Question

Find the 31st term of an Ap whose 11th term is 38 and the 16th term is 73.

Answer 1

Hint: An arithmetic progression can be given by a, (a+d), (a+2d), (a+3d), ……
a, (a+d), (a+2d), (a+3d),….. where a = first term, d = common difference.
a,b,c are said to be in AP if the common difference between any two consecutive number of the series is same ie \[b - a = c - b \Rightarrow 2b = a + c\]
Formula to consider for solving these questions
an=a+ (n− 1)d
Where d -> common difference
A -> first term
n-> term
an -> nth term

Complete step-by-step answer:
Given that, a11 = 38 and a16 = 73
If the first term = a and the common difference = d.
We know that, an = a + (n − 1) d, using this formula to find nth term of arithmetic progression,
\[\begin{array}{*{20}{l}}
  { \Rightarrow {a_{11}} = {\text{ }}a{\text{ }} + {\text{ }}\left( {11{\text{ }} - {\text{ }}1} \right){\text{ }}d} \\
  { \Rightarrow 38{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}10d{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ..\left( 1 \right)}
\end{array}\]
Similarly,
\[\begin{array}{*{20}{l}}
  { \Rightarrow {a_{16}} = {\text{ }}a{\text{ }} + {\text{ }}\left( {16{\text{ }} - {\text{ }}1} \right){\text{ }}d} \\
  { \Rightarrow 73{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}15d{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\left( 2 \right)}
\end{array}\]
These are equations consisting of two variables.
On subtracting (1) from (2), we obtain
\[\begin{array}{*{20}{l}}
  { \Rightarrow 35{\text{ }} = {\text{ }}5d} \\
  { \Rightarrow d{\text{ }} = {\text{ }}7}
\end{array}\]
Let us put the value of d in equation (1) to find the value of a.
\[\begin{array}{*{20}{l}}
   \Rightarrow 38{\text{ }} = {\text{ }}a{\text{ }} + {\text{ }}10{\text{ }}{\times}{\text{ 7}} \\
   \Rightarrow 38{\text{ }} - {\text{ }}70{\text{ }} = {\text{ }}a \\
  { \Rightarrow a{\text{ }} = {\text{ }} - 32}
\end{array}\]
Now we have a value of a and d. Let's find the term ${a_{31}}$:
\[\begin{array}{*{20}{l}}
  { \Rightarrow {a_{31}} = {\text{ }}a{\text{ }} + {\text{ }}\left( {31{\text{ }} - {\text{ }}1} \right){\text{ }}d} \\
  { \Rightarrow {a_{31}} = {\text{ }} - {\text{ }}32{\text{ }} + {\text{ }}30{\text{ }}\left( 7 \right)} \\
  { \Rightarrow {a_{31}} = {\text{ }} - {\text{ }}32{\text{ }} + {\text{ }}210} \\
  { \Rightarrow {a_{31}} = {\text{ }}178}
\end{array}\]

Therefore, the 31st term of AP is 178.

Note: To solve most of the problems related to AP, the terms can be conveniently taken as
3 terms: (a−d),a,(a+d)
4 terms: (a−3d),(a−d),(a+d),(a+3d)
5 terms: (a−2d),(a−d),a,(a+d),(a+2d)
${t_n}={S_n}-{S_{n-1}}$
If each term of an AP is increased, decreased, multiplied or divided by the same non-zero constant, the resulting sequence also will be in AP.
In an AP, the sum of terms equidistant from beginning and end will be constant.