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

2 Answers

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

Jun 24, 2018

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