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# The First Three of Four Given Numbers Are in G.P. and Their Last Three Are in A.P. with Common Difference 6. If First and Fourth Numbers Are Equal, Then the First Number is - Mathematics

MCQ

The first three of four given numbers are in G.P. and their last three are in A.P. with common difference 6. If first and fourth numbers are equal, then the first number is

#### Options

•  4

•  8

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#### Solution

8

$\text{ The first and the last numbers are equal } .$
$\text{ Let the four given numbers be p, q, r and p } .$
$\text{ The first three of four given numbers are in G . P } .$
$\therefore q^2 = p \cdot r . . . . . . . . \left( i \right)$
$\text{ And, the last three numbers are in A . P . with common difference 6 } .$
$\text{ We have }:$
$\text{ First term } = q$
$\text{ Second term } = r = q + 6$
$\text{ Third term } = p = q + 12$
$\text{ Also }, 2r = q + p$
$\text{ Now, putting the values of p and r in } \left( i \right):$
$q^2 = \left( q + 12 \right)\left( q + 6 \right)$
$\Rightarrow q^2 = q^2 + 18q + 72$
$\Rightarrow 18q + 72 = 0$
$\Rightarrow q + 4 = 0$
$\Rightarrow q = - 4$
$\text{ Now, putting the value of q in } p = q + 12:$
$p = - 4 + 12 = 8$


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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 20 Geometric Progression
Q 4 | Page 57
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