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प्रश्न
Find the seventh term of the G.P.:
`sqrt(3) + 1, 1, (sqrt(3) - 1)/2, .........`
योग
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उत्तर
Given G.P.: `sqrt(3) + 1, 1, (sqrt(3) - 1)/2, .........`
Here,
First term, a = `sqrt(3) + 1`
Common ratio, r = `1/(sqrt(3) + 1)`
Now, Tn = arn – 1
`\implies` T7 = `(sqrt(3) + 1) xx (1/(sqrt(3) + 1))^(7 - 1)`
= `(sqrt(3) + 1) xx (1/(sqrt(3) + 1))^6`
= `(sqrt(3) + 1)/1 xx 1/(sqrt(3) + 1)^6`
= `1/(sqrt(3) + 1)^5`
= `1/(sqrt(3) + 1)^5 xx (sqrt(3) - 1)^5/(sqrt(3) - 1)^5`
= `(sqrt(3) - 1)^5/[(sqrt(3) + 1)(sqrt(3) - 1)]^5`
= `(sqrt(3) - 1)^5/(3 - 1)^5`
= `(sqrt(3) - 1)^5/(2)^5`
= `(sqrt(3) - 1)^5/32`
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