The sum of the first n terms of an arithmetic sequence is given by formula Sum of n = 4n^2 -2n. Three terms of this sequence, u2, um and u32, are consecutive terms in a geometric sequence. Find m.
2x = m ; x=m/2 mx = 32 ; m(m/2) = 32 m^2 = 64 ; m=8
i confused myself, 2 m and 32 arent terms, they are positions ...
Sn = 4n^2 -2n S(2) = 12 S(32) = 4032
we knows it arithmetical seq. 2(a(1)+a(2)) ---------- = 12 ; a(1)+a(2) = 12 2 32(a(1)+a(32)) ------------- = 4032 ; 16(a(1)+a(32)) = 4032 2 hmm
right here; i subtracted an n instead of dividing it off: 8n^2 -4n = n (a(1)+a(n)) 8n^2 -5n = a(1) + a(n) to correct: 8n^2 -4n = n (a(1)+a(n)) 8n -4 = a(1) + a(n) ; and let n=1 8 -4 = a(1) + a(1) 4 = 2 a(1) a(1) = 2
1 2 3 4 5 6 7 8 2 10 18 26 +8 +8 +8 ....
16(2+a(32)) = 4032 2 + a(32) = 252 a(32) = 250 ... that should be better
lets try to put that into a formula for the sequence and test it a(n) = 2 + 8(n-1) a(1) = 2 + 8(0) = 2 a(2) = 2 + 8(1) = 10 ........ a(32) = 2 + 8(31) = 250 .. its matches
now we can do the first part i did, but with the real number lol
10n = m ; n = m/10 mn = 250 m(m/10) = 250 m^2 = 2500 m = 50
10*n = m ; 10n^2 = 250 n^2 = 250/10 n^2 = 25 n = 5 10*5^0 = 10 10*5^1 = 50 10*5^2 = 250 thats right, what was i worried about lol
that should clean it up :)
A(m) = 50 i should say, since we are going by placement and not value
A(m) = 2 + 8(m-1) 50 = 2 +8(m-1) 48 = 8(m-1) 6 = m-1 m = 7
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