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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
Find, which of the following sequence from a G.P. :
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If a, b and c are in A.P, a, x, b are in G.P. whereas b, y and c are also in G.P.
Show that : x2, b2, y2 are in A.P.
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Q 3.2
Q 3.3
The first term of a G.P. is –3 and the square of the second term is equal to its 4th term. Find its 7th term.
Find the sum of the sequence `-1/3, 1, -3, 9, ..........` upto 8 terms.
