Official solution
Let common difference of series
a1
, a2
, a3
,..., an
be d.
\[ \because \] a40 = a1 + 39d == --159 ...(i)
and a100 = a1 + 99d = --399 ...(ii)
From eqn. (ii) and (i)
d = --4 and a1
= --3.
The common difference of the A.P.
b1, b2, \ldots , bm
is 2 more than the common
difference of A.P. a1, a2, \ldots , an.
\[ \therefore \] Common difference of b1
, b2
, b3
, ..., be (--2).
\[ \because \] b100 = a70
\[ \therefore \] b1
+ 99(--2) = (--3) + 69(--4)
\[ \therefore \] b1
= 198 -- 279
\[ \therefore \] b1
= -- 81