Sum of 6th & 10th terms = 44

Find:

A.P.

Solution:We know that,nth term of an AP = a + (n - 1)d

Hence,

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

a + 7d + a + 3d = 24

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

Similarly,

a + 5d + a + 9d = 44

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

Subtract equation (1) from (2).

2a + 14d - (2a + 10d) = 44 - 24

2a + 14d - 2a - 10d = 20

4d = 20

d = 20/4

d = 5

Substitute the value of d in equation (1).

2a + 10(5) = 24

2a = 24 - 50

2a = 24 - 50

2a = - 26

a = - 26/2

a = - 13

We know,

General form of an AP is a , a + d , a + 2d...

Hence,

required AP = - 13 , - 13 + 5 , - 13 + 2(5)...

required AP = - 13 , - 8 , - 3....