Question

In an A.P. ${{t}_{10}}=43$ and $d=5$. Complete the following activity to find the first term of the A.P.
${{t}_{1}}=a+\left( n-1 \right)d$
$\therefore {{t}_{10}}=a+\left( .......-1 \right)......$
$\therefore \,..........=a+45$
$\therefore \,\,a=\,...............$

Answer 1

For this type of problem, observe the given activity, first try to compare the given equation with an unknown equation, then put the value of $n$. If we got then put in the blank and calculate it for the value $a$.

Complete step-by-step answer:
For this we should know ${{t}_{1}},{{t}_{2}}...........{{t}_{10}}$.
These are terms of the given A.P. ${{t}_{1}}$ means first and ${{t}_{10}}$ means tenth term.
In this question we have to find $a$ and the first equation of the given activity is
${{t}_{1}}=a+\left( n-1 \right)d$
Where $a=$ first term
$n=$ number of term
$d=$common difference
Similarly, if we take ${{t}_{10}}$, then we get
${{t}_{10}}=a+\left( n-1 \right)d$
Here $n=10$, then
${{t}_{10}}=a+\left( 10-1 \right)5$
If we subtract the bracket values and multiply with $5$, then we get
${{t}_{10}}=a+9\times 5$
${{t}_{10}}$=$43$ is given in the question, on putting in the above equation,
$43=a+45$.
On transferring number one side, we get
$a=43-45=-2$
Therefore it is the required value of $a$.

Note: A.P.- Arithmetic progressions-An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term. This fixed number is known as the common difference of the A.P.
General term of an AP: General term or nth term of an AP is written as
${{a}_{n}}=a+\left( n-1 \right)d$
In such types of problems students get confused between the general term and sum of first. General term is for the individual term and sum of n terms are for the sum of some numbers.
It's denoted by ${{S}_{n}}$.
${{S}_{n}}=\dfrac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$