Question

If ${{S}_{n}}$, the sum of first n terms of an A.P., is given by ${{S}_{n}}=5{{n}^{2}}+3n$, then find its ${{n}^{th}}$ term?

Answer 1

Hint:The sum of n terms of an A.P. is given as ${{S}_{n}}=5{{n}^{2}}+3n$ and we have to find ${{n}^{th}}$ term for that we require first term and the common difference which we can find by writing the series of an A.P. When you put $n=1$ in ${{S}_{n}}$ you will get the first term, then put $n=2$ in ${{S}_{n}}$ then you will get sum of the first two terms, as we have calculated the first term so subtracting sum of first two terms and the first term we get, the second term of the A.P. Similarly you can find the other terms of the A.P. also. Now, we know the first term (a) and subtracting any two consecutive terms of the A.P. will give you common difference (d) so substitute these values of “a” and “d” in the formula of general term of an A.P. i.e.${{T}_{n}}=a+\left( n-1 \right)d$ will give us the ${{n}^{th}}$ term of an A.P.

Complete step-by-step answer:
Sum of the first n terms of an A.P. is given by:
${{S}_{n}}=5{{n}^{2}}+3n$……….. Eq. (1)
Substituting $n=1$ in the above equation we get the first term of an A.P.
$\begin{align}
  & {{S}_{1}}=5{{\left( 1 \right)}^{2}}+3\left( 1 \right) \\
 & \Rightarrow {{S}_{1}}=5+3=8 \\
\end{align}$
From the above, the first term of an A.P. is 8.
Now, substituting $n=2$ in eq. (1) we get,
$\begin{align}
  & {{S}_{2}}=5{{\left( 2 \right)}^{2}}+3\left( 2 \right) \\
 & \Rightarrow {{S}_{2}}=20+6=26 \\
\end{align}$
The above summation is the sum of first two terms of an A.P. and we know the first term i.e. 8 so subtracting 8 from 26 we get the second term of an A.P.
${{S}_{2}}-{{S}_{1}}=26-8=18$
From the above second term of an A.P. is 18.
Substituting $n=3$ in eq. (1) we get,
$\begin{align}
  & {{S}_{3}}=5{{\left( 3 \right)}^{2}}+3\left( 3 \right) \\
 & \Rightarrow {{S}_{3}}=5\left( 9 \right)+9 \\
 & \Rightarrow {{S}_{3}}=45+9=54 \\
\end{align}$
The above summation is the summation of the first three terms so subtracting the summation of the first two terms from the above summation of the first three terms we will get the third term.
${{S}_{3}}-{{S}_{2}}=54-26=28$
From above, the first three terms of an A.P. are:
8, 18, 28
Similarly, we can find the other terms of an A.P. too.
Now, we are asked to find the ${{n}^{th}}$ term of an A.P. which we are going to calculate using the formula for general term of an A.P. i.e.,
${{T}_{n}}=a+\left( n-1 \right)d$……….. Eq. (2)
For the above formula, we need the value of “a” and “d” which we are going to calculate by the three terms of an A.P. that we have solved above.
8, 18, 28
The first term of an A.P. (a) is equal to 8 and for the common difference subtract any two consecutive terms so subtracting 8 from 18 we get 10. Hence, the common difference (d) is 10.
Substituting the values of “a” as 8 and “d” as 10 in eq. (2) we get,
$\begin{align}
  & {{T}_{n}}=8+\left( n-1 \right)10 \\
 & \Rightarrow {{T}_{n}}=8-10n-10 \\
 & \Rightarrow {{T}_{n}}=-10n-2 \\
\end{align}$
Hence, the ${{n}^{th}}$ term of an A.P. is $-10n-2$.

Note: You can stop after finding first two terms of an A.P. because we only need the first term and the common difference to calculate the ${{n}^{th}}$ term of an A.P. which can be found by the first two terms of an A.P. also. In the above solution, we have shown the other terms like third term to get you confidence that the common difference is correct that we have got from taking the difference of first two terms.
The first three terms of an A.P. that we have solved above is:
8, 18, 28
The difference of first two terms is:
$18-8=10$
The difference of last two terms is:
$28-18=10$
From the above steps, we can see that the difference is common when we take the difference of the first two terms and the last two terms i.e. 10.