 QUESTION

# In an arithmetic progression ${{T}_{7}}=18$ and ${{T}_{18}}=7$, obtain ${{T}_{25}}$.

Hint: In this question, we need to write the general form of the given terms and then find the first term and common difference of the series by using the two equations. Then again using the nth term of an A.P series we can find the 25th term.
${{T}_{n}}=a+\left( n-1 \right)d$

ARITHMETIC PROGRESSION: A sequence in which the difference of two consecutive terms is constant, is called Arithmetic Progression (AP).
nth TERM OF AN AP: If a is the first term, d is the common difference of an arithmetic progression.
Then the nth term of an arithmetic progression is given by:
${{T}_{n}}=a+\left( n-1 \right)d$
Where, a is the first term of the series and d is the common difference of the given series and n is number of the term respectively.
Now, in the given question that
${{T}_{7}}=18$
${{T}_{18}}=7$
Let us now use the nth term of an AP formula and write the given terms accordingly.
Let us assume that the first term of this series as ‘a’, common difference as ‘d’.
Now, we can write these terms as:
$\Rightarrow {{T}_{7}}=a+\left( 7-1 \right)d$
$\Rightarrow {{T}_{7}}=a+6d$
Now, on equating this to the respective value we get,
$\Rightarrow a+6d=18........\left( 1 \right)$
Let us now consider the other term and write its general form.
\begin{align} & \Rightarrow {{T}_{18}}=a+\left( 18-1 \right)d \\ & \Rightarrow {{T}_{18}}=a+17d \\ \end{align}
Now, on equating this to the respective value we get,
$\Rightarrow a+17d=7........\left( 2 \right)$
Let us now consider the equations (1) and (2) we get,
\begin{align} & a+6d=18 \\ & a+17d=7 \\ \end{align}
Now, on subtracting these two equations we get,
$\Rightarrow -11d=11$
Now, on multiplying both sides with -1 and dividing by 11 we get,
$\therefore d=-1$
Now, by substituting this value of d in any of the above two equations we get the value of a.
\begin{align} & \Rightarrow a+6\left( -1 \right)=18 \\ & \Rightarrow a-6=18 \\ \end{align}
Now, by adding 6 on both sides we get,
$\therefore a=24$
Now, by using these values of a and d we can get the 25th term accordingly,
$\Rightarrow {{T}_{25}}=a+\left( 25-1 \right)d$
Now, on substituting the respective values we get,
\begin{align} & \Rightarrow {{T}_{25}}=24+\left( 24\times -1 \right) \\ & \Rightarrow {{T}_{25}}=24-24 \\ & \therefore {{T}_{25}}=0 \\ \end{align}

Note:
Instead of using the formula for the nth term and then finding the values of a and d we can also use the difference between two consecutive sum of n terms that gives the nth term of that series. But, this method would be a little complicated and lengthy.
${{S}_{n}}-{{S}_{n-1}}={{T}_{n}}$
It is important to note that while calculating the values of a and d using the equations we need to be sure that appropriate equations are being used and while rearranging the terms we should not neglect any of them because it changes the result completely.