In the

In the arithmetic sequence a1 = 4.8, d = 0.4. How many consecutive members, starting with the first, need to be added so that the sum is greater than 170?

Correct answer:

n =  20

Step-by-step explanation:

a1=4.8 d=0.4 s=170  s=n (a1+an)/2 s=n (a1+(a1+(n1) d))/2  2s=n(a1+(a1+(n1)d))  2 170=n (4.8+(4.8+(n1) 0.4)) 0.4n29.2n+340=0 0.4n2+9.2n340=0  a=0.4;b=9.2;c=340 D=b24ac=9.2240.4(340)=628.64 D>0  n1,2=b±D2a=9.2±628.640.8 n1,2=11.5±31.3408678884 n1=19.8408678884 n2=42.8408678884   Factored form of the equation:  0.4(n19.8408678884)(n+42.8408678884)=0  n=n1=19.8409=20   Verifying Solution:  a20=a1+19 d=4.8+19 0.4=625=1225=12.4  s20=n (a1+a20)/2=20 (4.8+12.4)/2=172

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