**The sum of first 3 terms of a G.P is 16 and the sum of next 3 terms is 128. Find the sum of n terms of the G.P?**

__Question:__**Options:**

A. | 167(2n + 1) | |

B. | 167(2n - 1) | |

C. | 97(3n - 1) | |

D. | 167(3n + 1) |

**Answer: Option B**

: B

Conventional Method From the data given in the question S3=16=a(r3−1)r−1................................(1)

S6=16+128=144=a(r6−1)r−1=a(r3−1)(r3+1)r−1................................(2)

9 = (r3 + 1)

Hence r =2

Substituting in equation 1, we get

a = 167

We know that,

Sn=a(rn−1)r−1

Substituting the values of a and r found above, we get

Sn=167(2n−1)

Thus, answer is option (b)

Shortcut-

Substitution The answerto this question is in the question itself!

At n=3, the sum should be 16

Hence, in the correct answer option, at n=3, we should get 16

Option (a)= 167(23+ 1)≠16

Option(b)=167(23-1) = 16

Option (c)=97(33-1)≠16

Option (d)=167(33 -1)≠16

Options (a), (c) and (d) can never be the answer. Answer is option (b)

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## More Questions on This Topic :

**In how many ways can the number 100 be written as a sum of two or more consecutive positive integers?**

__Question 1.__- 2
- 3
- 5
- 6

**Answer: Option A**

: A

Conventional Approach To write a number as a sum of two or more consecutive positive integers, we need to form an AP with "n” terms with a common difference 1 Sum to "n” terms of an AP= n2(2a + (n - 1)d) where d=1 (as the numbers are consecutive) n(2a+(n-1))=200 thus, we need to write 200 as a product of 2 numbers, where one number is odd and the other is even this can be done in 2 ways 200= 5×40 200=25×8 Thus 100 can be expressed in 2 ways as a sum of two or more consecutive positive integers Shortcut:- The answer is just one step. The number of ways of writing any number as a sum of two or more consecutive positive integers= number of odd factors of that number-1 100 = 22×52 Number of odd factors=3 Number of ways of expressing 100 as a sum of two or more consecutive positive integers= 3 - 1=2

**Both the H.C.F and the difference of two numbers is 7. If the L.C.M of the two numbers is a three digit number, then what is the maximum possible value of the smaller number?**

__Question 2.__- 77
- 84
- 70
- 63

**Answer: Option A**

: A

Let the smaller number be 7x, then the larger number will be (7x+7) = 7(x+1) Since x and (x+1) are co-prime, the LCM would be 7×x×(x+1). Maximum value of x such that LCM is a three digit number is x=11. So, smaller number = 77.

- 56
- 28
- 76
- 01

**Answer: Option B**

: B

Shortcut of last two digits: When the question is about finding the remainder when you divide a number by 100, what we actually need to find out is the last two digits, for which we have a direct shortcut. For an indepth explanation, please refer the demo tutorial

287=(210)8×27 (210)evevpowerwillalwaysendin..76

27 ends in ...28

Therefore 76x...28=28

**1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x>y , how many integer values can x take?**

__Question 4.__- 19
- 20
- 21
- 13
- 15

**Answer: Option D**

: D

We know that (x+3y)4=60 or x+3y =240 and x>y As 0<x<100, let's start with x=99. When x= 99, y=47. When x=96, y=48 and so on. x has to be greater than y. Now in the limiting case, when x=60, y=60. So when x=63, y=59. So x can take the values of 63,66,69...... till 99, a total of 13 values.

**A tank has two pipes A and B. Pipe A is for filling the tank and Pipe B is for emptying the tank. Pipe A can fill the tank in 20 hours and Pipe B can empty the tank in 25 hours. How many hours will it take to completely fill a one-fourth empty tank?**

__Question 5.__- 10 hours
- 25 hours
- 50 hours
- 100 hours
- 5 hours

**Answer: Option B**

: B

In one hour pipe A can fill 10020 = 5% of the tank In one hour pipe B can empty 10025 = 4% of the tank (A+B)'s work = 5-4 = 1% To fill a one fourth of empty tank, it will take = 251 = 25 hours

**Check all Questions in this Topic : Click HERE**

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