Question

The ${11^{{\text{th}}}}$ term and the ${21^{{\text{st}}}}$ term of an A.P. are $16$ and $29$ respectively, then find the ${41^{{\text{st}}}}$ term of that A.P.

Answer 1

Hint: Use the formula of ${{\text{n}}^{{\text{th}}}}$ term of an A.P. to solve equations to find the first term and common difference of that A.P. Again use formula of ${{\text{n}}^{{\text{th}}}}$ term of an A.P. to find the ${41^{{\text{st}}}}$ term. The formula of ${{\text{n}}^{{\text{th}}}}$ term is
${t_n} = a + (n - 1)d$
Where, $a = $ first term of an A.P.
$d = $ Common difference of an A.P

Complete step-by-step answer:
Here A.P. means arithmetic progression it is the sequence of the given form
$a,a + d,a + 2d,.......,a + (n - 1)d$
Hence we do write $a + (n - 1)d$ as ${{\text{n}}^{{\text{th}}}}$ term of this sequence of A.P.
In mathematical term ${t_n} = a + (n - 1)d$
Here we can define the following terms
$a = $ first term of an A.P.
$d = $ Common difference of an A.P
${t_n} = {{\text{n}}^{{\text{th}}}}$ term of an A.P.
Given ${11^{{\text{th}}}}$ term of an A.P. is $16$
We can write it as following way
${t_{11}} = 16$
Applying above formula of ${t_n},$ we get
$a + (11 - 1)d = 16$
Simplifying further
$a + 10d = 16\; \ldots \ldots \left( 1 \right)$
Also ${21^{{\text{st}}}}$ term of an A.P. is $29$
We can write it as following way
${t_{21}} = 29$
Applying above formula of ${t_n},$ we get
$a + (21 - 1)d = 29$
Simplifying further
  $a + 20d = 29\; \ldots \ldots \left( 2 \right)$
Solving equations $\left( 1 \right)$ and $\left( 2 \right)$ by elimination method to find the values of $a$ and $d$
In the elimination method we multiply by any constant to make the same variables. After that we do add or subtract so that one variable gets cancelled and we will leave with another from this variable we do put in any two equations to get the other variable .
Here variable $a$ is same in equations first and second so we just subtract the given equations
$a - a + 10d - 20d = 16 - 29$
Simplifying further
$
   \Rightarrow - 10d = - 13 \\
   \Rightarrow d = \dfrac{{ - 13}}{{ - 10}} \\
   \Rightarrow d = \dfrac{{13}}{{10}} \\
 $
Putting value of $d$ in equation $\left( 1 \right)$, we get
$a + 10.\dfrac{{13}}{{10}} = 16$
After more simplifying, we get
$
   \Rightarrow a + 13 = 16 \\
   \Rightarrow a = 16 - 13 \\
   \Rightarrow a = 3 \\
 $
Therefore values of first term and common difference are $\dfrac{{13}}{{10}}$ and $3$ respectively
Now find the ${41^{{\text{st}}}}$ term of an A.P.
$
   \Rightarrow {t_{41}} = a + (41 - 1)d \\
   \Rightarrow {t_{41}} = a + 40d \\
 $
Putting values of $a$ and $d$, we get
$
   \Rightarrow {t_{41}} = 3 + 40 \times \left( {\dfrac{{13}}{{10}}} \right) \\
   \Rightarrow {t_{41}} = 3 + 4 \times 13 \\
   \Rightarrow {t_{41}} = 3 + 52 \\
   \Rightarrow {t_{41}} = 55 \\
 $

Hence ${41^{{\text{st}}}}$ term of the given A.P. is $55$

Note: The sum of $n$ terms of an A.P. is given by the formula:
$ \Rightarrow {S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$, where $n$ is the number of terms, $a$ is the first term and $d$ is the common difference of the A.P.
If the first term is 1 and the common difference is also 1, then the series boils down to the sum of first $n$ natural numbers. The formula in this case is resolved in to:
$ \Rightarrow S = \dfrac{{n\left( {n + 1} \right)}}{2}$