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Mathematics
ARITHMATIC PROGRESSION
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https://www.math10.com/en/algebra/arithmetic-progression.html
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The first term is 26 and common difference is -7. Then the Arithmatic progression is.
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Didn't know the answer
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calculate the 16th term of an arithmatic progression he its fifth term is 6 and its 12th is 41.
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a+4d = 6 a+11d = 41 7d = 35 d = 5 a = -14 -14 + 15(5) = 61
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The 9th term of an arithmetic progression is 4+5p and the sum of the four terms of the progression is 7p-10, where p is a
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a+8d = 4+5p d = 5, so a+40 = 4+5p I assume you mean the sum of the *first* 4 terms is 7p-10,so 4/2
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The 4th, 6th and 9th term of an arithmetical progression form the first three times of a geometric progression. If the first
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4 th ==> 3+3d = a 6 th ==> 3 + 5d = a r 9 th ==> 3 + 8 d = a r^2
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The first, the third and the seventh terms of an increasing arithmetic progression are three consecutive terms of a geometric
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AP: 10,10+2d,10+6d GP: (10+2d)/10 = (10+6d)/(10+2d) d = 5 check: AP: 10,15,20,25,30,35,40 GP:
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The third, fifth and seventeenth terms of an arithmetic progression are in geometric progression. Find the common ratio of the
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We can without loss of generality assume a=1, so (1+4d)/(1+2d) = (1+16d)/(1+4d) d = -5/8 The AP is
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three numbers are in harmonic progression. If the third number were decreased by 4 they would
be in arithmetic progression. If
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how do i first derive the equation?
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The first three terms of a geometric progression are also the first ninth and eleventh terms respectic of an arithmetic
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Can someone help me to solve it
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if 1, 2, 7 and 20, respectively, are added to the first terms of an arithmetic progression, a geometric progression of four
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let the first 4 terms of the AP be a-d, a , a+d, and a+2d new sequence is a-d+1, a+2, a+d + 7, and
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The three real,distint and non-zero numbers
a,b,c are such that a,b,c are in arithmetic progression and a,c,b are in geometric
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ripped off? Goodness Well the arith series tells us that b=a+d, and c=a+2d the geo series tells us
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