Question

The ninth term of an $AP$ is equal to seven times the second term and the twelfth term exceeds five times the third term by $2$. Find the first term and the common difference.

Answer 1

Hint: The given problem requires us to find the first term and common difference of an arithmetic progression given some conditions and relations between the different terms of an AP. So, we represent the terms of arithmetic progression in terms of the first term and the common difference between the terms using the formula for the general term of AP: ${a_n} = a + \left( {n - 1} \right)d$.

Complete step by step answer:
Now, we know that the general term of an arithmetic progression is given by ${a_n} = a + \left( {n - 1} \right)d$, where a is the first term of AP and d is the common difference of AP.So, we are given the ninth term is equal to seven times the second term of AP. We calculate the ninth term and second term of arithmetic progression.
So, ${a_2} = a + \left( {2 - 1} \right)d = a + d$
${a_9} = a + \left( {9 - 1} \right)d = a + 8d$
Now, following the condition given to us, we have,
${a_9} = 7{a_2}$
$ \Rightarrow \left( {a + 8d} \right) = 7\left( {a + d} \right)$
Simplifying the equation, we get,
$ \Rightarrow a + 8d = 7a + 7d$

Shifting all the terms consisting d to left side of the equation and terms consisting a to right side of equation, we get,
$ \Rightarrow 8d - 7d = 7a - a$
$ \Rightarrow d = 6a - - - - \left( 1 \right)$
Now, we are also given that the twelfth term exceeds five times the third term by $2$.
So, we find the twelfth and third term of AP.
So, ${a_{12}} = a + \left( {12 - 1} \right)d = a + 11d$
${a_3} = a + \left( {3 - 1} \right)d = a + 2d$
Now, following the condition given to us, we have,
${a_{12}} = 5{a_3} + 2$
$ \Rightarrow \left( {a + 11d} \right) = 5\left( {a + 2d} \right) + 2$
$ \Rightarrow a + 11d = 5a + 10d + 2$

Finding the value of d in terms of a, we get,
\[ \Rightarrow 11d - 10d = 5a - a + 2\]
\[ \Rightarrow d = 4a + 2 - - - - \left( 2 \right)\]
Now we solve the equations $\left( 1 \right)$ and $\left( 2 \right)$ to find the values of $a$ and $d$. Substituting the value of d from equation $\left( 1 \right)$ into equation $\left( 2 \right)$, we get,
\[ \Rightarrow 6a = 4a + 2\]
Shifting all the terms consisting of a to left side of equation, we get,
\[ \Rightarrow 2a = 2\]
\[ \therefore a = 1\]
So, we get the value of a as $1$.
Now, putting value of a in equation $\left( 1 \right)$
$ \therefore d = 6\left( 1 \right) = 6$

Therefore, the first term of AP is $1$ and the common difference of AP is $6$.

Note: Arithmetic progression is a series where any two consecutive terms have the same difference between them. The common difference of an arithmetic series can be calculated by subtraction of any two consecutive terms of the series. The nth term of arithmetic progression is of the form ${a_n} = a + \left( {n - 1} \right)d$. This formula helps to find any term of an AP. We must be careful while doing calculations.