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Question: If we are given \(\left( k-3 \right),\left( 2k+1 \right)\And \left( 4k+3 \right)\) three consecutive...

If we are given (k−3),(2k+1)&(4k+3)\left( k-3 \right),\left( 2k+1 \right)\And \left( 4k+3 \right) three consecutive terms of an A.P. Find the value of k?

Explanation

Solution

There is a relation between three terms say a, b, and c which is in A.P. is 2b=a+c2b=a+c so compare the three terms given in the question with a, b and c we get “a” as ‘k’–3‘k’ – 3 and “b” as 2k+12k + 1 and “c” as 4k+34k + 3 then substitute these values in the relationship that we have just shown between a, b and c. Solve the equation and you will get the value of k.

Complete step-by-step solution:
In the above problem, we have given the three terms which are in A.P. as:
(k−3),(2k+1)&(4k+3)\left( k-3 \right),\left( 2k+1 \right)\And \left( 4k+3 \right)
We know that, if three terms say a, b and c are in A.P. then the relation between these three terms as:
2b=a+c2b=a+c……….. Eq. (1)
Now, on comparing a, b, c with the terms given in the above problem (k−3),(2k+1)&(4k+3)\left( k-3 \right),\left( 2k+1 \right)\And \left( 4k+3 \right) the value of a, b and c are as follows:
a=k−3 b=2k+1 c=4k+3 \begin{aligned} & a=k-3 \\\ & b=2k+1 \\\ & c=4k+3 \\\ \end{aligned}
Substituting the above values of a, b and c in eq. (1) we get,
2b=a+c ⇒2(2k+1)=k−3+4k+3 \begin{aligned} & 2b=a+c \\\ & \Rightarrow 2\left( 2k+1 \right)=k-3+4k+3 \\\ \end{aligned}
Multiplying 2 by 2k+12k + 1 on the left hand side of the above equation we get,
4k+2=5k4k+2=5k
Subtracting 4k4k on both the sides of the above equation we get,
2=k2=k
From the above solution, we have got the value of k as 2.
Hence, the value of k is 2.

Note: We can cross check whether the value of k that we are getting is correct or not by substituting this value of k in (k−3),(2k+1)&(4k+3)\left( k-3 \right),\left( 2k+1 \right)\And \left( 4k+3 \right).
Substituting the value of k as 2 in (k−3),(2k+1)&(4k+3)\left( k-3 \right),\left( 2k+1 \right)\And \left( 4k+3 \right) we get,
(2−3),(2(2)+1),(4(2)+3) =−1,5,11 \begin{aligned} & \left( 2-3 \right),\left( 2\left( 2 \right)+1 \right),\left( 4\left( 2 \right)+3 \right) \\\ & =-1,5,11 \\\ \end{aligned}
The above three terms should satisfy the condition of three terms in A.P. which is 2b=a+c2b=a+c where a is -1, b is 5 and c is 11 substituting these values in the relation between a, b and c we get,
2(5)=−1+11 ⇒10=10 \begin{aligned} & 2\left( 5 \right)=-1+11 \\\ & \Rightarrow 10=10 \\\ \end{aligned}
As you can see that L.H.S = R.H.S so the value of k that we have solved above is correct.