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Question

In a sequence of 21 terms, the first 11 terms are in AP with common difference 2 and the last 11 terms are in GP with common ratio 2. If the middle term of an AP is equal to the middle term of the GP, then the middle term of the entire sequence is


A

-1031

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B

1031

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C

3231

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D

-3132

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Solution

The correct option is A

-1031


Explanation for the correct option.

Step 1. Evaluate the AP terms.

The first 11 terms are in AP.

Let the first term be a and the common difference is 2.

So the 11th term of the AP is

a11=a+(11-1)×2=a+20

Now, the middle term of the AP is the sixth term, so the middle term is:

a6=a+6-1×2=a+10

Step 2. Evaluate the GP terms.

The next eleven terms in the sequence is in GP.

Let the first term of the GP be band its common ratio is 2.

So, the middle term of the GP series is:

b6=b26-1=32b

Step 3. Form two equations.

The last term of the AP will be the same as the first term of the GP. So,

a+20=b...(1)

It is also given that the middle term of the AP is the same as the middle term of the GP that is a6=b6 and so

a+10=32b...(2)

Step 4. Solve for a and b.

Using equation 1 substitute a+20 for b in equation 2.

a+10=32a+20a+10=32a+640-31a=630a=-63031

Now, substitute -63031 for a in equation 1.

b=-63031+20=-630+62031=-1031

Step 5. Find the middle term of the sequence.

For the sequence of 21 terms, the middle term is the 11th term. So it is the last term of the AP or the first term of the GP.

The first term of the GP is b=-1031.

So the middle term of the sequence is -1031.

Hence, the correct option is A.


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