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Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.?
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Find a thee digit no. whose consecutive digit form a G.P. If wE subtra...
**Finding the three-digit number that forms a G.P.**
To find the three-digit number whose consecutive digits form a geometric progression (G.P.), let's assume the number is "abc", where a, b, and c are the digits of the number.
The G.P. can be written as a, ar, and ar^2, where r is the common ratio.
So, we have the equations:
b = ar
c = ar^2
Since a, b, and c are digits, we know that they can only take values from 0 to 9.
To find the possible values of a, b, and c, let's substitute b and c in terms of a and r:
ar = ar^2
Dividing both sides by a:
r = r^2
Simplifying the equation, we get:
r^2 - r = 0
Factorizing the equation, we have:
r(r - 1) = 0
So, r = 0 or r = 1.
**Case 1: r = 0**
If r = 0, the G.P. becomes a, a(0), and a(0)^2.
This implies that b = 0 and c = 0.
But since the number "abc" is a three-digit number, a cannot be equal to 0.
Therefore, r cannot be 0.
**Case 2: r = 1**
If r = 1, the G.P. becomes a, a(1), and a(1)^2.
This implies that b = a and c = a.
So, the three-digit number becomes aaa.
Now, let's find the possible values of a.
Since a, b, and c are digits, their sum should be less than or equal to 9 + 9 + 9 = 27.
If we substitute a = 1, the number becomes 111.
If we substitute a = 2, the number becomes 222.
If we substitute a = 3, the number becomes 333.
If we substitute a = 4, the number becomes 444.
If we substitute a = 5, the number becomes 555.
If we substitute a = 6, the number becomes 666.
If we substitute a = 7, the number becomes 777.
If we substitute a = 8, the number becomes 888.
If we substitute a = 9, the number becomes 999.
Therefore, the three-digit number whose consecutive digits form a G.P. is either 111, 222, 333, 444, 555, 666, 777, 888, or 999.
**Subtracting 792 from the number**
Let's subtract 792 from the number to get a new number consisting of the same digits written in reverse order.
If the original number is aaa, subtracting 792 will give us aaa - 792.
Now, let's evaluate the possible values of aaa - 792:
111 - 792 = -681 (not a three-digit number)
222 - 792 = -570 (not a three-digit number)
333 - 792 = -459 (not a three-digit number)
444 - 792 = -348 (not a three-digit number)
555 - 792 = -237 (not a three-digit number)
666 - 792 = -126 (not a
To find the three-digit number whose consecutive digits form a geometric progression (G.P.), let's assume the number is "abc", where a, b, and c are the digits of the number.
The G.P. can be written as a, ar, and ar^2, where r is the common ratio.
So, we have the equations:
b = ar
c = ar^2
Since a, b, and c are digits, we know that they can only take values from 0 to 9.
To find the possible values of a, b, and c, let's substitute b and c in terms of a and r:
ar = ar^2
Dividing both sides by a:
r = r^2
Simplifying the equation, we get:
r^2 - r = 0
Factorizing the equation, we have:
r(r - 1) = 0
So, r = 0 or r = 1.
**Case 1: r = 0**
If r = 0, the G.P. becomes a, a(0), and a(0)^2.
This implies that b = 0 and c = 0.
But since the number "abc" is a three-digit number, a cannot be equal to 0.
Therefore, r cannot be 0.
**Case 2: r = 1**
If r = 1, the G.P. becomes a, a(1), and a(1)^2.
This implies that b = a and c = a.
So, the three-digit number becomes aaa.
Now, let's find the possible values of a.
Since a, b, and c are digits, their sum should be less than or equal to 9 + 9 + 9 = 27.
If we substitute a = 1, the number becomes 111.
If we substitute a = 2, the number becomes 222.
If we substitute a = 3, the number becomes 333.
If we substitute a = 4, the number becomes 444.
If we substitute a = 5, the number becomes 555.
If we substitute a = 6, the number becomes 666.
If we substitute a = 7, the number becomes 777.
If we substitute a = 8, the number becomes 888.
If we substitute a = 9, the number becomes 999.
Therefore, the three-digit number whose consecutive digits form a G.P. is either 111, 222, 333, 444, 555, 666, 777, 888, or 999.
**Subtracting 792 from the number**
Let's subtract 792 from the number to get a new number consisting of the same digits written in reverse order.
If the original number is aaa, subtracting 792 will give us aaa - 792.
Now, let's evaluate the possible values of aaa - 792:
111 - 792 = -681 (not a three-digit number)
222 - 792 = -570 (not a three-digit number)
333 - 792 = -459 (not a three-digit number)
444 - 792 = -348 (not a three-digit number)
555 - 792 = -237 (not a three-digit number)
666 - 792 = -126 (not a
Community Answer
Find a thee digit no. whose consecutive digit form a G.P. If wE subtra...
Answer is 931. Because in the question it is given that after subtracting 792 from the original number , the number obtained will be reverse of the original number and as the number will be <= 999="" (three="" digit="" number)="" ,="" also="" the="" number="" is="" in="" g.p.="" (obviously="" not="" an="" increasing="" one).="" so="" the="" first="" term="" will="" have="" to="" be="" 8="" or="" 9="" and="" the="" ratio="" will="" have="" to="" be="" from="" 1,2/3,1/2="" and="" 1/3.so="" the="" 5="" possible="" numbers="" are="" :999="" (a="9," r="1" but="" couldn't="" satisfy="" the="" condition)888="" (a="8," r="1" but="" couldn't="" satisfy="" the="" condition)842="" (a="8," r="1/2" but="" couldn't="" satisfy="" the="" condition)964="" (a="9," r="2/3" but="" couldn't="" satisfy="" the="" condition)931="" (a="9," r="1/3" and="" satisfies="" the="" condition)hence,="" 931="" is="" the="" correct="" answer.="" hope="" you="" understand="" it.="" 999="" (three="" digit="" number)="" ,="" also="" the="" number="" is="" in="" g.p.="" (obviously="" not="" an="" increasing="" one).="" so="" the="" first="" term="" will="" have="" to="" be="" 8="" or="" 9="" and="" the="" ratio="" will="" have="" to="" be="" from="" 1,2/3,1/2="" and="" 1/3.so="" the="" 5="" possible="" numbers="" are="" :999="" (a="9," r="1" but="" couldn't="" satisfy="" the="" condition)888="" (a="8," r="1" but="" couldn't="" satisfy="" the="" condition)842="" (a="8," r="1/2" but="" couldn't="" satisfy="" the="" condition)964="" (a="9," r="2/3" but="" couldn't="" satisfy="" the="" condition)931="" (a="9," r="1/3" and="" satisfies="" the="" condition)hence,="" 931="" is="" the="" correct="" answer.="" hope="" you="" understand="" it.="">
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Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.?
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Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.?.
Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.?.
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Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.?, a detailed solution for Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.? has been provided alongside types of Find a thee digit no. whose consecutive digit form a G.P. If wE subtract 792 from this no. we get a no. consisting of the same digits written in the reverse order. Now if we increase the second digit of the required no. by 2 ,then the resulting digits will form an A.P. -from A. P. and G.P.? theory, EduRev gives you an
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