The sum of 4th and 8th term of an ap is 24 and the sum of 6th and 10th term of AP is 34 find the first term and the common difference of AP
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6
let the first term of AP is 'a' and common difference is 'd' then the
4th term is a+3d
8th term is a+7d
6th term is a+5d
10th term is a+9d
given that sum of 4th and 8th is 24 so
a+3d+a+7d=24 ,2a+10d=24
sum of 6th and 10th term is 34
a+5d+a+9d=34, 2a+14d=34
on solving the two equations we get d=2.5
a=-0.5
the answer is d=2.5 a=-0.5
Answered by
6
Answer:
a = - 13
d = 5 / 2
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 = 34
2 a + 14 d = 34
a + 7 d = 17
a = 17 - 7 d ... ( ii )
From ( i ) and ( ii )
12 - 5 d = 17 - 7 d
7 d - 5 d = 17 - 12
2 d = 5
d = 5 / 2
We have :
a = 12 - 5 / 2
a = 19 / 2
a = - 13
d = 5 / 2
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