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------------------------------------------------ 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....

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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....

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