SOLUTION: How many terms of the A.P 26,21,16,11..........are needed to give the sum 11?

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Question 768612: How many terms of the A.P 26,21,16,11..........are needed to give the sum 11?
Found 2 solutions by oscargut, MathLover1:
Answer by oscargut(2103) About Me  (Show Source):
You can put this solution on YOUR website!
How many terms of the A.P 26,21,16,11..........are needed to give the sum 11?

the rest are: 6,1,-4,-9,-14,-19,-24
Answer: 11 terms
You can ask me more at: mthman@gmail.com
Thanks

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

26,21,16,11..........are needed to give the sum 11
26-21=5
21-16=5
16-11=5
so, d=-5
Sn+=+n%2F2%282a+%2B+%28+n-1+%29d+%29 , where ‘n’ is the number of terms, ‘a’ is first term, ‘d’ is the common difference
a=26,+Sn=11,+d=-5
11+=+n%2F2%282%2A26+%2B+%28+n+-1+%29%28-5%29+%29 ....solve for n
11+=+n%2F2%2852+-+5n+%2B5++%29
11%2A2+=+n%2857+-5n++%29
22+=+57n+-5n%5E2++
+5n%5E2-57n+%2B22=0....solve for n using quadratic formula
n+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
n+=+%28-%28-57%29+%2B-+sqrt%28+%28-57%29%5E2-4%2A5%2A22+%29%29%2F%282%2A5%29+
n+=+%2857+%2B-+sqrt%28+3249-440+%29%29%2F10+
n+=+%2857+%2B-+sqrt%28+2809+%29%29%2F10+
n+=+%2857+%2B-+53%29%2F10+
+n+=+11 and n+=+2%2F5

so, we need only whole number +n+=+11

26,21,16,11,6,1,-4,-9,-14,-19,-24
26%2B21%2B16%2B11%2B6%2B1%2B%28-4%29%2B%28-9%29%2B%28-14%29%2B%28-19%29%2B%28-24%29=11