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Question 1163889: Question: The first term of an arithmetic sequence is 2; and the first, third, and eleventh terms are also the first three terms of a geometric sequence. Find the sum of the first eleven terms of the arithmetic sequence.
How far I got:
The third term of AP: 2+2d which is equal to second term of GP 2r giving
equation (1)...... 2 + 2d - 2r = 0
Likewise, the 11th term of AP: 2+10d is equal to third term of GP 2r^2 giving
equation (2)...... 2 + 10d - 2r^2 = 0
Solving the system I get, r = 1 and 4 (discard 1 as it is GP) and d = 3
So the sum of AP is (34/2)= 17.
Obviously I am wrong. Can you please point it out.
Please dont give me the answer. Just give me the broad steps to solve the problem.
Thank you
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Your steps are fine.
Your initial equations are fine.
Your solution to the system of equations is not right.
Since you didn't show the work you did to solve the system of equations, we can't point out your error. You should be able to find it yourself.
Don't try to look at your previous work and see what was wrong with it; when you try to do that, you often end up making the same mistake.
Start over with your two equations an try solving them anew.
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I misread your solution; your answers r=4 and d=3 are correct.
What is wrong is your final answer to the question as to the sum of the first 11 terms of the arithmetic sequence.
That sum is the average of the first and eleventh terms (which you show), MULTIPLIED BY the number of terms (11).
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