Question

If the sum of first n terms of A.P. is $ \dfrac{1}{2}(3{n^2} + 7n) $ then find its $ {n^{th}} $ term and hence 20th term.

Answer 1

Hint: The sum of the first term that is $ {S_1} $ will give us the value of the first term. The difference between $ {S_n} $ and $ {S_{n - 1}} $ gives us the value of $ {n^{th}} $ term. Using this relation, we can find out the common difference and then using the formula for finding the value of $ {a_n} $ , we can find the value of $ {n^{th}} $ term and 20th term.

Complete step-by-step answer:
Sum of n terms of an A.P. is given by $ {S_n} = \dfrac{1}{2}(3{n^2} + 7n) $
 $ \Rightarrow {S_1} = \dfrac{1}{2}[3{(1)^2} + 7(1)] = \dfrac{1}{2}(3 + 7) = \dfrac{{10}}{2} = 5 $
It means the first term of the A.P. is 5.
 $\Rightarrow {S_2} = \dfrac{1}{2}[3{(2)^2} + 7(2)] = \dfrac{1}{2}(12 + 14) = \dfrac{{26}}{2} = 13 $
It means the sum of the first two terms is 13, to find the second term we have to subtract the first term from their sum, so the second term is $ {a_2} = {S_2} - {S_1} = 13 - 5 = 8 $
\[{n^{th}}\] term of A.P. is represented $ {a_n} = a + (n - 1)d $
where $ {a_n} = $ the nth term, $ a = $ the first term and $ d = $ the common difference between two consecutive terms.
 $
\Rightarrow {a_2} = 8 \\
\Rightarrow 5 + (2 - 1)d = 8 \\
\Rightarrow d = 3 \;
  $
Now, $ {S_3} = \dfrac{1}{2}[3{(3)^2} + 7(3)] = \dfrac{1}{2}(27 + 21) = \dfrac{{48}}{2} = 24 $
The third term is $ {a_3} = {S_3} - {S_2} = 24 - 13 = 11 $
 $ \Rightarrow d = {a_3} - {a_2} = 11 - 8 = 3 $
Hence, it is verified that the common difference is 3 and the first term is 5.
Putting the values of a and d in the formula for calculating the $ {n^{th}} $ term, we get –
 $
\Rightarrow {a_n} = a + (n - 1)d \\
\Rightarrow {a_n} = 5 + (n - 1)3 \\
\Rightarrow {a_n} = 5 + 3n - 3 \\
\Rightarrow {a_n} = 3n + 2 \;
  $
And hence 20th term is $ {a_{20}} = 5 + (20 - 1)3 = 5 + 57 = 62 $
So, the correct answer is “62”.

Note: An arithmetic progression is a progression or sequence of numbers such that the difference between any two consecutive numbers is constant.
For example, the sequence 2, 4, 6, 8…. Is an arithmetic progression or arithmetic sequence whose common difference can be obtained by subtracting one term from its next term, in this example the common difference is 2.