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

Answer: 15

Mean $\left(X_1,X_2\dots X_n\right)=\frac{\frac{n}{2}\left[2X_1+\left(n-1\right)d\right]}{n}=5$
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 $\left(y_{P+1},y_{P+2},y_{P+n}\right)$
$\frac{\frac{n}{2}\left[2X_6+\left(n-1\right)d\right]}{n}=\frac{y_{P+1}\left({\left(16\right)}^n-1\right)}{15n}=\frac{y_1{\left(16\right)}^P\left({\left(16\right)}^n-1\right)}{15n}$
$X_6+\left(\frac{n-1}{2}\right)d=\frac{{\left(16\right)}^P\left(75n\right)}{15n}=5\times {\left(16\right)}^P$
$X_1+5d+\left(\frac{n-1}{2}\right)d=5\times {\left(16\right)}^P$
$5-\left(\frac{n-1}{2}\right)d+5d+\left(\frac{n-1}{2}\right)d=5\times {\left(16\right)}^P\Rightarrow d={\left(16\right)}^P-1$
$\because d$ is $2$ digit natural number $\Rightarrow P=1,d=15$