The sum 4th & 8th term of an ap is 24 and the sum of 6th & 10th term 44 .find the 5th term of ap
Answers
ANSWER:
Given that
4th term( t4 ) + 8th term ( t8 ) = 24
6th term ( t6 ) + 10th term ( t10 ) = 44
Using,
tn = a + (n - 1)d
Similarly
t4 = a + (4 - 1)d = a + 3d
t8 = a + (8 - 1)d = a + 7d
t6 = a + (6 - 1)d = a + 5d
t10 = a + (10 - 1)d = a + 9d
a + 3d + a + 7d = 24 ( .°. Given )
2a + 10d = 24 -----------( 1 )
a + 5d + a + 9d = 44 ( .°. Given )
2a + 14d = 44 -----------( 2 )
Subracting equation ( 2 ) from ( 1 )
2a + 10d = 24
- 2a + 14d = 44
_______________
-4d = - 20
d = -20/-4
Common difference ( d ) = 5
TO FIND :
5th term ( t5 ) = a + (5 - 1)d = a + 4d
Substitute d = 5 in equation ( 1 )
2a + 10(5) = 24
2a = 24 - 50
2a = - 26
a = -26/2 = - 13
a = -13
We know that
‘a’ means first term ( t1 )
Now,
t5 = a + 4d
t5 = - 13 + 4(5)
t5 = - 13 + 20
t5 = 7
Hence 5th term (t5) = 7
GIVEN:
Sum of 4th & 8th term of an ap is 24
Sum of 6th & 10th term = 44
TO FIND:
Fifth term of AP
SOLUTION:
4th term: a + 3d ; 8th term: a + 7d ; 6th term: a + 5d and 10th term: a + 9d
Sum of 4th and 8th term = 24
a + 3d + a + 7d = 24
2a + 10 = 24 -------(1)
Sum of 6th and 10th term = 44
a + 5d + a + 9d = 44
2a + 14d = 44 -----(2)
After subtracting both eq - (1) and (2)
==> -4d = -20
==> d = 20/4
==> d = 5
Common Difference = 5
Substitute d in eq - (1) to find first term (a).
2a + 10d = 24
2a + 10(5) = 24
2a + 50 = 24
2a = 24 - 50
==> 2a = -26
==> a = -16/2
==> a = -13
First term = -13
Fifth term of AP = a + 4d = (-13) + 4(5) = -13 + 20 = 7.
Therefore, fifth term of AP is 7.