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SEQUENCES AND SERIES — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSEQUENCES AND SERIES3 Marks
Question
The sum of three numbers in G.P. is $56$. If we subtract $1, 7, 21$ from these numbers in that order, we obtain an arithmetic progression. Find the numbers.
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Answer

Let $a, a r, a r^2$ be three numbers in G.P., therefore, $a+a r+a r^2=56 \Rightarrow a\left(1+r+r^2\right)=56$ . . . . .(i)
According to question, $a-1$, ar $-7, a r^2-21$ are in A.P.
$\therefore(a r-7)-(a-1)=\left(a r^2-21\right)-(a r-7)$
$\Rightarrow a r-7-a+1=a r^2-21-a r+7$
$\Rightarrow a r-a-6=a r^2-a r-14$
$\Rightarrow a r^2-2 a r+a=8$
$\Rightarrow a\left(r^2-2 r+1\right)=8 \ldots . . . . .(i i)$
Dividing eq. (i) by eq. (ii), $\frac{a\left(1+r+r^2\right)}{a\left(r^2-2 r+1\right)}=\frac{56}{8}$
$\Rightarrow 1+r+r^2=7 r^2-14 r+7$
$\Rightarrow 6 r^2-15 r+6=0$
$\Rightarrow 2 r^2-5 r+2=0$
$\Rightarrow r = \frac { - ( - 5 ) \pm \sqrt { ( - 5 ) ^ { 2 } - 4 \times 2 \times 2 } } { 2 \times 2 }$
$= \frac { 5 \pm \sqrt { 25 - 16 } } { 4 } = \frac { 5 \pm \sqrt { 9 } } { 4 } = \frac { 5 \pm 3 } { 4 }$=
$\Rightarrow r = \frac { 5 + 3 } { 4 } = \frac { 8 } { 4 } = 2$or $r = \frac { 5 - 3 } { 4 } = \frac { 2 } { 4 } = \frac { 1 } { 2 }$
Putting r = 2 in eq. (i), a$(1 + 2 + 2^2) = 56$
$\Rightarrow a = \frac { 56 } { 7 } = 8$
Then the required numbers are 8, 16, 32.
Putting $r = \frac { 1 } { 2 }$ in eq. (i), $a \left( 1 + \frac { 1 } { 2 } + \frac { 1 } { 4 } \right) = 56$
$\Rightarrow a \times \frac { 7 } { 4 } = 56$
$\Rightarrow$ a = 32
Then the required numbers are $32, 16, 8$.

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