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

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Math Expert
Joined: 02 Sep 2009
Posts: 58402
If each number in a sequence is three more than the previous number, a  [#permalink]

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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?
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Joined: 13 Nov 2016
Posts: 4
If each number in a sequence is three more than the previous number, a  [#permalink]

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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?
Manager
Joined: 05 Nov 2015
Posts: 62
Location: India
Re: If each number in a sequence is three more than the previous number, a  [#permalink]

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01 Jan 2018, 12:26
1
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
VP
Joined: 07 Dec 2014
Posts: 1222
If each number in a sequence is three more than the previous number, a  [#permalink]

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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
Manager
Joined: 23 Oct 2017
Posts: 61
Re: If each number in a sequence is three more than the previous number, a  [#permalink]

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01 Jan 2018, 21:25
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)
Senior Manager
Status: love the club...
Joined: 24 Mar 2015
Posts: 269
Re: If each number in a sequence is three more than the previous number, a  [#permalink]

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21 Mar 2018, 01:04
d = 3

a1 + 5d = 32

=) a1 = 17

and, a100 = a1 + 99d

=) a100 = 314

thanks
Re: If each number in a sequence is three more than the previous number, a   [#permalink] 21 Mar 2018, 01:04
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