Algebra

Find the sum of 35 terms of an arithmetic series of which the first term is a and the fifteenth term is 9a.

\(t_1=35 \qquad t_{15}=9a\\ t_n=t_1+(n-1)d\\ t_{15}=t_1+(15-1)d=9a\\ a+14d=9a\\ 14d=8a\\ 7d=4a d=\frac{4a}{7}\\ S_n=\frac{n}{2}(2a+(n-1)d)\\ S_{35}=\frac{35}{2}(2a+(34)\frac{4a}{7})\\ S_{35}=\frac{35}{1}(1a+17*\frac{4a}{7})\\ S_{35}=35(a+\frac{68a}{7})\\ S_{35}=35a+5*68a\\ S_{35}=35a+340a\\ S_{35}=375a\\ \)

(F + L] / 2 x N =, where F=1a, L=9a, N=15

[1 + 9] / 2 x 15 =75a

75 =15/2[2*1 + (15 - 1)*D], solve for D

D=4/7

Sum(35) =35/2[2*1 + (35 - 1)*4/7]

Sum(35) =17.5 * (21 3/7)

Sum(35) =375a

Find the sum of 35 terms of an arithmetic series
of which the first term is a and the fifteenth term is 9a.

Let t₁ = a

Let t₁₅ = 9a

Formula:

\(\begin{array}{|lrcll|} \hline (1) & t_x &=& t_1 + (x-1)\cdot d \\ (2) & t_y &=& t_1 + (y-1)\cdot d \\ (3) & t_z &=& t_1 + (z-1)\cdot d \\ \hline I = (1) - (2) & t_x-t_y &=& d\cdot (x-y) \\ II = (1) - (3) & t_x-t_z &=& d\cdot (x-z) \\ \hline II / I & \frac{t_x-t_z}{t_x-t_y} &=& \frac{d\cdot (x-z)}{d\cdot (x-y)} \\ & \frac{t_x-t_z}{t_x-t_y} &=& \frac{x-z}{x-y} \quad & | \quad \cdot (t_x-t_y) \\ & t_x-t_z &=& (t_x-t_y)\cdot \frac{x-z}{x-y} \\ & t_z &=& t_x-(t_x-t_y)\cdot \frac{x-z}{x-y} \\ & t_z &=& t_x- t_x\cdot (\frac{x-z}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) \\ & t_z &=& t_x\cdot (1-\frac{x-z}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) \\ & t_z &=& t_x\cdot (\frac{x-y-x+z}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) \\ &\mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ \hline \end{array}\)

t₃₅ = ?

\(\begin{array}{|lrclrcl|} \hline x = 1 \quad & t_x &=& t_1 = a \\ y = 15 \quad & t_y &=& t_{15} = 9a \\ z = 35 \quad & t_z &=& t_{35} = \ ? \\ &&&& \mathbf{ t_z } &\mathbf{=}& \mathbf{ t_x\cdot (\frac{z-y}{x-y}) +t_y\cdot (\frac{x-z}{x-y}) } \\ &&&& t_{35} & = & a\cdot ( \frac{35-15}{1-15}) + 9a\cdot (\frac{1-35}{1-15}) \\ &&&& t_{35} & = & a\cdot (\frac{20}{-14}) + 9a\cdot (\frac{-34}{-14}) \\ &&&& t_{35} & = & -a\cdot (\frac{10}{7}) + 9a\cdot (\frac{17}{7}) \\ &&&& t_{35} & = & 9a\cdot (\frac{17}{7}) -a\cdot (\frac{10}{7}) \\ &&&& t_{35} & = & a\cdot (\frac{9\cdot 17}{7}) -a\cdot (\frac{10}{7}) \\ &&&& t_{35} & = & a\cdot (\frac{153}{7}) -a\cdot (\frac{10}{7}) \\ &&&& t_{35} & = & a\cdot (\frac{153-10}{7}) \\ &&&& \mathbf{ t_{35} } &\mathbf{=}& \mathbf{ \frac{143}{7}a }\\ \hline \end{array} \)

s₃₅ = ?

\(\begin{array}{|rcll|} \hline s_{35} &=& \left(\frac{t_1+t_{35}}{2} \right)\cdot 35 \quad & | \quad t_1 = a \quad t_{35} = \frac{143}{7}a \\ s_{35} &=& \left(\frac{a+\frac{143}{7}a}{2} \right)\cdot 35 \\ s_{35} &=& \left(\frac{1+\frac{143}{7} }{2} \right)\cdot 35 a\\ s_{35} &=& \left( \frac{150}{14} \right)\cdot 35 a\\ s_{35} &=& 150\cdot \frac{5}{2} a\\ s_{35} &=& 75\cdot 5 a\\ \mathbf{ s_{35} } &\mathbf{=}& \mathbf{375 a}\\ \hline \end{array} \)