+956 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.
+476 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
+128408
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
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
