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प्रश्न
Find the sum of 10 terms of the series `1 + sqrt3 + 3` + .....
बेरीज
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उत्तर
a = a, ar = b, ar2 = c ⇒ G.P
`b/a = c/a`
`sqrt3/1 = sqrt3`
`3/sqrt3 = (sqrt3 xx sqrt3)/sqrt3`
Sn = `(a(r^n - 1))/(r - 1)`
r > 1
`S_10 = (1 xx (sqrt3)^10 - 1)/(sqrt3 - 1)`
= `(3^5 - 1)/(sqrt3 - 1)`
= `242/(sqrt3 - 1) xx (sqrt3 + 1)/(sqrt3 + 1)`
= `(242(sqrt3 + 1))/(3 - 1)`
= `(242(sqrt3 + 1))/2`
= `121(sqrt3 + 1)`
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