In an A.P., first term is -5 and last term is 45. If sum of all the numbers in the A.P. is 120, then how many terms are there? What is the common difference.

Answer

There are 6 terms in the A.P. the common difference`=d=10`.

Explanation

Let the common difference be d and the number of terms be `n`
Here,`a=-5and t_(n)=45,
S_(n)=120`
`S_(n)=(n)/(2)(t_(1)+t_(n))` …(Formula)
`therefore120=(n)/(2)(-5+45)` …(Substituting the values)
`therefore120=(n)/(2)xx40`
`therefore120=20n " "therefore20n=120 " " thereforen=(120)/(20)`
`thereforen=6` ...(1)
`therefore` there are terms in A.P.
`t_(n)=t_(6)=45` ...(2)
`t_(n)=a+(n-1)d` ...(Formula)
`therefore45=-5+(6-1)d` ...(Substituting the values)
`therefore45+5=5d`
`therefore5d=50`
`therefored=(50)/(5)`
`therefored=10`