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

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

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New post 01 Jan 2018, 10:56
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If each number in a sequence is three more than the previous number, a  [#permalink]

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New post 01 Jan 2018, 10:57
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]

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New post 01 Jan 2018, 11:26
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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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If each number in a sequence is three more than the previous number, a  [#permalink]

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New post 01 Jan 2018, 20:20
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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Re: If each number in a sequence is three more than the previous number, a  [#permalink]

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New post 01 Jan 2018, 20:25
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)
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Re: If each number in a sequence is three more than the previous number, a  [#permalink]

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New post 21 Mar 2018, 00:04
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 &nbs [#permalink] 21 Mar 2018, 00:04
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