# Find an A.P. in which sum of any number of terms is always three times the squared number of these terms.

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Hint: Here, we will show the given condition mathematically using variables. Then, by assuming different values of the variable, we will find the sum of the first term, first two terms, and first three terms and so on. Then, we will apply the fact that in an A.P. the first term and the sum of first term is one and the same thing, also, the second term in A.P. is equal to the difference between the sum of first three and first two terms respectively and so on. This will give us the required arithmetic progression satisfying the given condition.

An Arithmetic Progression or A.P. is a sequence in which the difference between two consecutive terms is the same. This difference is known as the common difference and we find it by subtracting each term by its preceding term respectively.
Let the number of terms be $n$.
It is given that the sum of these $n$ terms is equal to three times the squared number of these terms.
Hence, we can write this mathematically as:
${S_n} = 3{n^2}$
Now, let $n = 1$
$\Rightarrow {{S}_{1}}=3{{\left( 1 \right)}^{2}}=3$
Also, let $n = 2$
$\Rightarrow {{S}_{2}}=3{{\left( 2 \right)}^{2}}=3\times 4=12$
Similarly, let $n = 3$
$\Rightarrow {{S}_{3}}=3{{\left( 3 \right)}^{2}}=3\times 9=27$
And so on.
Now, we know that in an A.P., the sum of one term is the first term itself.
Hence, ${a_1} = {S_1} = 3$, where ${a_1}$ is the first term of the A.P.
Also, the second term is the difference between the sum of first two terms and the first term.
Hence, ${a_2} = {S_2} - {S_1} = 12 - 3 = 9$
Similarly, the third term is equal to the difference between the sum of first three terms and the first two terms.
Hence, ${a_3} = {S_3} - {S_2} = 27 - 12 = 15$
And so on.
Hence, the required A.P is:
$3,9,15,...$
Here, first term is 3 and the common difference, $d = 9 - 3 = 15 - 9 = 6$
Therefore, an A.P. in which sum of any number of terms is always three times the squared number of these terms is $3,9,15,...$.