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Given a three digit number whose digit are the three successive terms of a G.P. If we subtract 792 from it,we get a number written by the same digit in reverse order.Now if we subtract four from the hundred's digit of the initial number and leave the other digit unchanged,we get a number whose digit are successive terms of an A.P. Find the number.
Let our number's digits be ar² ar aSo, we have our number asar²(100) + ar(10) + aNow, if we subtract 792 from this, we get our digits reversed, soar²(100) + ar(10) + a - 792 = a(100) + ar(10) + ar²100ar² + 10ar + a - 792 = 100a + 10ar + ar²99ar² - 792 = 99aar² - 8 = aa(r² - 1) = 8....(1) now if we subtract 4 frm the hundred digit ar² , ar – 4, a are in A.P. 2ar – 8 = a(1+r²) use this and the earier equation to find a and r.
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