 • Add a Discussion  Number Series and Progressions

Pls help me solve these problems . Also give an optimized solution for it . × Surya Pratap Rana 1 year ago Pls help me solve these problems . Pls give an optimized solution with the answer. × shubham bharti 1 year ago
Option D Approach pls 1 Comment
× Tushar 1 year ago
11 If sum of first 4 terms of an AP is one-fifth the sum of the first 9 terms of the AP, what is the ratio of the first term of the AP to its common difference?

1. A 20:11 9%
2. B 25:11 31%
3. C 50:11 7%
4. D 06:11 52%
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1 Thanks 54 Attempts 3 Comments
× Ankit Singh 1 year ago
Sn = n/2[2a+(n-1)d]

S4= 2[2a + 3d] = 4a+6d
S9= 9/2[2a + 8d]

S4=1/5 S9 (given)

4a+6d = 1/5 x (9/2) x [2a + 8d]

40a + 60d = 9(2a + 8d)
40a + 60d = 18a + 72d
22a = 12d
a/d = 6/11 solution for this plz. × Abhijith 1 year ago
but it does not wen x =2 Question 20 × himanshu 1 year ago
let the nos be a and b now GM(a,b)=root(ab) and HM(a,b)=2/(1/a+1/b)=2ab/(a+b). Now root(ab)/(2ab/(a+b)) or 1/2*(root(a/b)+root(b/a))=13/12 or root(a/b)+root(b/a)=13/6 so its clear that option C fits the given condition i.e a/b=4:9. If the square of the 7th term of an arithmetic progression with positive common difference equals the product of the 3rd and 17th terms, then the ratio of the first term to the common difference is

1. A 2:3 69%
2. B 3:2 17%
3. C 3:4 11%
4. D 4:3 3%
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2 Thanks 35 Attempts 1 Comment
× 2:3 #QuestionoftheDay

A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is pi*r during the first 30 seconds, pi*r/2 during next one minute, pi*r/4 using next 4 minutes, and so on.
What is the ratio of the time taken for the nth round to that for previous round?

1. A 4 35%
2. B 8 24%
3. C 16 35%
4. D 32 6%
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1 Thanks 17 Attempts 3 Comments
× Vaanya Gupta 1 year ago
Jatin Kudha the question has the incomplete series.

Circumference of ground is 2pir

In 0.5min distance covered is pir/2

In 1min distance covered is 1* pir/2 = pir/2

In 2min distance covered is 2* pir/4 = pir/2

In 4min distance covered is 4* pir/8 = pir/2

So in a total of 7.5min, a distance of 4*pir/2 = 2pir or 1 round is covered.

Time taken for next round is 8+16+32+64=120mins

Ratio of time taken = 120/7.5 = 16:1 #QuestionoftheDay

The integers 1, 2, …, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two
numbers, say a and b, currently on the blackboard are erased and a new number a + b–1 is written. What will be the number left on the board at the end?

1. A 820 11%
2. B 821 11%
3. C 781 58%
4. D 819 21%
5. E 780 0%
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× Vaanya Gupta 1 year ago
Kakoli

Total sum of the numbers written on the blackboard = 40*41/2 = 820

when two numbers a and b are erased and replaced by a new number a+b-1, the total sum of the numbers written on the blackboard is reduced by 1.

The operation will be done 39 times. Thus the sum will be reduced by 1*39=39.

So 820-39=781 Open in App