Question

An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.

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
${a_n}=a+ (n− 1)d$
Where d -> common difference
A -> first term
n-> term
an -> nth term

Complete step-by-step answer:
Given that, ${a_3} = 12$ and ${a_{50}} = 106$
If the first term = a and the common difference = d.
We know that, ${a_n} = a + (n − 1) d$, using this formula to find nth term of arithmetic progression,
\[\begin{array}{*{20}{l}}
  { \Rightarrow {a_3} = {\text{ }}a{\text{ }} + {\text{ }}\left( {{\text{3 }} - {\text{ }}1} \right){\text{ }}d} \\
  { \Rightarrow 12{\text{ }} = {\text{ }}a{\text{ }} + {\text{ 2}}d{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots ..\left( 1 \right)}
\end{array}\]
Similarly,
\[\begin{array}{*{20}{l}}
  { \Rightarrow {a_{50}} = {\text{ }}a{\text{ }} + {\text{ }}\left( {{\text{50 }} - {\text{ }}1} \right){\text{ }}d} \\
  { \Rightarrow 106{\text{ }} = {\text{ }}a{\text{ }} + {\text{ 49}}d{\text{ }} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots .\left( 2 \right)}
\end{array}\]
These are equations consisting of two variables.
On subtracting (1) from (2), we obtain
\[\begin{array}{*{20}{l}}
  { \Rightarrow 94{\text{ }} = {\text{ 47}}d} \\
  { \Rightarrow d{\text{ }} = {\text{ 2}}}
\end{array}\]
Let us put the value of d in equation (1) to find the value of a.
\[\begin{array}{*{20}{l}}
   \Rightarrow 12{\text{ }} = {\text{ }}a{\text{ }} + {\text{ 2 }}{\times}{\text{ 2}} \\
   \Rightarrow 12{\text{ }} - {\text{ 4 }} = {\text{ }}a \\
  { \Rightarrow a{\text{ }} = {\text{ 8}}}
\end{array}\]
Now we have a value of a and d. Let's find the term ${a_{31}}$:
\[\begin{array}{*{20}{l}}
  { \Rightarrow {a_{29}} = {\text{ }}a{\text{ }} + {\text{ }}\left( {{\text{29 }} - {\text{ }}1} \right){\text{ }}d} \\
  { \Rightarrow {a_{29}} = {\text{ 8 }} + {\text{ 28 }}\left( 2 \right)} \\
  { \Rightarrow {a_{29}} = {\text{ }}8{\text{ }} + {\text{ }}56} \\
  { \Rightarrow {a_{31}} = {\text{ 64}}}
\end{array}\]

Therefore, the 29th term of AP is 64.

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.