Question

The third term of an AP is p and the fourth term of the AP is q. find the tenth term of an AP and the general term?

Answer 1

In the given question, we have been asked the general term and the tenth term of an AP. In order to find, we first need to find the common difference of an AP by subtracting the fourth term and the third term as they are given in the question. Then we find the first term of an AP and after that put the value of ‘a’ and ‘d’, we get ${{n}^{th}}term$ and the tenth term of an AP.

Formula used:
Any ${{n}^{th}}$term of an AP i.e. arithmetic progression is,
${{a}_{n}}=a+\left( n-1 \right)d$, where
$a=$ First term of an AP
$n=\ {{n}^{th}}$Term of an AP
d = common difference between two consecutive term of an AP

Complete step-by-step solution:
We have given the first term of the AP and the fourth term of an AP i.e.
First term = ${{a}_{1}}=p$
Fourth term = ${{a}_{4}}=q$
Any ${{n}^{th}}$term of an AP i.e. arithmetic progression is,
${{a}_{n}}=a+\left( n-1 \right)d$, where
$a=$ First term of an AP
$n=\ {{n}^{th}}$term of an AP
d = common difference between two consecutive term of an AP
Thus,
We have given the third and the fourth term of an AP, so
Difference between third term and the fourth term = ${{a}_{4}}-{{a}_{3}}=q-p$
Therefore,
Common difference= $q-p$
Now,
Third term= ${{a}_{3}}=a+\left( n-1 \right)d$, where
n = number of term i.e. equal to 3 here.
$d=q-p$
$\therefore {{a}_{3}}=a+\left( 3-1 \right)\left( q-p \right)$
$\Rightarrow p=a+\left( 3-1 \right)\left( q-p \right)$
Simplifying the above equation, we get
$\Rightarrow p=a+2\left( q-p \right)$
$\Rightarrow p=a+2q-2p$

Solving this equation for the value of $a$i.e. the first term
$\Rightarrow -a=2q-2p-p$
Simplifying it further, we get
$\Rightarrow -a=2q-3p$
$\Rightarrow a=3p-2q$
Now,
$Tenth\ term={{a}_{10}}=a+\left( n-1 \right)d=a+\left( 10-1 \right)d=a+9d$
Putting the value of $a\ and\ d$, we get
$\Rightarrow a+9d=\left( 3p-2q \right)+9\times \left( q-p \right)=3p-2q+9q-9p=-6p+7q=7q-6p$
$\therefore {{a}_{10}}=7q-6p$
$General\ term={{n}^{th}}term=a+\left( n-1 \right)d$
Putting the value of $a\ and\ d$, we get
$\Rightarrow a+\left( n-1 \right)d=\left( 3p-2q \right)+\left( n-1 \right)\left( q-p \right)$
$\therefore {{n}^{th}}term=\left( 3p-2q \right)+\left( n-1 \right)\left( q-p \right)$

Note: When the numbers have been arranged in a particular manner or order, then the numbers are said to be in sequence. AP i.e. arithmetic progression is a sequence when we add a fixed number to a number to get the next consecutive number. For example if the fixed number is ‘d’ and ‘a’ is the term in the sequence, to get the number next to ‘a’, we will simply add ‘d’ to ‘a’ i.e. a+d is the next term to ‘a’.