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

Question

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

MARKS App · Accepted Answer

Mean X 1 , X 2 … X n = n 2 2 X 1 + n - 1 d n = 5 Mean of y 1 , y 2 . . . . y n = y 1 16 n - 1 15 n = 5 2 X 1 + n - 1 d = 10 . . . 1 y 1 16 n - 1 = 75 n . . . 2 Mean of X 6 , X 7 . . . . X n + 5 = Mean of y P + 1 , y P + 2 , y P + n n 2 2 X 6 + n - 1 d n = y P + 1 16 n - 1 15 n = y 1 16 P 16 n - 1 15 n X 6 + n - 1 2 d = 16 P 75 n 15 n = 5 × 16 P X 1 + 5 d + n - 1 2 d = 5 × 16 P 5 - n - 1 2 d + 5 d + n - 1 2 d = 5 × 16 P ⇒ d = 16 P - 1 ∵ d is 2 digit natural number ⇒ P = 1 , d = 15

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