Orange phase part 2

Solutions to all questions of Orange phase part 2.

Sum of first 12 terms of a GP is equal to the sum of the first 14 terms in the same GP. Sum of the first 17 terms is 92, what is the third term in the GP?
92 -92 46 231
Choice A. 92
DETAILED SOLUTION
Sum of first 12 terms is equal to sum of first 14 terms.
Sum of first 14 terms = Sum of first 12 terms + 13th term + 14th term
=> 13th term + 14th term = 0
Let us assume 13th term = k, common ratio = r. 14th term will be kr.
k + kr = 0
k (1 + r) = 0
=> r = -1 as k cannot be zero

Common ratio = -1.
Now, if the first term of this GP is a, second term would be -a, third would be a and so on
The GP would be a, -a, a, -a, a, -a,...
Sum to even number of terms = 0
Sum to odd number of terms = a
Sum to 17 terms is 92 => a = 92
Third term = a = 92
Correct Answer: 92

ARITHMETIC PROGRESSION

If 4 times the 4th term of an A.P. is equal to 9 times the 9th term of the A.P., what is 13 times the 13th term of this A.P.?
7 times the 13th term 0 13 times the 7th term 4 times the 4th term + 9 times the 9th term
DETAILED SOLUTION
4t 4 = 9t 9 , we need to find t13
4(a + 3d) = 9(a + 8d)
4a + 12d = 9a + 72d

=> 5a + 60d = 0
=> a + 12d = 0
=> t13 = 0
=> 13 × t"13" = 0
As a simple rule, if n × tn = m × tm, then tm+n = 0. See, if you can prove this.
Answer choice (b).

Correct Answer: 0

Q.2: Find the smallest number that leaves a remainder of 4 on division by 5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9 2519 5039 1079 979
DETAILED SOLUTION

LCM (5, 6, 7, 8, 9) - 1 = 2519.
Choice (A)

Correct Answer: 2519

.7: The sum of two non co–prime numbers added to their HCF gives us 91. How many such pairs are possible?
2 4
3 6

DETAILED SOLUTION
Let HCF of the numbers be h. The numbers can be taken as ha + hb, where a, b are coprime.
h + ha + hb = 91
h(1 + a + b) = 91
h ≠ 1
h = 7
=> 1 + a + b = 13 a + b = 12

h = 13
=> 1 + a + b = 7
=> a + b = 6

Case 1: h = 7, a + b = 12
(1, 11), (5, 7) => Only 2 pairs are possible as a, b have to be coprime.

Case 2: h = 13, a + b = 6
(1, 5) only one pair is possible as a, b have to be coprime.

Overall, 3 pairs of numbers are possible – (7, 77) (35, 49) and (13, 65)
Answer choice (c)