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Solution
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Solution provided by AtoZmath.com
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| Geometric Progression |
10. Find 6 arithmetic mean between 3 and 24 .
Let ` A_1, A_2, A_3, A_4, A_5, A_6` be the `6` arithmetic mean between `3` and `24`.
`:. 3, A_1, A_2, A_3, A_4, A_5, A_6, 24` are in AP .
Here First term `a = 3` and Last term `b = 24`
Comman difference `d = (b-a)/(n+1) = (24-3)/(6+1) = 21/7 = 3`
`:.` Required Arithemetic Means,
`A_1 = T_2 = a + d = 3 + 3 = 6`
`A_2 = T_3 = a + 2d = 3 + 2(3) = 9`
`A_3 = T_4 = a + 3d = 3 + 3(3) = 12`
`A_4 = T_5 = a + 4d = 3 + 4(3) = 15`
`A_5 = T_6 = a + 5d = 3 + 5(3) = 18`
`A_6 = T_7 = a + 6d = 3 + 6(3) = 21`
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