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# @Hecitar @CPhill @ANYONE??

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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
+458
0

U1=543

Un=Un-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
+90027
+1

First one

t16   =  543 - 12 (15)   =  363

t53  =    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  :

an   =   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
#4
+90027
+2

Yuck!!....I screwed up the second one....

But...I'm still sticking by my answer  to the first....!!!!

CPhill  Jun 2, 2017

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