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

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Answer 1 · 2018-06-24T19:40:17

#10# & #314#

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 · 2018-06-24T19:57:55

First term =10 and Last term =304

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#