Question

Let $X_{1}, \, X_{2}, \, X_{3}....$ are in arithmetic progression with a common difference equal to $d$ which is a two digit natural number. $y_{1}, \, y_{2}, \, y_{3}....$ are in geometric progression with common ratio equal to $16$ . Arithmetic mean of $X_{1},X_{2}....X_{n}$ is equal to the arithmetic mean of $y_{1},y_{2}....y_{n}$ which is equal to $5$ . If the arithmetic mean of $X_{6},X_{7}....X_{n + 5}$ is equal to the arithmetic mean of $y_{P + 1},y_{P + 2}....y_{P + n}$ , then $d$ is equal to

Answer 1

Mean $\left(X_{1} , X_{2} \ldots X_{n}\right)=\frac{\frac{n}{2} \left[2 X_{1} + \left(n - 1\right) d\right]}{n}=5$
Mean of $\left(y_{1} , y_{2} . . . . y_{n}\right)=\frac{\frac{y_{1} \left(\left(16\right)^{n} - 1\right)}{15}}{n}=5$
$2X_{1}+\left(n - 1\right)d=10$ … $\left(1\right)$
$y_{1}\left(\left(16\right)^{n} - 1\right)=75n$ … $\left(2\right)$
Mean of $\left(X_{6} , X_{7} . . . . X_{n + 5}\right)=$ Mean of $\left(y_{P + 1} , y_{P + 2} , y_{P + n}\right)$
$\frac{\frac{n}{2} \left[2 X_{6} + \left(n - 1\right) d\right]}{n}=\frac{y_{P + 1} \left(\left(16\right)^{n} - 1\right)}{15 n}=\frac{y_{1} \left(16\right)^{P} \left(\left(16\right)^{n} - 1\right)}{15 n}$
$X_{6}+\left(\frac{n - 1}{2}\right)d=\frac{\left(16\right)^{P} \left(75 n\right)}{15 n}=5\times \left(16\right)^{P}$
$X_{1}+5d+\left(\frac{n - 1}{2}\right)d=5\times \left(16\right)^{P}$
$5-\left(\frac{n - 1}{2}\right)d+5d+\left(\frac{n - 1}{2}\right)d=5\times \left(16\right)^{P}\Rightarrow d=\left(16\right)^{P}-1$
$\because d$ is $2$ digit natural number $\Rightarrow P=1,d=15$