find AP whose 4th and 8th term are 43 and 95 respectively also find 12th term
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Answered by
6
a + 3d = a4
a + 3d = 43 ---> (1)
a + 7d = a8
a + 7d = 95 ---> (2)
Subtracting (1) and (2) we get:
-4d = -52
=> d = 13
From equation (1)
=> a + 3(13) = 43
=> a + 39 = 43
=> a = 4
a12 = a + 11d
=> a12 = 4 + 11(13)
=> a12 = 4 + 143
=> a12 = 147
#BAL#answerwithquality
Answered by
1
Answer:
T4 =a+3d
T8 = a+7d
Here t4 =43
And t8 = 95
43 = a+3d
95 = a+7d
-52=-2d
D=26
Now, 43=a+3d
Putting value of D to find A
43=a+3×26
43=a+72
A=43-72
A=-39
So t12=a+11d
T12 = -39+11×26
T12=-39+286
T12=247
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