Question

find the sum of all old numbar from 1 to 250 arithmetic progression 10th​

Answer 1

Step-by-step explanation:

1+3+5+........+249(nth term)

using A(n)=a+(n-1)d

249=1+(n-1)2

n=125

Using S(n)=n/2(a+l)

S(125)=125/2(1+249)=125×125=15625

Answer 2

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here is your answer.......

Number series: 1,3,5,7.........499

First term a=1

Common difference d=2

Total number of terms n=250

sum=n/2×(a+tn)

250/2×(1+499)

(250×500)/2

125000/2

62500

hope helps you...

please mark brainliest.....