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प्रश्न
Find the 8 th term of the G.p. `sqrt3, 1/sqrt3, 1/(3sqrt3)`, ....
योग
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उत्तर
a = `sqrt3`
r = `(1/sqrt3)/sqrt3`
= `1/3`
`a_n = ar^(n - 1)`
`a_8 = ar^(8 - 1)`
a8 = ar7
a8 = `sqrt3 xx (1/3)^7`
a8 = `sqrt3 xx 1/sqrt3^14`
a8 = `1/(729sqrt3)`
a8 = `sqrt3/2187`
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