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The 26th, 11th and the last terms of an ...

The 26th, 11th and the last terms of an AP are, 0, 3 and `-(1)/(5)`,respectively. Find the common difference and the number of terms.

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Let the first term, common difference and the number of terms of an A.P. are a, d and n respectively.
Then `T_(26)=0`
`rArr a+25d=0 " " ...(1)`
`T_(11)=3`
`rArr a+10d=3 " " ...(2)`
Subtracting equation (2) from equation (1), we get
`{:(" "a+25d=""0),(underset(-)" "a underset(-)+10d=""underset(-)3),(bar(" "15d=-3" ")):}`
`rArr" " d=-(1)/(5)`
From equation (1), we get
`a=-25d`
`=-25xx(-(1)/(5))=5`
Now, the last term `=-(1)/(5)`
`rArr a+(n-1)d=-(1)/(5)`
`rArr 5+(n-1)(-(1)/(5))=-(1)/(5)`
`rArr 25-n+1=-1`
`rArr n=27`
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