Question

The third term of an A.P is \[7\] and its \[{7^{th}}\] term is two more than thrice of its \[{3^{rd}}\] term. Find the common difference.

Answer 1

Hint: Here we use the formula for \[{n^{th}}\] term of an A.P and write the third term according to that formula. Then we form an equation for the seventh term using the relation given in the question.
* In an A.P the \[{n^{th}}\] term is given by ${a_n} = a + (n - 1)d$
Where, a = First term of an A.P
              n = number of terms
              d = common difference
* Whenever a term is more than the other term we add the value to the other term and equate it with a term that is more in quantity.

Complete step-by-step answer:
First we solve for the third term.
Given that the third term is 7. From the formula for the \[{n^{th}}\] term
${a_n} = a + (n - 1)d$
Here \[n = 3\]
Therefore, we can write
$
  {a_3} = a + (3 - 1)d \\
  {a_3} = a + 2d \\
 $
Now we substitute the value of third term as \[{a_3}\]
$a + 2d = 7$ ... (1)
Now Find the second relation between a and d.
Now we are given 7th term is 2 more than thrice of the third term
So we solve stepwise, thrice of the third term means 3 times the third term \[ \Rightarrow 3 \times {a_3}\].
Now seventh term is two more than thrice of third term, so we keep seventh term on LHS and add 2 to thrice of third term
 ${a_7} = 3{a_3} + 2$
Substitute the value of ${a_3} = 7$
 $
   \Rightarrow {a_7} = 3 \times 7 + 2 \\
   \Rightarrow {a_7} = 21 + 2 \\
   \Rightarrow {a_7} = 23 \\
 $
Also, from formula for the \[{n^{th}}\] term
${a_n} = a + (n - 1)d$
Here \[n = 7\]
Therefore, we can write
$
  {a_7} = a + (7 - 1)d \\
  {a_7} = a + 6d \\
 $
Equating both the values of the seventh term
$a + 6d = 23$ ... (2)
Now solve the two equations (1) and (2) to find values of a and d
From equation (1) $a + 2d = 7$
So, $a = 7 - 2d$
Substitute the value of $a = 7 - 2d$ in equation (2)
\[
  7 - 2d + 6d = 23 \\
  7 + 4d = 23 \\
 \]
Shift all the constant values to one side of the equation
\[
  4d = 23 - 7 \\
  4d = 16 \\
 \]
Divide both sides by 4
\[
  \dfrac{{4d}}{4} = \dfrac{{16}}{4} \\
  d = 4 \\
 \]
Therefore, the common difference of the given AP is 4.

Note: Students can many times make mistake of forming the equation with more than part wrong because they think that value is added to the part which is given more but students should keep in mind that the value is always added to the lesser part and equate it to the part which is given more.