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805+ (Hard)|   Sequences|      
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twobagels
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1/(a+b), 1/(b+5) and 1/(a+5) are three terms of an arithmetic 19 Feb 2021, 12:09
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95% (hard)

Question Stats:

47% (02:46) correct 53% (02:34) wrong based on 119 sessions

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\(\frac{1}{(a+b)}\), \(\frac{1}{(b+5)}\) and \(\frac{1}{(a+5)}\) are three terms of an arithmetic progression. Which of the following represents three consecutive terms in the progression in question?

A. 25, 2a, 2b
B. 2a, 25, 2c
C. \(a^2\), \(5^2\), \(b^2\)
D. \(a^2\), \(b^2\), \(5^2\)
E. \(5^2\), \(a^2\), \(b^2\)
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E
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1/(a+b), 1/(b+5) and 1/(a+5) are three terms of an arithmetic 19 Feb 2021, 20:43
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twobagels
\(\frac{1}{(a+b)}\), \(\frac{1}{(b+5)}\) and \(\frac{1}{(a+5)}\) are three terms of an arithmetic progression. Which of the following represents three consecutive terms in the progression in question?

A. 25, 2a, 2b
B. 2a, 25, 2c
C. \(a^2, 5^2,b^2\)
D. \(a^2,b^2,5^2\)
E. \(5^2,a^2,b^2\)
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Start with - Difference in consecutive terms in an AP is the same

\(\frac{1}{(a+b)}, \frac{1}{(b+5)},\frac{1}{(a+5)}\) are in AP.

\(\frac{1}{(a+b)}+\frac{1}{(a+5)}=2*\frac{1}{(b+5)}\)

\((a+5+a+b)(b+5)=2*(a+b)(a+5)\)

\((2a+b+5)(b+5)=(a+b)(2a+10)\)

\(2ab+10a+b^2+10b+25=2a^2+2ab+10a+10b\)

\(b^2+25=2a^2\)

\(25-a^2=a^2-b^2\)

So \(25,a^2,b^2\) are consecutive terms in an AP as the difference between the two consecutive terms is the same : \(25-a^2=a^2-b^2\)

E
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1/(a+b), 1/(b+5) and 1/(a+5) are three terms of an arithmetic 25 Aug 2021, 07:46
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chetan2u
twobagels
\(\frac{1}{(a+b)}\), \(\frac{1}{(b+5)}\) and \(\frac{1}{(a+5)}\) are three terms of an arithmetic progression. Which of the following represents three consecutive terms in the progression in question?

A. 25, 2a, 2b
B. 2a, 25, 2c
C. \(a^2, 5^2,b^2\)
D. \(a^2,b^2,5^2\)
E. \(5^2,a^2,b^2\)


Start with - Difference in consecutive terms in an AP is the same

\(\frac{1}{(a+b)}, \frac{1}{(b+5)},\frac{1}{(a+5)}\) are in AP.

\(\frac{1}{(a+b)}+\frac{1}{(a+5)}=2*\frac{1}{(b+5)}\)

\((a+5+a+b)(b+5)=2*(a+b)(a+5)\)

\((2a+b+5)(b+5)=(a+b)(2a+10)\)

\(2ab+10a+b^2+10b+25=2a^2+2ab+10a+10b\)

\(b^2+25=2a^2\)

\(25-a^2=a^2-b^2\)

So \(25,a^2,b^2\) are consecutive terms in an AP as the difference between the two consecutive terms is the same : \(25-a^2=a^2-b^2\)

E
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It is mentioned these terms are 3 terms of AP, not the consecutive terms, can you explain why you have taken it like that
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1/(a+b), 1/(b+5) and 1/(a+5) are three terms of an arithmetic 24 Jun 2022, 20:26
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Same question as asked above.
It wasn't mentioned that these are consecutive terms, then how are we assuming this? I wasted a lot of time only because I couldn't understand how to start.
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1/(a+b), 1/(b+5) and 1/(a+5) are three terms of an arithmetic 25 May 2026, 04:21
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Method 1: In an AP a, a+d, a+2d..we know that some of 1st and 3rd term will be a+(a+2d)= 2a+2d = 2(a+d) which 2 times the middle term..(a+d)

MEthod 2: you can also do (a+d)-a= (a+2d)-(a+d) to prove an AP or play with numbers in an AP.

Method 1 is what Chetan has used in the 1st post as method to the answer.

Method 2:

1/(b+5) - 1/(a+b) = 1/(a+5) - 1/(b+5)

we get (a+b-b-5)/(b+5)(a+b) =(b+5-a-5)/(a+5)(b+5)

we get (a-5)(a+5)=(a+b)(b-a)

\( a^2-25= b^2-a^2\)

\(2a^2=b^2+25\) Thsi is of the form AP 2times middle term is sum of terms before and after in an AP as per method 1 above

Hence the AP is \( b^2,a^2, 25 or 25, a^2, b^2\) Hence e is the answer
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1/(a+b), 1/(b+5) and 1/(a+5) are three terms of an arithmetic 25 May 2026, 05:01
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\(\frac{1}{(a+b)}\), \(\frac{1}{(b+5)}\) and \(\frac{1}{(a+5)}\) are three terms of an arithmetic progression.

Which of the following represents three consecutive terms in the progression in question?

\(\frac{2}{b+5} = \frac{1}{a+b} + \frac{1}{a+5}\)

\(\frac{2}{b+5} = \frac{(a+b+a+5)}{(a+b)(a+5)} = \frac{(2a+b+5)}{(a+b)(a+5)}\)
\(2(a+b)(a+5) = (b+5)(2a+b+5)\)
\(2a^2+2ab+10a+10b = 2ab+b^2+5b+10a+5b+25\)
\(2a^2 = b^2 + 25 = b^2 + 5^2\)
\(2a^2 = b^2 + 5^2\)

a^2 is 2nd term (middle term) and b^2 & 5^2 are extremes ends of 3 term arithmetic progression

IMO E
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1/(a+b), 1/(b+5) and 1/(a+5) are three terms of an arithmetic 25 May 2026, 08:14
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twobagels
\(\frac{1}{(a+b)}\), \(\frac{1}{(b+5)}\) and \(\frac{1}{(a+5)}\) are three terms of an arithmetic progression. Which of the following represents three consecutive terms in the progression in question?

A. 25, 2a, 2b
B. 2a, 25, 2c
C. \(a^2\), \(5^2\), \(b^2\)
D. \(a^2\), \(b^2\), \(5^2\)
E. \(5^2\), \(a^2\), \(b^2\)
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Question wording is suspect since in the second part, it is clearly mentioned "three consecutive terms in the progression" while "consecutive" is not mentioned in the first part. Though we have no other choice but to assume that they are in order.
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