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10 + 5 = (5/2)(2)(3) 15 + 10 + 5 = (5/2)(3)(4) 20 + 15 + 10 + 5 = (5
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14 Mar 2022, 03:00
14 Mar 2022, 03:00
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Question Stats:
85% (01:46) correct
14% (02:19) wrong based on 21 sessions
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\(10 + 5 = (\frac{5}{2})(2)(3) \)
\(15 + 10 + 5 = (\frac{5}{2})(3)(4) \)
\(20 + 15 + 10 + 5 = (\frac{5}{2})(4)(5) \)
For each of the three sums shown above, let n represent the number of multiples of 5 being added. Which of the following is an expression for each of the three sums in terms of n?
(A) 2n + 1
(B) 5n
(C) n(n - 1)
(D) (5/2)(n)(n + 1)
(E) n(n + 1)
\(15 + 10 + 5 = (\frac{5}{2})(3)(4) \)
\(20 + 15 + 10 + 5 = (\frac{5}{2})(4)(5) \)
For each of the three sums shown above, let n represent the number of multiples of 5 being added. Which of the following is an expression for each of the three sums in terms of n?
(A) 2n + 1
(B) 5n
(C) n(n - 1)
(D) (5/2)(n)(n + 1)
(E) n(n + 1)
Re: 10 + 5 = (5/2)(2)(3) 15 + 10 + 5 = (5/2)(3)(4) 20 + 15 + 10 + 5 = (5
[#permalink]
14 Mar 2022, 10:24
14 Mar 2022, 10:24
2
Carcass wrote:
\(10 + 5 = (\frac{5}{2})(2)(3) \)
\(15 + 10 + 5 = (\frac{5}{2})(3)(4) \)
\(20 + 15 + 10 + 5 = (\frac{5}{2})(4)(5) \)
For each of the three sums shown above, let n represent the number of multiples of 5 being added. Which of the following is an expression for each of the three sums in terms of n?
(A) 2n + 1
(B) 5n
(C) n(n + 1)
(D) (5/2)(n)(n + 1)
(E) n(n + 1)
\(15 + 10 + 5 = (\frac{5}{2})(3)(4) \)
\(20 + 15 + 10 + 5 = (\frac{5}{2})(4)(5) \)
For each of the three sums shown above, let n represent the number of multiples of 5 being added. Which of the following is an expression for each of the three sums in terms of n?
(A) 2n + 1
(B) 5n
(C) n(n + 1)
(D) (5/2)(n)(n + 1)
(E) n(n + 1)
for 1st expression 5 + 10 ---> in term of n = n*1 + n*2
Similarly 2nd expression ----> n*1 + n*2 +n*3
So ans should be D, as we can't ignore 5/2, unless I'm missing on any difference in option C & E
Re: 10 + 5 = (5/2)(2)(3) 15 + 10 + 5 = (5/2)(3)(4) 20 + 15 + 10 + 5 = (5
[#permalink]
01 Nov 2022, 08:22
01 Nov 2022, 08:22
1
Remembering the sum of an AP: n(first + last) / 2
1: 2 * (5 + 10) / 2 => 2 * 5 * (1 + 2) / 2
2: 3 * (5 + 15) / 2 => 3 * 5 * (1 + 3) / 2
3: 4 * (5 + 20) / 2 => 4 * 5 * (1 + 4) / 2
we can see the pattern: n * 5 * (1 + n) / 2 => (5/2) * n * (n+1)
hence option D
1: 2 * (5 + 10) / 2 => 2 * 5 * (1 + 2) / 2
2: 3 * (5 + 15) / 2 => 3 * 5 * (1 + 3) / 2
3: 4 * (5 + 20) / 2 => 4 * 5 * (1 + 4) / 2
we can see the pattern: n * 5 * (1 + n) / 2 => (5/2) * n * (n+1)
hence option D
