Question 30

Consider an arithmetic progression of positive terms with the first term as $$\alpha$$. Let $$S_n$$ denote the sum of the first n terms of this arithmetic progression and let $$\frac{S_m}{S_n} = \frac{m^2}{n^2}$$ for m ≠ n. Then the $$50^{th}$$ term is

Solution

$$\frac{m^2}{n^2}$$ = $$\frac{\frac{m*(2\alpha+ (m-1)d)}{2}}{\frac{n*(2\alpha+ (n-1)d)}{2}}$$

$$\frac{m}{n}$$ = $$\frac{(2\alpha+ (m-1)d)}{(2\alpha+ (n-1)d)}$$

$$m(2\alpha+ (n-1)d) = n(2\alpha+ (m-1)d))$$

$$(n-m)d=(n-m)2\alpha\ $$

d = 2$$\alpha$$

$$50^{th}$$ term = $$\alpha$$+49*2$$\alpha$$ = 99$$\alpha$$

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