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705-805 (Hard)|   Sequences|      
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Bunuel
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A club consists of members whose ages are in arithmetic progression (a 03 Dec 2020, 08:00
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E
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75% (hard)

Question Stats:

64% (03:09) correct 36% (03:14) wrong based on 148 sessions

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A club consists of members whose ages are in arithmetic progression (an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant). The common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250, then number of members in the club are:

A. 18

B. 20

C. 22

D. 24

E. 25
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E
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A club consists of members whose ages are in arithmetic progression (a 03 Dec 2020, 08:17
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Bunuel
A club consists of members whose ages are in arithmetic progression (an arithmetic progression is a sequence of numbers such that the difference between the consecutive terms is constant). The common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250, then number of members in the club are:

A. 18

B. 20

C. 22

D. 24

E. 25
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Sum of an AP = \(\frac{n}{2}* [2a + (n - 1)d]\)

Converting all values into months

Sum = 250 * 12 = 3000, a = 7 * 12 = 84, d = 3

Substituting, we get 3000 = \(\frac{n}{2}* [2*84 + (n - 1)3]\)

6000 = n[168 + 3n - 3]

\(3n^2 + 165n - 6000 = 0\)

\(n^2 + 55n - 2000 = 0\)

\(n^2 + 80n - 25n - 2000 = 0\)

n(n + 80) - 25(n + 80) = 0

(n + 80)(n - 25) = 0

n = 25 or - 80


Option E

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BKD5050
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A club consists of members whose ages are in arithmetic progression (a 24 May 2025, 11:12
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how to find 80 and 25 no quickly?
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SamplingCoast
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A club consists of members whose ages are in arithmetic progression (a 24 May 2025, 22:39
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BKD5050
how to find 80 and 25 no quickly?
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Once we have the quadratic equation, we can use the list of options as a clue. Two numbers whos PRODUCT -> -2000 SUM-> 55

Prime Factoriz. of 2000

2000 = 2^4 * 5^3

(A) 18 not possible with just 5s and 2s (Theres at least one 3 in there) so WRONG

(B) 20 = 5^1 * 2^2; Whats left: 5^2 and 2^2 = 25 * 4 = 100. Difference (100-20) not 55 WRONG
(C) & (D) similar to option (A), other prime factors, other than 2 and 5 are lurking in each, so not possible.

(E) 25 = 5^2; Whats left: 5^1 * 2^4 = 5*16 = 80. 80-25 = 55 CORRECT.
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ParthSunger
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A club consists of members whose ages are in arithmetic progression (a 17 Jul 2025, 01:52
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I don't think there is a need to solve the whole equation.

Since D = 3 months or 0.25 of a year, in the equation

(n/2)∗[2a+(n−1)d], "D" would be 1/4 which implies that (N-1) should be divisible by 4(Number of members cannot be in decimals).

Out of all the options, only E - 1 is divisible by 4.

Hence the answer should be E.
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