Question

Find the sum of first 50 terms of an A.P. whose second term is 14 and 5th term is 26​

Answer 1

2nd term is 14

an=a+(n-1)d

14=a+d

an=a+(n-1)d

26=a+4d

a+4d=26

a+d. =14

- -. -

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3d = 12

d=4

a=14-12

a=2

Sn=n/2(2a+(n-1)d)

Sn=50/2(2(2)+(50-1)4)

Sn=25(4+196)

Sn=25(200)

S50=5000

Answer 2

a2 = 14 and a3 = 18

Common difference = a3 - a2 = 18 - 14 = 4 = d

Now

a2 = a+d=14

a+4=14

a = 10

Now, sum of 51 terms

={51(2a+(50)d)}/2

={51(20+200)}/2

={51×220}/2

=51×110=5610

Therefore sum of 51 terms is 5610

Hope this helps;)