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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.
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Rewritten by : Barada
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....
