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IPMAT Indore 2024

Algebra
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Progression & Series

Medium

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

✅ Correct Option: 2
Let's break this down so it's crystal clear!
We have an infinite geometric progression where: \newline - Sum of infinite GP = 8080 \newline - Sum of first 22 terms = 3535 \newline - We need to find nn where sum of first nn terms is closest to 100100
For an infinite GP: Sum = a1−r\dfrac{a}{1-r} where ∣r∣<1|r| < 1
So: a1−r=80\dfrac{a}{1-r} = 80 ... (equation 1)
Sum of first 22 terms = a+ar=a(1+r)=35a + ar = a(1+r) = 35 ... (equation 2)
From equation 2: a=351+ra = \dfrac{35}{1+r}
Substituting in equation 1:
351+r÷(1−r)=80\dfrac{35}{1+r} \div (1-r) = 80
35(1+r)(1−r)=80\dfrac{35}{(1+r)(1-r)} = 80
351−r2=80\dfrac{35}{1-r^2} = 80
35=80(1−r2)35 = 80(1-r^2)
35=80−80r235 = 80 - 80r^2
80r2=4580r^2 = 45
r2=916r^2 = \dfrac{9}{16}
r=±34r = \pm\dfrac{3}{4}
Since the infinite sum exists and is positive, we need ∣r∣<1|r| < 1. Both ±34\pm\dfrac{3}{4} satisfy this.
If r=34r = \dfrac{3}{4}: a=351+34=3574=20a = \dfrac{35}{1+\dfrac{3}{4}} = \dfrac{35}{\dfrac{7}{4}} = 20
If r=−34r = -\dfrac{3}{4}: a=351−34=3514=140a = \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=34a = 20, r = \dfrac{3}{4} → First 22 terms: 20+15=3520 + 15 = 35 ✓
Case 2: a=140,r=−34a = 140, r = -\dfrac{3}{4} → First 22 terms: 140+(−105)=35140 + (-105) = 35 ✓
Using the formula: Sn=a(1−rn)1−rS_n = \dfrac{a(1-r^n)}{1-r}
Case 1: a=20,r=34a = 20, r = \dfrac{3}{4}
Sn=20(1−(34)n)1−34=80(1−(34)n)S_n = \dfrac{20(1-(\dfrac{3}{4})^n)}{1-\dfrac{3}{4}} = 80(1-(\dfrac{3}{4})^n)
Case 2: a=140,r=−34a = 140, r = -\dfrac{3}{4}
Sn=140(1−(−34)n)1−(−34)=80(1−(−34)n)S_n = \dfrac{140(1-(-\dfrac{3}{4})^n)}{1-(-\dfrac{3}{4})} = 80(1-(-\dfrac{3}{4})^n)
We want Sn≈100S_n \approx 100:
Case 1: 80(1−(34)n)≈10080(1-(\dfrac{3}{4})^n) \approx 100 → (34)n≈−0.25(\dfrac{3}{4})^n \approx -0.25
This is impossible since (34)n(\dfrac{3}{4})^n is always positive!
Case 2: 80(1−(−34)n)≈10080(1-(-\dfrac{3}{4})^n) \approx 100 → (−34)n≈−0.25(-\dfrac{3}{4})^n \approx -0.25
Since (−34)n(-\dfrac{3}{4})^n is negative when nn is odd, nn must be odd.
Testing odd values:
n=1n = 1: (−34)1=−0.75(-\dfrac{3}{4})^1 = -0.75, S1=80(1−(−0.75))=140S_1 = 80(1-(-0.75)) = 140
n=3n = 3: (−34)3≈−0.422(-\dfrac{3}{4})^3 \approx -0.422, S3=80(1−(−0.422))≈113.8S_3 = 80(1-(-0.422)) \approx 113.8
n=5n = 5: (−34)5≈−0.237(-\dfrac{3}{4})^5 \approx -0.237, S5=80(1−(−0.237))≈99.0S_5 = 80(1-(-0.237)) \approx 99.0
n=7n = 7: (−34)7≈−0.133(-\dfrac{3}{4})^7 \approx -0.133, S7=80(1−(−0.133))≈90.6S_7 = 80(1-(-0.133)) \approx 90.6
n=5n = 5 gives the sum closest to 100100 (approximately 99.099.0)
Therefore, n=5n = 5

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