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प्रश्न
Find the 9th term of the series :
1, 4, 16, 64, ...............
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उत्तर
Given sequence: 1, 4, 16, 64, ...............
Now,
`4/1 = 4, 16/4 = 4, 64/16 = 4`
Since `4/1 = 16/4 = 64/16 = ...... = 4,` the given sequence is a G.P. with the first term, a = 1 and common ratio, r = 4.
Now, Tn = arn – 1
`\implies` T9 = 1 × 49 – 1
= 1 × 48
= 48
= 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4
= 65536
संबंधित प्रश्न
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.
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Find the third term from the end of the G.P.
`2/27, 2/9, 2/3, .........,162.`
For the G.P. `1/27, 1/9, 1/3, ........., 81`; find the product of fourth term from the beginning and the fourth term from the end.
If for a G.P., pth, qth and rth terms are a, b and c respectively; prove that : (q – r) log a + (r – p) log b + (p – q) log c = 0
Q 5
Q 6
Find the sum of G.P. :
0.3 + 0.03 + 0.003 + 0.0003 + ........... to 8 items.
How many terms of the geometric progression 1 + 4 + 16 + 64 + …….. must be added to get sum equal to 5461?
Find the sum of G.P. : 3, 6, 12, .........., 1536.
