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CAT 2017 Set-1 | Question: 97

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If the square of the $7^{\text{th}}$ term of an arithmetic progression with positive common difference equals the products of the $3^{\text{rd}}$ and $17^{\text{th}}$ terms, then the ratio of the first term to the common difference is 

  1. $2:3$
  2. $3:2$
  3. $3:4$
  4. $4:3$
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Given that, the square of the $7^{\text{th}}$ term is equal to the products of the $3^{\text{rd}}$ and $17^{\text{th}}$ terms.

The $n^{\text{th}}$ term of an arithmetic progression $= a + (n-1)d$

Where, $a = $ first term, and $d = $common difference.

Now, ${(a+6d)}^{2} = (a + 2d)(a + 16d)$
                               
$\Rightarrow a^{2} + 36d^{2} + 12ad  = a^{2} + 16ad + 2ad + 32d^{2}$  

$\Rightarrow 4d^{2}  = 6ad$

$\Rightarrow 2d  = 3a$

$\Rightarrow 3a = 2d$

$\Rightarrow \frac{a}{d} = \frac{2}{3}$

$\therefore$ The ratio of the first term to the common difference is $2:3.$

Correct Answer $:\text{A}$
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