Question

change into pasive or active voice..​

Answer 1

Answer:

I don't know the answer sorry

Answer 2

Answer:

We have to find sum of even natural numbers between 300 and 400.

Given series is in A.P. as difference between consecutive terms is constant.

a

1

=302 a n

=398

d=304−302=2

nth term of A.P. is given by,

a n

=a 1+(n−1)d

∴398=302+(n−1)2

∴2(n−1)=398−302

∴2(n−1)=96

∴(n−1)=48

∴n=49

Now, Sum of n terms of A.P. is,

∑a n = 2n [2a+(n−1)d]

∴∑a n = 249 [(2×302)+(49−1)×2]

∴∑a n = 249 [604+(48×2)]

∴∑a n = 249 [604+96]

∴∑a n = 249 [700]

∴∑a n =49×350

∴∑a n =17150