Question

3 + 9 + 15 + 21 + ... + (6n - 3) = 3n²​

Answer 1

Step-by-step explanation:

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Answer 2

The sum of an AP series with first term 'a =3' and common difference 'd=6' till term '6n-3' is $3n^{2}$.

Given :

The first term of AP series : 3

The sum of AP terms = 3$n^{2}$

To Find :

The value of summation of AP terms

Solution :

Firstly we find the difference between each pair of consecutive terms in the series.

                $d_{1} = a_{2} - a_{1} = 9-3=6$

                $d_{2} = a_{3} - a_{2} = 15-9=6$

                $d_{4} = a_{4} - a_{3} = 21-15=6$

Thus, as the difference between two consecutive terms is the given series is constant, it is an Arithmetic Progression i.e. AP with :

First term : $a_{1}=3$

Common difference : $d=6$

Xth term : $a_{X} = 6n-3$

Sum of X terms : $S_{X} = 3n^{2}$

The formula for finding the Xth term in the AP is : ( Equation 1 )

                     $a_{X} = a + (X-1)d\\6n-3 = a+(X-1)*6\\6n-3 = 3+6X-6\\6n-3=6X-3\\6n=6X\\n=X$

Then, we apply the formula for the sum of X terms in an AP series :

(Equation 2) :      

               $S_{X} = X/2[2a + (X - 1) *d]$

                     $= X/2[2(3) + (X - 1) *6]$

                     $= X/2[6 + 6X - 6]$

                     $= 3X^{2}$

From Equation 1 we get that: $n = X$. Applying it to Equation 2 :

              $S_{X} = 3X^{2} \\S_{X} = 3n^{2} [ As, n=X]$

Hence, it is proved that the sum of an AP series with first term 'a =3' and common difference 'd=6' till term '6n-3' is $3n^{2}$.

         

To learn more about Arithmetic Progression, visit

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