AP - consecutive members

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:

$$\begin{gathered} a_1=4.8 \\ d=0.4 \\ s=170 \\ s=n\cdot (a_1+an)/2 \\ s=n\cdot (a_1+(a_1+(n-1)\cdot d))/2 \\ 2\cdot s=n\cdot (a_1+(a_1+(n-1)\cdot d)) \\ 2\cdot 170=n\cdot (4.8+(4.8+(n-1)\cdot 0.4)) \\ -0.39999999999998n^2-9.2n+340=0 \\ 0.39999999999998n^2+9.2n-340=0 \\ a=0.4;b=9.2;c=-340 \\ D=b^2-4ac=9.2^2-4\cdot 0.4\cdot (-340)=628.64 \\ D>0 \\ {n}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-9.2\pm \sqrt{628.64}}{0.8} \\ {n}_{1,2}=-11.5\pm 31.340868 \\ {n}_1=19.840867888 \\ {n}_2=-42.840867888 \\ n=\lceil n_1\rceil =\lceil 19.8409\rceil =20 \\ \text{ Verifying Solution: } \\ {a}_{20}=a_1+19\cdot d=4.8+19\cdot 0.4=\frac{62}{5}=12\frac{2}{5}=12.4 \\ {s}_{20}=n\cdot (a_1+a_{20})/2=20\cdot (4.8+12.4)/2=172 \end{gathered}$$

Our quadratic equation calculator calculates it.

Try another example