if the 19th term of anAP is equal to the three of its sixth term if the 9th term is 19then find the first three terms of an AP
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a (19)=3a (6)
a+18d=3 [a+5d]
a+18d=3a+15d
-2a=-3d
a=3/2d.........(i)
a (9)=19
a+8d=19
3/2d+8d=19
19d=38
D=2
From i equation. ..we get....
a=3/2×2
a=3.....
So the first three terms are.....3,5,7
a+18d=3 [a+5d]
a+18d=3a+15d
-2a=-3d
a=3/2d.........(i)
a (9)=19
a+8d=19
3/2d+8d=19
19d=38
D=2
From i equation. ..we get....
a=3/2×2
a=3.....
So the first three terms are.....3,5,7
Answered by
0
As Given
A19=3A6
a+18d=3(a+5d)
a+18d=3a+15d
2a-3d=0
Also given that
A9=19
a+8d=19
a=19-8d
putting this value of a into 2a-3d=0 we get d=2 so a will be 3
so the first three terms of AP are 3, 5, 7
A19=3A6
a+18d=3(a+5d)
a+18d=3a+15d
2a-3d=0
Also given that
A9=19
a+8d=19
a=19-8d
putting this value of a into 2a-3d=0 we get d=2 so a will be 3
so the first three terms of AP are 3, 5, 7
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