Question

How many terms of the sequence tn = 130 - Sn are included for the sum of the series to equal 1690?

Answer 1

$$t_1=130-S_1, S_1=t_1$$

$$2t_1=130=>t_1=65$$

$$S_2=65+t_2, t_2=130-S_2=130-(65+t_2)$$

$$t_2=\dfrac{1}{2}(65)$$

$$S_3=65+\dfrac{1}{2}(65)+t_3,$$

$$t_3=130-S_2=130-(65+\dfrac{1}{2}(65)+t_3 )$$

$$t_3=(\dfrac{1}{2})^2(65)$$

$t_1=65, q=\dfrac{1}{2}$

$$t_n=t_1q^{n-1}$$

$$S_n=\displaystyle\sum_{i=1}^nt_i=\displaystyle\sum_{i=1}^n65\big(\dfrac{1}{2}\big)^n$$

$$0<\dfrac{1}{2}<1, S=\displaystyle\sum_{i=1}^{\infin}t_i=\dfrac{1}{1-\dfrac{1}{2}}=2$$ $$S_n=\displaystyle\sum_{i=1}^n65\big(\dfrac{1}{2}\big)^n<65(2)=130$$

The sum of the series is not equal 1690 for $n\in\N.$

Therefore there are no solutions.