Question

Show that ${a_1},{a_2},...,{a_n}$ form an AP where ${a_n}$ is defined as below:
(i) ${a_n} = 3 + 4n$
(ii) ${a_n} = 9 - 5n$
Also find the sum of the first 15 terms in each case.

Answer 1

Firstly, calculate the value of ${a_1},{a_2},...,{a_n}$ by substituting the value of n $n = 1,2,3,...$. After that get a series. Use the formula of the sum of n terms in Arithmetic progression that is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$ where, a initial term of the AP and d is the common difference of successive numbers. Substitute the values n, a, and d and then calculate the sum of the AP ${S_n}$.

Complete step by step answer:
(i) The equation is ${a_n} = 3 + 4n$.
Now, substitute $n = 1$ in the expression ${a_n} = 3 + 4n$.
$\Rightarrow {a_1} = 3 + 4\left( 1 \right)$
$= 7$
Now, substitute $n = 2$ in the expression ${a_n} = 3 + 4n$.
$\Rightarrow {a_2} = 3 + 4\left( 2 \right)$
$= 11$
Now, substitute $n = 3$ in the expression ${a_n} = 3 + 4n$.
$\Rightarrow {a_3} = 3 + 4\left( 3 \right)$
$= 15$
Hence, the series is $7,11,15,...$.
We can see the difference is same in the above series. It means the series show the AP series.
We know about the formula of the sum of n terms in Arithmetic progression that is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
Now, calculate the value of ${S_n}$. Substitute the values $n = 15,a = 7$ and $d = 4\left( {11 - 7} \right)$ in the expression ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
$\Rightarrow {S_{15}} = \dfrac{{15}}{2}\left[ {2\left( 7 \right) + \left( {15 - 1} \right)4} \right]$
$= 7.5\left[ {14 + 56} \right]$
$= 7.5\left( {70} \right)$
$= 525$

Hence, the sum of the first 15 term is ${S_{15}} = 525$.

(ii) The equation is ${a_n} = 9 - 5n$.
Now, substitute $n = 1$ in the expression ${a_n} = 9 - 5n$.
$\Rightarrow {a_1} = 9 - 5\left( 1 \right)$
$= 4$
Now, substitute $n = 2$ in the expression ${a_n} = 9 - 5n$.
$\Rightarrow {a_2} = 9 - 5\left( 2 \right)$
$= - 1$
Now, substitute $n = 3$ in the expression ${a_n} = 9 - 5n$.
$\Rightarrow {a_3} = 9 - 5\left( 3 \right)$
$= - 6$
Hence, the series is $4, - 1, - 6,...$.
We can see the difference is same in the above series. It means the series show the AP series.
we know about the formula of the sum of n terms in Arithmetic progression that is ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
Now, calculate the value of ${S_n}$. Substitute the values $n = 15,a = 4$ and $d = - 5\left( { - 1 - 4} \right)$ in the expression ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.
$\Rightarrow {S_{15}} = \dfrac{{15}}{2}\left[ {2\left( 4 \right) + \left( {15 - 1} \right)\left( { - 5} \right)} \right]$
$= 7.5\left[ {8 - 70} \right]$
$= 7.5\left( { - 62} \right)$
$= - 465$

Hence, the sum of the first 15 term is ${S_{15}} = - 465$.

Note:
The general formula of the Arithmetic progression is $a, a + d, a + 2d, a + 3d,...$, where a is the initial term of the AP and d is the common difference of successive numbers. The definition of the arithmetic progression (A.P.) is the sequence of numbers with a common difference between any two consecutive numbers. For example: $1,2,3,4,...$ and $1,3,5,7,...$ both are arithmetic progression because of the difference difference any two consecutive numbers same.