the sum of the 4th and 8th is it 24 the sum of the 6th and 10th term is 44 find the first three terms of the AP
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As we know that,
a n = a + (n − 1) d
a 4 = a + (4 − 1) d
a 4 = a + 3d
Similarly,
a 8 = a + 7d
a 6 = a + 5d
a 10 = a + 9d
Sum of 4th and 8th term = 24 (Given)
a 4 + a 8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12.................... (i)
Sum of 6th and 10th term = 44 (Given)
a 6 + a 10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ......................(ii)
Solving (i) and (ii), we get,

From equation (i), we get,
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a 2 = a + d = − 13 + 5 = −8
a 3 = a 2 + d = − 8 + 5 = −3
a n = a + (n − 1) d
a 4 = a + (4 − 1) d
a 4 = a + 3d
Similarly,
a 8 = a + 7d
a 6 = a + 5d
a 10 = a + 9d
Sum of 4th and 8th term = 24 (Given)
a 4 + a 8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12.................... (i)
Sum of 6th and 10th term = 44 (Given)
a 6 + a 10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 ......................(ii)
Solving (i) and (ii), we get,

From equation (i), we get,
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a 2 = a + d = − 13 + 5 = −8
a 3 = a 2 + d = − 8 + 5 = −3
APrajput8755:
right
Answered by
2
Answer:
- 13 , - 8 , - 3 .
Step-by-step explanation:
Let the first term a and common difference be d.
We know :
t_n = a + ( n - 1 ) d
t_4 = a + 3 d
t_8 = a + 7 d
We have given :
t_4 + t_8 = 24
2 a + 10 d = 24
a + 5 d = 12
a = 12 - 5 d ....( i )
t_6 = a + 5 d
t_10 = a + 9 d
: t_6 + t_10 = 44
2 a + 14 d = 44
a + 7 d = 22
a = 22 - 7 d ... ( ii )
From ( i ) and ( ii )
12 - 5 d = 22 - 7 d
7 d - 5 d = 22 - 12
2 d = 10
d = 5
We have :
a = 12 - 5 d
a = 12 - 25
a = - 13
Now required answer as :
- 13 , - 8 , - 3 .
Finally we get answer.
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