Question -
Answer -
Given that,
3rd┬аterm, a3┬а= 4
and 9th┬аterm, a9┬а= тИТ8
We know that,
an┬а=┬аa+(nтИТ1)d
Therefore,
a3┬а=┬аa+(3тИТ1)d
4 =┬аa+2d┬атАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж┬а(i)
a9┬а=┬аa+(9тИТ1)d
тИТ8 =┬аa+8d┬атАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАжтАж┬а(ii)
On subtracting equation┬а(i)┬аfrom┬а(ii), we will get here,
тИТ12 = 6d
d┬а= тИТ2
From equation┬а(i), we can write,
4 =┬аa+2(тИТ2)
4 =┬аaтИТ4
a┬а= 8
Let┬аnth┬аterm ofthis A.P. be zero.
an┬а=┬аa+(nтИТ1)d
0 = 8+(nтИТ1)(тИТ2)
0 = 8тИТ2n+2
2n┬а= 10
n┬а= 5
Hence, 5th┬аtermof this A.P. is 0.