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Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10

In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term - Mathematics

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Question

In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term

Sum

Solution

Given, 9th term = 32805

a.rn–1 = `1/2817`

t9 = 32805 [tn = arn–1]

a.r8 = 32805 …(1)

6th term = 1215

a.r5 = 1215 …(2)

Divide (1) by (2)

`("ar"^8)/("ar"^5) = 32805/1215`

⇒ r3 = `6561/243`

= `2187/81`

= `729/27`

= `243/9`

= `81/3`

r3 = 27

⇒ r3 = 3

r = 3

Substitute the value of r = 3 in (2)

a.35 = 1215

a × 243 = 1215

a = `1215/243` = 5

Here a = 5, r = 3, n = 12

t12 = `5 xx 3^((12 - 1))`

= 5 × 311

∴ 12th term of a G.P. = 5 × 311

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Chapter 2: Numbers and Sequences - Exercise 2.7 [Page 73]

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Samacheer Kalvi Mathematics [English] Class 10 SSLC TN Board
Chapter 2 Numbers and Sequences
Exercise 2.7 | Q 6 | Page 73
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