Find the
a.P whose third term is 16 and the differences of the 9th term from 11th term is 12
Answers
Step-by-step explanation:
here we have given,
t3 = 16
t11 - t9 = 12
we know that,
tn = a + (n - 1) d
therefor,
t3 = a + (3 - 1)d
a + 2d = 16 __________(1)
Now ,
t11 - t9 = 12
(a + (11 - 1)d)-(a + (9-1)d) = 12
(a + 10d) - (a + 8d) = 12
a + 10d - a - 8d = 12
10d - 8d = 12
2d = 12
d = 12 /2
d = 6
put the value of d in equation (1)
a + 2d = 16
a + 2(6)=16
a + 12 = 16
a = 16 - 12
a = 4
The terms of Ap are ,
t1 = a = 4
t2 = a + d = 4 + 6 = 10
t3 = t2 + d = 10 + 6 = 16
t4 = t3 + d = 16 + 6 = 22
t5 = t4 + d = 22 + 6 = 28
therefor, AP is 4 , 10 , 16 , 22 , 28 ..
GIVEN :
Third term of an AP = 16
a + 2d = 16 ----(1)
Difference between 9th and 11th term = 12
(a + 10d) - (a + 8d) = 12
a + 10d - a - 8d = 12
10d - 8d = 12
2d = 12
d = 12/2
d = 6
Difference (d) = 6
Substitute d in eq - (1) to find first term (a)
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
a = 16 - 12
a = 4
First term (a) = 4
Second term = a + d = 4 + 6 = 10
Third term = a + 2d = 4 + 12 = 16
Fourth term = a + 3d = 4 + 18 = 22
Therefore, AP is 4, 10, 16, 22,......