Question

The first three terms of an arithmetic sequence are $ 2k + 3,\;5k - 2 $ and $ 10k - 15, $ how do you show that $ k = 4 $ ?

Answer 1

Hint: Arithmetic progression is the sequence in which there is the common difference between any two consecutive terms. Here first of all we will set terms in AP and then will follow the given conditions to simplify and get the resultant value.

Complete step-by-step answer:
We know that the arithmetic sequence has the common difference “d” in the consecutive terms. So, the given terms can be written as-
 $ 5k - 3 = (2k + 3) + d $
Simplify the above equation –
When you move any term from one side to the opposite side then the sign of the terms also changes. Positive terms become negative and vice-versa.
 $
  5k - 2k - 2 - 3 = + d \\
  3k - 5 = d\;{\text{ }}....{\text{ (A)}} \;
  $
And $ 10k - 15 = (5k - 2) + d $
Simplify the above equation-
 $
  10k - 15 - 5k + 2 - d = 0 \\
  5k - 13 = d{\text{ }}.....{\text{ (B)}} \;
  $
Equation (A) and (B) since right hand side of both the equations are equal
 $ 3k - 5 = 5k - 13 $
Move all constants on one side and variables on opposite side –
 $
   - 5 + 13 = 5k - 3k \\
  8 = 2k \\
  k = 4 \;
  $
So, the correct answer is “ k = 4”.

Note: An Arithmetic Progression (AP) is the sequence of numbers in which the difference of two successive numbers is always constant.
The standard formula for Arithmetic Progression is – $ {a_n} = a + (n - 1)d $
Where $ {a_n} = $ nth term in the AP
 $ a = $ First term of AP
 $ d = $ Common difference in the series
 $ n = $ Number of terms in the AP