How many terms of the arithmetic progression 1, 4, 7 … are needed to make the sum 51?

Hint: The common difference of an arithmetic progression can be found out by just subtracting the first term from the second term that is
\[{{a}_{2}}-{{a}_{1}}=d\]
The formula for writing \[{{n}^{th}}\] term of an arithmetic progression is
\[{{n}^{th}}\ term=a+(n-1)d\] (where ‘a’ is the first term and‘d’ is the common difference of the arithmetic progression)
The formula for writing the sum of first n terms of an arithmetic progression is
$S_n$ \[=\dfrac{n}{2}\left( 2a+(n-1)d \right)\]

Complete step-by-step answer:
As mentioned in the question, it is given that the first term of the arithmetic progression is 1, therefore a is equal to 1.
Now, using the formula for finding the common difference of the arithmetic progression which is given in the hint is as follows
\[\begin{align}
  & {{a}_{2}}-{{a}_{1}}=d \\
 & 4-1=d \\
 & d=3 \\
\end{align}\]
Now, as we know the first term that is ‘a’ and the common difference that is ‘d’ of the arithmetic progression, therefore, we can use the formula for calculating the sum of the arithmetic progression which is given in the hint as follows
\[=\dfrac{n}{2}\left( 2\times 1+(n-1)3 \right)\]
Now, we know that the value of the sum should be 51 as mentioned in the question, so, we can write the above equation as follows
\[\begin{align}
  & 51=\dfrac{n}{2}\left( 2\times 1+(n-1)3 \right) \\
 & n\left( 2\times 1+(n-1)3 \right)=102 \\
 & 3{{n}^{2}}-n-102=0 \\
\end{align}\]
On solving this quadratic equation using the quadratic formula, we get the value of n as 6.
Hence, 6 terms of the arithmetic progression would be required for getting the sum as 51.

NOTE: The students can make an error in evaluating the common difference of the arithmetic progression that is ‘d’ as it is not given straight away in the question rather one has to use the formula given in the hint to calculate the common difference of the arithmetic progression.