Question

If k,k+1,2k+3 are three consequtive terms in ap then find the value of k

Answer 1

Solution :

Let a1 = k , a2= k + 1 ,

a3=2k+3 are three

consecutive terms .

We know that ,

a2 - a1 = a3 - a2

=> k+1-k = 2k+3-(k+1)

=> 1 = 2k +3-k-1

=> 1 = k + 2

=> 1 - 2 = k

=> -1 = k

Therefore ,

k = -1

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Answer 2

As there is common difference in AP,

$(k + 1) - k = (2k+3)-(k+1) \\ \\ k+1-k=2k+3-k-1 \\ \\ 1=k+2 \\ \\ k=1-2 \\ \\ k=\bold{-1}$

OR

Among three consecutive terms in an AP, the sum of first and third terms is equal to twice the second term.

Therefore,

$k+(2k+3)=2(k+1) \\ \\ k+2k+3=2k+2 \\ \\ 3k+3=2k+2 \\ \\ 3k-2k=2-3 \\ \\ k=\bold{-1}$

∴ k = -1

 

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Oh, no! I found a thing from the second method.

Whenever you're asked to prove '3 = 2' trickily, we can use the second method, as the following:

$k+(2k+3)=2(k+1) \\ \\ k+2k+3=2k+2 \\ \\ 3k+3=2k+2 \\ \\ 3(k+1)=2(k+1) \\ \\ 3=2$

How is this possible?!

We found earlier that k = -1 here. Therefore, k + 1 becomes 0.

At last step, the sides are divided by k + 1 to get 3 = 2. As dividing any number by 0 is invalid, dividing the sides by k + 1 is impossible.

So this is the contradiction!!!

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Thank you. Have a nice day. :-)

#adithyasajeevan