The sum of a given infinite geometric progression is 80 and the sum of its first two terms is 35. Then the value of $n$ for which the sum of its first $n$ terms is closest to 100, is

Let's break this down so it's crystal clear!

We have an infinite geometric progression where:

• Sum of infinite GP = $80$
  • Sum of first $2$ terms = $35$
  • We need to find $n$ where sum of first $n$ terms is closest to $100$

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For an infinite GP: Sum = $\dfrac{a}{1-r}$ where $|r| < 1$

So: $\dfrac{a}{1-r} = 80$ ... (equation 1)

Sum of first $2$ terms = $a + ar = a(1+r) = 35$ ... (equation 2)

From equation 2: $a = \dfrac{35}{1+r}$

Substituting in equation 1:

$\dfrac{35}{1+r} \div (1-r) = 80$

$\dfrac{35}{(1+r)(1-r)} = 80$

$\dfrac{35}{1-r^2} = 80$

$35 = 80(1-r^2)$

$35 = 80 - 80r^2$

$80r^2 = 45$

$r^2 = \dfrac{9}{16}$

$r = \pm\dfrac{3}{4}$

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Since the infinite sum exists and is positive, we need $|r| < 1$. Both $\pm\dfrac{3}{4}$ satisfy this.

If $r = \dfrac{3}{4}$: $a = \dfrac{35}{1+\dfrac{3}{4}} = \dfrac{35}{\dfrac{7}{4}} = 20$

If $r = -\dfrac{3}{4}$: $a = \dfrac{35}{1-\dfrac{3}{4}} = \dfrac{35}{\dfrac{1}{4}} = 140$

Let's verify both cases satisfy our conditions:

Case 1: $a = 20, r = \dfrac{3}{4}$ → First $2$ terms: $20 + 15 = 35$ ✓

Case 2: $a = 140, r = -\dfrac{3}{4}$ → First $2$ terms: $140 + (-105) = 35$ ✓

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Using the formula: $S_n = \dfrac{a(1-r^n)}{1-r}$

Case 1: $a = 20, r = \dfrac{3}{4}$

$S_n = \dfrac{20(1-(\dfrac{3}{4})^n)}{1-\dfrac{3}{4}} = 80(1-(\dfrac{3}{4})^n)$

Case 2: $a = 140, r = -\dfrac{3}{4}$

$S_n = \dfrac{140(1-(-\dfrac{3}{4})^n)}{1-(-\dfrac{3}{4})} = 80(1-(-\dfrac{3}{4})^n)$

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We want $S_n \approx 100$:

Case 1: $80(1-(\dfrac{3}{4})^n) \approx 100$ → $(\dfrac{3}{4})^n \approx -0.25$

This is impossible since $(\dfrac{3}{4})^n$ is always positive!

Case 2: $80(1-(-\dfrac{3}{4})^n) \approx 100$ → $(-\dfrac{3}{4})^n \approx -0.25$

Since $(-\dfrac{3}{4})^n$ is negative when $n$ is odd, $n$ must be odd.

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Testing odd values:

$n = 1$: $(-\dfrac{3}{4})^1 = -0.75$, $S_1 = 80(1-(-0.75)) = 140$

$n = 3$: $(-\dfrac{3}{4})^3 \approx -0.422$, $S_3 = 80(1-(-0.422)) \approx 113.8$

$n = 5$: $(-\dfrac{3}{4})^5 \approx -0.237$, $S_5 = 80(1-(-0.237)) \approx 99.0$

$n = 7$: $(-\dfrac{3}{4})^7 \approx -0.133$, $S_7 = 80(1-(-0.133)) \approx 90.6$

$n = 5$ gives the sum closest to $100$ (approximately $99.0$)

Therefore, $n = 5$