An arithmetic progression (A.P) consists of $12$ terms whose sum is $354$. The ratio of the sum of the even terms to the sum of the odd terms is $32 : 27$. Find the common difference of the progression.
Let the first term of A.P $= t_1 = a$ and the common difference $= d$
Sum to $n$ terms of an A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$
Given: Number of terms of A.P $= n = 12$; $\;\;$ $S_n = 354$
i.e. $\;$ $\dfrac{12}{2} \times \left[2a + 11d\right] = 354$
i.e. $\;$ $2a + 11d = 59$ $\;\;\; \cdots \; (1)$
$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right) d$
Even terms of A.P $= t_2, \; t_4, \; t_6, \cdots, t_{12}$ $\;$ i.e. $\;$ $6$ terms.
For even terms, first term $= t_2 = a + d$ $\;$ and last term $= t_{12} = a + 11d$
$\therefore \;$ Sum of even terms of A.P $= S_{\text{even}} = \dfrac{t_2 + t_{12}}{2}$
i.e. $\;$ $S_{\text{even}} = \dfrac{a + d + a + 11d}{2} = a + 6d$
Odd terms of A.P $= t_1, \; t_3, \; t_5, \cdots, t_{11}$ $\;$ i.e. $\;$ $6$ terms.
For odd terms, first term $= t_1 = a$ $\;$ and last term $= t_{11} = a + 10d$
$\therefore \;$ Sum of odd terms of A.P $= S_{\text{odd}} = \dfrac{t_1 + t_{11}}{2}$
i.e. $\;$ $S_{\text{odd}} = \dfrac{a + a + 10d}{2} = a + 5d$
Given: $\dfrac{S_{\text{even}}}{S_{\text{odd}}} = \dfrac{32}{27}$
i.e. $\;$ $\dfrac{a + 6d}{a + 5d} = \dfrac{32}{27}$
i.e. $\;$ $27a + 162 d = 32a + 160d$
i.e. $\;$ $2d = 5a$ $\implies$ $a = \dfrac{2d}{5}$
Substituting $a = \dfrac{2d}{5}$ in equation $(1)$ gives
$2 \times \dfrac{2d}{5} + 11d = 59$
i.e. $\;$ $\dfrac{59 d}{5} = 59$ $\implies$ $d = 5$
i.e. $\;$ The common difference of the given A.P is $d = 5$