Let the first term, common difference and number of terms of an AP are a, d and n, respectively.
We know that, if last term of an AP is known, then
`l=a+(n-1)d " " …(i)`
and nth term of an AP is
`T_(n)=a+(n-1)d " " ...(ii)`
Given that, 26th term of an AP = 0
`implies T_(26)=a+(26-1)d=0 " " `[from Eq. (i)]
`implies a+25d=0 " " ...(iii)`
11th term of an AP = 3
`implies T_(11)=a+(11-1)d=3 " " `[from Eq. (ii)]
`implies a+10d=3 " " ...(iv)`
and last term of an AP = - 1/5
`implies l=a+(n-1)d " " `[from Eq. (i)]
`implies -1//5=a+(n-1)d " " ...(v)`
Now, subtracting Eq. (iv) from Eq. (iii),
`({:(a,+,25d=0),(a,+,10d=3),(-,-," "-):})/({:(15d=-3),(" "d=-(1)/(5)):})`
Put the value of d in Eq. (iii), we get
`a+25(-(1)/(5))=0`
`implies a-5=0impliesa=5`
Now, put the value of a,d in Eq. (v), we get
`-1//5=5+(n-1)(-1//5)`
`implies -1=25-(n-1)`
`implies -1=25-n+1`
`implies n=25+2=27`
Hence, the common difference and number of terms are -1/5 and 27, respectively.
