If the 10th term of an ap is 13, then the sum of 19 Terms
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Answer:
t10=a+9d
13=a+9d
a=13-9d
s19= n/2(2a+(n-1)d)
=19/2(2*(13-9d) +18d)
=19/2(26-18d)+18d
=19/2*26
=19*13
=247
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