Question

Determine the first and last terms of an arithmetic series with 50 terms, a common difference of 6, and a sum of 7850. Help?

Answer 1

`10` & `314`

Explanation:

Let `a` & `l` be the first & last terms respectively of an A.P.

Now, the last 50th term of series having `n=50` terms with a common difference `d=6` will be `l`

`\therefore l=a+(n-1)d`

`l=a+(50-1)6`
`l-a=294 \ .......(1)`

Now, the sum of AP with `n=50` terms is 7850

`\therefore 7850=\frac{n}{2}(a+l)`

`7850=\frac{50}{2}(a+l)`
`a+l=314\ .......(2)`

Adding (1) & (2), we get

`2l=608`

`l=304` &

`a=314-l=314-304=10`

hence, the first & last terms of given AP. are `10` & `304`

Answer 2

First term =10 and Last term =304

Explanation:

We have No. of terms=50;C.D=6;`S_n`=7850

using `S_n = n*(2*a+(n-1)*d)/2`
now putting the values

`50*(2*a+(50-1)6)/2=7850`

`50*a+7350=7850`
`50a=500`
`a=10`
now again using `S_n = n*(a+l)/2` where l=last term

`25*(10+l)=7850`
`10+l=314`
l=last term =`304`