In a sequence of 21 terms the first 11 t...

In a sequence of 21 terms the first 11 terms are in A.P. with common difference 2. and the lastterms are in G.P. with common ratio 2. If the middle tem of the A.P. is equal to themiddle term of the G.P., then the middle term of the entire sequence is

`-10/31`

`10/31`

`32/31`

`-31/32`

Text Solution

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The correct Answer is:

Since , the first 11 terms are in AP, d= 2
` :. A_(11) a+ 10d = a+ 20 `
The middle term of AP is
`T_(6)=a+5d = a+10`
For the next terms which are in GP, r = 2
Hence, the middle term of GP is `b(2)^(5)` , where b is the first term of a GP which is the last term of AP
` :. b(2)^(5) = (a +20)32`
According to the given condition
`a+10= (a+20)32`
` rArr 31a = 10- 640 rArr a = -630/31 `
So, middle term of entire sequence is 11th term
Now, ` T_(11) = -630/31 +10xxd = -630/31 + 10xx 2 = -10/31`

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