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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
A. 312
B. 313
C. 314
D. 315
E. 316
01 Jan 2018, 11:57
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Bunuel
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?
01 Jan 2018, 12:26
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Bunuel
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
