This problem involves two Arithmetic Progressions (APs), let's call them AP 'a' and AP 'b'. We are given relationships between their common differences and specific terms, and we need to find the first term ($a_1$) of AP 'a'.
Determining AP 'b' Common Difference ($d_b$)
We are given two terms from AP 'b': $b_{31} = -277$ and $b_{43} = -385$. The formula for the difference between two terms in an AP is $T_n - T_k = (n-k)d$. Using this:
$b_{43} - b_{31} = (43 - 31)d_b$
Substitute the given values:
$-385 - (-277) = (12)d_b$
$ -385 + 277 = 12d_b $
$ -108 = 12d_b $
Now, solve for $d_b$:
$ d_b = \frac{-108}{12} $
$ d_b = -9 $
Calculating AP 'a' Common Difference ($d_a$)
The problem states that the common difference of AP 'a' ($d_a$) is 13 more than the common difference of AP 'b' ($d_b$).
$ d_a = d_b + 13 $
Substitute the value of $d_b$ we found:
$ d_a = -9 + 13 $
$ d_a = 4 $
Finding the First Term ($a_1$) of AP 'a'
We know the 78th term of AP 'a' is $a_{78} = 327$, and we just found its common difference $d_a = 4$. The formula for the n-th term of an AP is $a_n = a_1 + (n-1)d$. Applying this to $a_{78}$:
$ a_{78} = a_1 + (78 - 1)d_a $
Substitute the known values:
$ 327 = a_1 + (77)(4) $
$ 327 = a_1 + 308 $
Solve for $a_1$:
$ a_1 = 327 - 308 $
$ a_1 = 19 $
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