1+5+3+9+5+13+7+.................20terms = ?

1 + 5 + 3 + 9  + 5  + 13 + 7 ...........

dividing this series in two series

S = [1 + 3 + 5  ........10 terms] + [5 + 9 + 13 .........10terms]

now we have two AP series , sum of each will be n/2 [ 2a + (n-1)d]

  = 5[2+9*2] + 5[10 + 9*4]

  =5 [ 20 + 46 ]

  = 330

this is the required ans

Approved

thank you very much .!