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प्रश्न
Find the seventh term from the end of the series :
`sqrt(2), 2, 2sqrt(2), ........., 32.`
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उत्तर
Given series: `sqrt(2), 2, 2sqrt(2), ........., 32.`
Now, `2/sqrt(2) = (2sqrt(2))/2 = sqrt(2)`
So, the given series is a G.P. with common ratio, r = `sqrt(2)`
Here, last term, l = 32
∴ 7th term from an end = `1/r^6`
= `32/(sqrt(2))^6`
= `32/8`
= 4
संबंधित प्रश्न
Which term of the G.P.:
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The product of 3rd and 8th terms of a G.P. is 243. If its 4th term is 3, find its 7th term.
Second term of a geometric progression is 6 and its fifth term is 9 times of its third term. Find the geometric progression. Consider that each term of the G.P. is positive.
Q 7
If each term of a G.P. is raised to the power x, show that the resulting sequence is also a G.P.
If a, b and c are in A.P. and also in G.P., show that : a = b = c.
Find the sum of G.P. :
1 + 3 + 9 + 27 + .......... to 12 terms.
Find the sum of G.P. :
`sqrt(3) + 1/sqrt(3) + 1/(3sqrt(3)) + ..........` to n terms.
Q 7
Find the 5th term of the G.P. `5/2, 1, .........`
