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.

Explanation

Let the first term, common difference and number of terms in the A.P. be a, d and n respectively.
Given that `a_(26)=0`
`rArr a+25d=0`
`rArr a=-25d " " ...(1)`
`a_(11)=3 rArr a+10d=3 `
`rArr -25d+10d=3 " "`[from (1)]
`rArr -15d=3`
`rArr d=-(1)/(5)`
put `d=-(1)/(5)` in equation (1), we get
`a=-25(-(1)/(5))=5 " " and " " a_(n)=-(1)/(5)`
`rArr 5+(n-1)(-(1)/(5))=-(1)/(5) rArr (n-1)(-(1)/(5))=-(1)/(5)-5=-(26)/(5)`
`rArr n-1=26 rArr n=27`
Now, common difference `=-(1)/(5)`
No. of terms in the A.P. = 27.