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If each number in a sequence is three more than the previous number, a

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Bunuel
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If each number in a sequence is three more than the previous number, a [#permalink] New post   01 Jan 2018, 11:56
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If each number in a sequence is three more than the previous number, and the sixth number is 32, what is the 100th number?

A. 312
B. 313
C. 314
D. 315
E. 316
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sohinide
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Re: If each number in a sequence is three more than the previous number, a [#permalink] New post   01 Jan 2018, 11:57
Bunuel wrote:
If each number in a sequence is three more than the previous number, and the sixth number is 32, what is the 100th number?


If each number in a sequence is three more than the previous number, and the
sixth number is 32, what is the 100th number?

The answer says -
From the sixth to the one hundredth term, there are 94 “jumps” of 3. Since 94 x 3 = 282, there is an
increase of 282 from the sixth term to the one hundredth term:
32 + 282 = 314.

Can someone please help me understand how the 94 figure is arrived at?
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Re: If each number in a sequence is three more than the previous number, a [#permalink] New post   01 Jan 2018, 12:26
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sohinide wrote:
Bunuel wrote:
If each number in a sequence is three more than the previous number, and the sixth number is 32, what is the 100th number?


If each number in a sequence is three more than the previous number, and the
sixth number is 32, what is the 100th number?

The answer says -
From the sixth to the one hundredth term, there are 94 “jumps” of 3. Since 94 x 3 = 282, there is an
increase of 282 from the sixth term to the one hundredth term:
32 + 282 = 314.

Can someone please help me understand how the 94 figure is arrived at?



6th no -- 32 ....7th - 35 .....100th --?

sp 100 - 6 = 94 terms between 6 n 100

each term is three more than the previous number

so all 94 Ts X 3 = 282

94 terms + 6 terms = 100th term
282 + 32 = 314
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gracie
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Re: If each number in a sequence is three more than the previous number, a [#permalink] New post   01 Jan 2018, 21:20
Bunuel wrote:
If each number in a sequence is three more than the previous number, and the sixth number is 32, what is the 100th number?


32-5*3=first number=17
17+99*3=100th number=314
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luffy_ueki
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Re: If each number in a sequence is three more than the previous number, a [#permalink] New post   01 Jan 2018, 21:25
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Since there is a constant difference of 3 between terms in the series => Arithmetic progression
d=3 ;
sinc 6th term =32
tn = a+(n-1)d
t6 = a+(5)*3
a=17

For 100th term
t100 = a+99*d
=17+99*3
=17+297
=314 (100th term)
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Re: If each number in a sequence is three more than the previous number, a [#permalink] New post   21 Mar 2018, 01:04
d = 3

a1 + 5d = 32

=) a1 = 17

and, a100 = a1 + 99d

=) a100 = 314

thanks
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Re: If each number in a sequence is three more than the previous number, a [#permalink] New post   31 May 2020, 08:22
each number in a sequence is three more than the previous number
{ m,m+3, m+6 , m+9,........}

Above set is in A.P., so
6th term, a6 = 32 = a+(n-1)d
m+(n-1)d = 32
m+(6-1)3 = 32
m+15 = 32
m = 32-15 = 17

100th term will be , = m + (100-1)3
= 17+99*3
= 17 + 297
= 314

Bunuel, please suggest
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Re: If each number in a sequence is three more than the previous number, a [#permalink] New post   08 Feb 2022, 09:10
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Re: If each number in a sequence is three more than the previous number, a [#permalink]
08 Feb 2022, 09:10
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