Question

Determine A.P. whose 4th term is 18 and the difference of the ninth term from \[{a_{15}}\] term is 30.

Answer 1

Hint: Here we will use the formula of the nth term of an A.P. to find the 4th , 9th, and 15th term of the AP and then solve the equations so formed to find the values of a and d and then finally find the AP.
The nth term of an AP is given by:-
\[{T_n} = a + \left( {n - 1} \right)d\] where a is the first term , d is the common difference and n is the total number of terms.
The general AP is given by:-
\[a,a + d,a + 2d,a + 3d..............\]

Complete step-by-step answer:
We know that
nth term of an AP is given by:-
\[{T_n} = a + \left( {n - 1} \right)d\]
Where a is the first term, d is the common difference and n is the total number of terms.
Applying this formula for finding 4th term we get:-
Putting \[n = 4\] in above formula we get:-
\[{T_4} = a + \left( {4 - 1} \right)d\]
\[{T_4} = a + 3d\]
Now it is given that 4th term is 18
Hence,
\[18 = a + 3d\]……………………………(1)
Now it is given that the difference of the ninth term from \[{a_{15}}\] term is 30.
Hence,
\[{T_{15}} - {T_9} = 30\]
Applying the formula of nth term we get:-
\[a + \left( {15 - 1} \right)d - \left[ {a + \left( {9 - 1} \right)d} \right] = 30\]
Simplifying it we get:-
\[
  a + 14d - \left[ {a + 8d} \right] = 30 \\
   \Rightarrow a + 14d - a - 8d = 30 \\
 \]
Cancelling the terms we get:
\[6d = 30\]
Solving for d we get:-
\[
   \Rightarrow d = \dfrac{{30}}{6} \\
   \Rightarrow d = 5 \\
 \]
Now putting this value in (1) we get:-
\[18 = a + 3\left( 5 \right)\]
\[ \Rightarrow 18 = a + 15\]
Solving for a we get:-
\[
  a = 18 - 15 \\
   \Rightarrow a = 3 \\
 \]
Hence the first term of the AP is 3 and common difference is 5
Now we know that general AP is given by:-
\[a,a + d,a + 2d,a + 3d..............\]
Hence putting the values we get:-
\[3,3 + 5,3 + 2\left( 5 \right),3 + 3\left( 5 \right)..............\]
Solving it further we get:-
\[3,8,13,18,....................\]
Hence the required AP is
\[3,8,13,18,....................\]

Note: Students should take a note that an AP can be increasing as well as decreasing.
In an increasing AP the common difference is always positive while in decreasing AP the common difference is always negative.