Question

if 3k,(4k+2),(6k+3) are three consecutive term of an AP then find the value of k is​

Answer 1

Given :

From the given series , we get :

• No. of terms (n) = 3

• First term (a) = 3k

• Last term (tn) = (6k + 3)

To find :

The value of k in the AP.

Solution :

First let us find the Common Difference of the AP in terms of k !!

We know the formula for Common Difference of an AP :

$\boxed{\bf{d = a_{n} - a_{(n - 1)}}}$

Where :

• d = Common Difference
• a = Any term of the AP

Now using the formula for Common Difference of an AP and substituting the values in it, we get :

$:\implies \bf{d = a_{n} - a_{(n - 1)}}$

$:\implies \bf{d = (6k + 3) - (4k + 2)}$

$:\implies \bf{d = 6k + 3 - 4k - 2}$

$:\implies \bf{d = 2k + 1}$

$\boxed{\therefore \bf{d = 2k + 1}}$

Hence the common difference of the AP is (2k + 1).

Here we are provided with the nth term of the AP (i.e, 6k + 3) , first term of the AP (i.e, 3k) and we found the value of Common Difference (i.e, 2k + 1) , so by using the formula for nth term of the AP we can find the value of k.

Formula for nth term of the AP :

$\boxed{\bf{t_{n} = a_{1} + (n - 1)d}}$

Where :

• tn = nth term of the AP
• a1 = First term of the AP
• n = No. of terms
• d = Common Difference

Now using the above formula and substituting the values in it, we get :

$:\implies \bf{t_{n} = a_{1} + (n - 1)d}$

$:\implies \bf{(6k + 3) = 3k + (3 - 1)(2k + 1)}$

$:\implies \bf{(6k + 3) = 3k + 2(2k + 1)}$

$:\implies \bf{(6k + 3) = 3k + 4k + 2}$

$:\implies \bf{(6k + 3) = 7k + 2}$

$:\implies \bf{6k + 3 = 7k + 2}$

$:\implies \bf{3 - 2 = 7k - 6k}$

$:\implies \bf{3 - 2 = 7k - 6k}$

$:\implies \bf{1 = 1k}$

$:\implies \bf{1 = k}$

$\boxed{\therefore \bf{k = 1}}$

Hence the value of k is 1.

Answer 2

Answer:

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