Arithmetic progression Questions in English

(D) Let the arithmetic means be $A_1, A_2, \dots, A_{39}$ between $a = 59$ and $b = 159$.
The common difference $d$ is given by $d = \frac{b - a}{n + 1} = \frac{159 - 59}{39 + 1} = \frac{100}{40} = 2.5$.
The $k$-th arithmetic mean is $A_k = a + k \cdot d = 59 + 2.5k$.
We need to find the mean of $A_{25}, A_{28}, A_{31}, A_{36}$:
$\text{Mean} = \frac{A_{25} + A_{28} + A_{31} + A_{36}}{4} = \frac{(59 + 25d) + (59 + 28d) + (59 + 31d) + (59 + 36d)}{4}$.
$\text{Mean} = \frac{4 \cdot 59 + (25 + 28 + 31 + 36)d}{4} = 59 + \frac{120d}{4} = 59 + 30d$.
Substituting $d = 2.5$,we get $\text{Mean} = 59 + 30(2.5) = 59 + 75 = 134$.

(A) Let the first term be $a = \frac{10}{3}$ and the number of terms be $n = 30$.
The last term $L$ is given by $L = a + (n-1)d = \frac{10}{3} + 29d$.
The sum of the $A$.$P$. is $S_{30} = \frac{n}{2}(a + L) = \frac{30}{2}(\frac{10}{3} + L) = 15(\frac{10}{3} + L) = 50 + 15L$.
Given that $S_{30} = L^3$,we have $L^3 = 15L + 50$,which implies $L^3 - 15L - 50 = 0$.
By testing values,if $L = 5$,then $5^3 - 15(5) - 50 = 125 - 75 - 50 = 0$.
Thus,$L = 5$ is a root.
Substituting $L = 5$ into the expression for the last term: $\frac{10}{3} + 29d = 5$.
$29d = 5 - \frac{10}{3} = \frac{15 - 10}{3} = \frac{5}{3}$.
Therefore,$d = \frac{5}{3 \times 29} = \frac{5}{87}$.