Question

7th term of an ap is 40 then the sum of first thirteen terms is.

Answer 1

Answer:

t7=a+6d

a+6d=40...(1)

t13=a+12d

a+12d=2(6d)+a

a+12d=2(40-a)+a....from1

a+12d=80-a

d=80/12

d=20/3

therefore a=0...substitute d in 1

t13=0+80=80

s13=80×13/2

s13=40×13=520

Answer 2

Answer :

Sum of thirteen terms is 520

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