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Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18. - Mathematics

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प्रश्न

Find four numbers forming a geometric progression in which third term is greater than the first term by 9, and the second term is greater than the 4th by 18.

योग
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उत्तर

Let the geometric series be a, ar, ar2, ar3,...

Third term = ar2, first term = a

∴ ar2 – a = 9 …........(i)

Second term = ar, fourth term = ar3

ar – ar3 = 18 ….........(ii)

Dividing equation (i) by (ii), we get

`("a"("r"^2 - 1))/("a"("r" - "r"^3))`

= `9/18`

= `1/2`

or 2(r2 − 1) = r − r3

∴ r3 + 2r2 − r − 2 = 0

or (r − 1) (r + 1) (r + 2) = 0

or r = 1, −1, −2 if r = −2,

From equation (i), a(4 − 1) = 9

∴ a = 3

∴ 4th terms of the geometric progression 3, −6, 12, −24.

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अध्याय 9: Sequences and Series - Exercise 9.3 [पृष्ठ १९३]

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एनसीईआरटी Mathematics [English] Class 11
अध्याय 9 Sequences and Series
Exercise 9.3 | Q 21 | पृष्ठ १९३

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