Find the nth term of this number sequence 1, 10, 19, 28, ... Pls help

Mathematics

2 answers:

dalvyx [7]3 years ago

8 0

Answer:

The nth term = 9n - 8.

Step-by-step explanation:

This is an Arithmetic Sequence with common difference d =

10 - 1 = 19 - 10 = 28 - 19 = 9.

The nth term an = a1 + (n - 1)d where a1 = the first term and d = the common difference.

So the nth term = 1 + (n - 1)9

= 1 + 9n - 9

= 9n - 8.

kherson [118]3 years ago

5 0

Answer:

9n - 8

Step-by-step explanation:

The sequence is arithmetic since the difference between consecutive terms is common

d = 10 - 1 = 19 - 10 = 28 - 19 = 9

The n th term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

where a₁ is the first term and d the common difference

here a₁ = 1 and d = 9, thus

$a_{n}$ = 1 + 9(n - 1) = 1 + 9n - 9 = 9n - 8, hence

$a_{n}$ = 9n - 8

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blagie [28]

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3 years ago

For an arithmetic sequence, a35=57. If the common difference is 5. Find a1 and the sum of the first 61 terms.

Flauer [41]

$\displaystyle\bf\\a_{35}=57\\r=5\\----\\a_1=?\\S_{61}=?\\----\\Solve:\\1)\\a_1=a_{35}-(35-1)\times r=57-34\times 5=57-170=-113\\\boxed{\bf a_1=-113}\\\\$

$\displaystyle\bf\\2)\\A_{61}=a_1+(61-1)\times r=-113+60\times5=-113+300=187\\\\S_{61}=(-113)+(-108)+(-103)+...+(-8)+(-3)+2+7+12+...+187\\\\S_{61}=SA+SB\\\\SA=(-113)+(-108)+(-103)+...+(-8)+(-3)=-(3+8+103+108+113)\\\\n=\frac{113-8}{5}+1=\frac{105}{5}+1=21+1=22~terms\\SA=-\Big( \frac{n(113+3)}{2} \Big)=-\Big( \frac{22\times116}{2} \Big)=-11\times116 =-1276$

$\displaystyle\bf\\SB=2+7+12+...+187\\\\n=\frac{187-2}{5}+1=\frac{185}{5}+1=37+1=38~terms\\\\SB=\frac{n(187+2)}{2}=\frac{38\times189}{2}=19\times189=3591\\\\S_{61}=SA+SB=-1276+3591=2315\\\\\boxed{\bf S_{61}=2315}$