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Sequences and Series — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSequences and Series3 Marks
Question
Find four numbers forming a geometric progression in which the third term is greater than the first term by $9$ and the second term is greater than by $4^{th}$ by $18.$
✓

Answer

Let the four numbers in G.P. be $a, ar, ar^2, ar^3$
$\therefore ar^2 = a + 9$ and $ar = ar^3 + 18$
Now, $ar^2 - a = 9$
$\Rightarrow a(r^2 - 1) = 9 ...(i)$
And $ar - ar^3 = 18$
$\Rightarrow ar(1 - r^2) = 18+$
$\Rightarrow -ar(r^2 - 1) = 18 ...(ii)$
Dividing eq. $(ii)$ by eq. $(i),$ we have
$\frac { - a r \left( r ^ { 2 } - 1 \right) } { a \left( r ^ { 2 } - 1 \right) } = \frac { 18 } { 9 }$
$\Rightarrow r = -2$
Putting value of r in eq. (i), we get
$a(4 - 1) = 9$
$\Rightarrow a = 3$
$\therefore ar = 3 \times (-2) = -6$
$ar^2 = 3 \times (-2)^2 = 12a r ^ { 3 }$
$= 3 \times (-2)^3 = -24$
Therefore, the required numbers are $3, -6, 12, -24$

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