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Step-by-Step Solution
Step 1: Express the given terms of the A.P. in a standard form
Let the common difference of the A.P.
$a_1, a_2, \ldots$ be $d$. Then the $n$th term of this A.P. is given by:
$ a_n = a_1 + (n-1)\,d. $
Step 2: Form equations using the given conditions for the A.P. $a_n$
We are given:
$ a_{40} = -159, \quad a_{100} = -399. $
Using the general form of the $n$th term:
$ a_{40} = a_1 + 39\,d = -159 \quad \ldots \text{(i)}$
$ a_{100} = a_1 + 99\,d = -399 \quad \ldots \text{(ii)}$
Step 3: Solve the system of equations to find $a_1$ and $d$
Subtract equation (i) from equation (ii):
$ (a_1 + 99\,d) - (a_1 + 39\,d) = -399 - (-159). $
Simplifying,
$ 99\,d - 39\,d = -399 + 159 $
$ 60\,d = -240 $
$ d = -4. $
Substitute $d = -4$ into equation (i):
$ a_1 + 39(-4) = -159 $
$ a_1 - 156 = -159 $
$ a_1 = -3. $
Step 4: Identify the common difference of the second A.P. $b_n$
We are told the common difference of
$ b_1, b_2, \ldots $
is 2 more than the common difference of
$ a_1, a_2, \ldots .$
Since $d_{\text{(for a)}} = -4,$
the common difference for $b_n$ is
$ d_{\text{(for b)}} = d_{\text{(for a)}} + 2 = -4 + 2 = -2. $
Step 5: Use the condition $b_{100} = a_{70}$
The general term of $b_n$ is
$ b_n = b_1 + (n - 1)\,(-2). $
So,
$ b_{100} = b_1 + 99 \times (-2). $
Also,
$ a_{70} = a_1 + 69\,d = -3 + 69 \times (-4). $
Since $b_{100} = a_{70},$ equate these:
$ b_1 + 99(-2) = -3 + 69 \times (-4). $
Calculate the right side:
$ 69 \times (-4) = -276, $
so
$ -3 + (-276) = -279. $
Therefore,
$ b_1 + 99(-2) = -279. $
$ b_1 - 198 = -279 $
$ b_1 = -279 + 198 = -81. $
Step 6: State the final result
Thus, the first term of the A.P.
$ b_1 = -81.
