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If the square of the 7th term of an arithmetic progression with positi

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Bunuel
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If the square of the 7th term of an arithmetic progression with positi [#permalink] New post  13 May 2020, 02:22
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If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is:

A. 2 : 3
B. 3 : 2
C. 3 : 4
D. 4 : 3
E. 5 : 3


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Re: If the square of the 7th term of an arithmetic progression with positi [#permalink] New post  13 May 2020, 02:51
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The nth term of arithmetic progression can be written as \(= a+(n-1)d\)

a= first term; d = common difference

7th term of AP \(= a+6d\)

3rd term of AP \(= a+2d\)

17th term of AP \(= a+16d\)


\((a+6d)^2 = (a+2d)(a+16d)\)

\(a^2+12ad+36d^2 = a^2+18ad + 32d^2\)

\(6ad= 4d^2\)

Since d>0

\(6a = 4d\)

\(\frac{a}{d} = \frac{2}{3}\)


Bunuel wrote:
If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is:

A. 2 : 3
B. 3 : 2
C. 3 : 4
D. 4 : 3
E. 5 : 3


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Re: If the square of the 7th term of an arithmetic progression with positi [#permalink] New post  13 May 2020, 02:56
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(A+6D)^2=(A+2D)(A+16D) given
A^2+36D^2+12AD=A^2+32D^2+18AD
4D^2=6AD
2D=3A
A/D=2/3

Answer is A

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Re: If the square of the 7th term of an arithmetic progression with positi [#permalink] New post  13 May 2020, 04:51
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Bunuel wrote:
If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is:

A. 2 : 3
B. 3 : 2
C. 3 : 4
D. 4 : 3
E. 5 : 3



Solution


    • Let us assume that \(a_1, d,\) and \(a_n\) denotes first term, common difference and nth term of the given A.P. respectively.
      o Where n is a positive integer.
    • \(d > 0\)
    • We need to find \(\frac{a_1}{d}\)
    • Now, according to the questions
      \((a_7)^2 = a_3*a_{17}\)
      \(⟹ (a_1+ 6d)^2 = (a_1 + 2d)(a_1+16d)\)
      \(⟹(a_1)^2 + 12a_1*d + 36d^2 = (a_1)^2 + 18a_1*d + 32d^2\)
      \(⟹ 6a_1*d – 4d^2 = 0\)
      \(⟹ d(6a_1 – 4d) = 0 \)
      \(⟹ \frac{a_1 }{ d} = \frac{4}{6 }= \frac{2}{3}\)
Thus, the correct answer is Option A.
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Re: If the square of the 7th term of an arithmetic progression with positi [#permalink] New post  13 May 2020, 22:40
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Re: If the square of the 7th term of an arithmetic progression with positi [#permalink] New post  18 May 2020, 09:19
Bunuel wrote:
If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is:

A. 2 : 3
B. 3 : 2
C. 3 : 4
D. 4 : 3
E. 5 : 3


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Asked: If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is:

Let a be first term and d be common difference of the arithmetic progression.

(a + 6d)^2 = (a+2d)(a+16d)
a^2 + 12ad + 36d^2 = a^2 + 18ad + 32d^2
6ad = 4d^2; 6a = 4d; a:d = 4:6 = 2:3

IMO A
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Re: If the square of the 7th term of an arithmetic progression with positi [#permalink] New post  09 Aug 2021, 18:38
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