Question

If the $6th$ term of an A.P is equal to four times its first term and the sum of first six terms is $75$ , find the first term and common difference.

Answer 1

In this question we will first apply the general form of an A.P i.e. ${a_n} = a + (n - 1)d$. So we will first find out the sixth term or ${a_6}$. So we have the nth term as $6$ and the first term is $a$. We can also write this as the sum of the first term and the previous term i..e fifth term : $a + (6 - 1)d = a + 5d$. After this we will apply the sum formula of an arithmetic progression. We can find the sum of the ${n^{th}}$ term of AP by the formula: ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.

Complete step by step answer:
We know that Arithmetic progression or AP is a sequence of numbers in order in which the difference of any two consecutive numbers is a constant value.
The formula for finding the ${n^{th}}$ term of an AP is:
${a_n} = a + (n - 1)d$,
Where we know that $a$ is the first term, $d$ is the common difference, $n =$number of terms and ${a_n} = {n^{th}}term$.
According to the question we have been given that $6th$ term of an A.P is equal to four times its first term, it can be written as:
${a_6} = 4a$

Now we know the formula of ${a_n}$ is
${a_n} = a + (n - 1)d$,
Here we have $n = 6$, so by putting this in the left hand side of the equation we can write that,
$a + (6 - 1)d = 4a$
It gives us new expression i.e.
$a + 5d = 4a$
Now we can bring the similar terms together and then we have;
$5d = 4a - a$
It gives us the value of the first term i.e.
$3a = 5d \Rightarrow a = \dfrac{{5d}}{3}$ .
Now we will apply the sum formula , we have been given the sum of six terms i.e.
${S_n} = 75,n = 6,a = \dfrac{{5d}}{3}$
So by substituting the values in the formula we get:
$75 = \dfrac{6}{2}\left( {2 \times \dfrac{{5d}}{3} + (6 - 1)d} \right)$

On simplifying the values we have;
$75 = 3\left( {\dfrac{{10d}}{3} + 5d} \right)$
We can add the values inside the bracket by taking LCM in the right hand side of the equation:
$75 = 3\left( {\dfrac{{10d + 15d}}{3}} \right) \Rightarrow 75 = 3\left( {\dfrac{{25d}}{3}} \right)$
On further simplifying we have:
$75 = 25d \Rightarrow d = \dfrac{{75}}{{25}}$
It gives us the value of the common difference i.e.
$d = 3$
Now from the above solution, we have first term as
$a = \dfrac{{5d}}{3}$
So by putting the value of $d$ in the above, it gives:
$\therefore a = \dfrac{{5 \times 3}}{3} = 5$

Hence we have the value of first term and common difference, $a = 5,d = 3$.

Note: We should note that we get the value of the first term and common difference, so we can get the second term of the A.P by adding the common difference to the first term:
${a_2} = a + d$
So we can put the values, and we have:
${a_2} = 5 + 3 = 8$
Similarly we can get the third term by adding the common difference to the second term;
${a_3} = 8 + 3 = 11$
So we have now the sequence of an A.P i.e. $5,8,11...$