t9 + t22 = 61

OR [ a + 8d ] + [ a + 21d] = 61 [ Because tn = a + ( n-1 ) d where ‘a’ is the first- that is 2a + 29d = 61 ---(1) - term and ‘d’ the common difference. ]

Now sum of n terms of AP is given by Sn = n/2 [ 2a + ( n-1)d]

Thus s30 = 15 [ 2a+ 29d] = 15 x 61 Using equation (1)

Therefore Sum of the first 30 terms = 15 x 61 = 915