Calculate the sum between and including the terms t16 and t53 of an arithmetic sequence with first term 543 and common difference –12.

SOOO Confused, if anyone can help, I'll appreciate it a lot! Thanks

The sum of a geometric sequence 2 – 6 + 18 – 54 + ... – tn = –29 524. Find the number of terms.

Julius
Jun 2, 2017

#1**0 **

U_{1}=543

U_{n}=U_{n-1}-12 n>1

__First Term: 543__

__2:__ 531

__3:__ 519

__4:__ 507

...

__16:__ 363

...

__53:__ -81

-81

Σ((n-1)-12)

i=543

Sum=136250

*not sure if this is completely right, verify with CPhill or someone else to make sure

ZZZZZZ
Jun 2, 2017

#2**+1 **

First one

t_{16} = 543 - 12 (15) = 363

t_{53} = 543 - 12(52) = -81

Sum of the terms = [ first term in series + last term in series] [ number of terms] / 2

**The number of terms is figured as [ 53 - 16 + 1 ] = [ 38]

So....the sum is : [ 363 + - 81 ] [ 38 ] / 2 = [282] [38] / 2 = 5358

Second one

The nth term of series can be expressed as :

a_{n} = 2 (-3)^{n-1}

So

-29524 = 2 (-3)^{ n-1 }

Notice here that - 29524 is not a term in this series :

https://www.wolframalpha.com/input/?i=2(-3)%5E(n+-+1)++++from+n++%3D+1++to+n+%3D+10

CPhill
Jun 2, 2017

#3**+1 **

The second one:

2*[1-(-3)^n] / [1 -(-3)]=-29,524

Solve for n:

1/2 (1 - (-3)^n) = -29524

Multiply both sides by 2:

1 - (-3)^n = -59048

Subtract 1 from both sides:

-(-3)^n = -59049

Multiply both sides by -1:

(-3)^n = 59049

Take the logarithm of both sides, ignoring the - sign:

Log(59,049)/ Log(3)

**Answer: n = 10**

P.S. Regarless of -3, you can still take the positive log of both sides. It will still give you an accurate n.

The first one:

I shall consider the 16th term as the 1st. term =543

53 - 16 + 1 =38 number of terms.

Common difference =-12

Sum =n/2[2a + (n -1)d]

Sum =38/2[2*543 + (38 - 1)*(-12)]

Sum = 19[1,086 + (-444)]

Sum = 19[642]

**Sum =12,198**

Guest Jun 2, 2017