Question

The sum of the first six terms of an arithmetic progression is 42. The ratio of its 10th and 30th term is 1:3. Calculate the first and thirteenth term of an A.P.

Answer 1

Hint: Applying the formula of sum of n terms in A.P. to find the sum of first six terms of an A.P. We know that when the first term is “a” and common difference is “d” then the sum of first n terms of an A.P. is ${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$. Now, substituting the value of n as 6 and then equating this summation expression to 42 we get $\dfrac{6}{2}\left( 2a+\left( 6-1 \right)d \right)=42$.Write the 10th term and 30th term of an A.P. using the formula for nth term of A.P. which is equal to ${{T}_{n}}=a+\left( n-1 \right)d$ and then take the ratio of 10th and 30th term and equate it to 1:3. Now, solve the summation equation and the ratio equation to get the value of “a” and “d” and then using the value of “a” and “d” we can find the first and the thirteenth term of A.P.

We know that the formula for summation of first n terms of an A.P. having the first term as “a” and the common difference as “d”.
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
Using this formula, we are going to write the sum of first six terms of an A.P. by plugging the value of n equal to 6.
${{S}_{6}}=\dfrac{6}{2}\left( 2a+\left( 6-1 \right)d \right)$
It is given that the sum of first six terms of an A.P. is equal to 42 so equating the above expression to 42.
$\begin{align}
  & \dfrac{6}{2}\left( 2a+\left( 6-1 \right)d \right)=42 \\
 & \Rightarrow 3\left( 2a+5d \right)=42 \\
\end{align}$
Dividing 3 on both the sides of the above equation we get,
$2a+5d=14$……. Eq. (1)
Now, to write the 10th and the 30th term of an A.P. we are going to use the formula for the general term of an A.P.
The general term for an A.P. having first term as “a” and the common difference as “d”.
${{T}_{n}}=a+\left( n-1 \right)d$
Writing 10th term and 30th term of an A.P. we get,
$\begin{align}
  & {{T}_{10}}=a+\left( 10-1 \right)d=a+9d \\
 & {{T}_{30}}=a+\left( 30-1 \right)d=a+29d \\
\end{align}$
It is given in the question that the ratio of 10th and 30th term is 1:3 so dividing the above 10th and 30th term of an A.P. and then equates it to $\dfrac{1}{3}$.
$\dfrac{{{T}_{10}}}{{{T}_{30}}}=\dfrac{a+9d}{a+29d}=\dfrac{1}{3}$
On cross-multiplying the above equation we get,
$\begin{align}
  & 3\left( a+9d \right)=a+29d \\
 & \Rightarrow 3a+27d=a+29d \\
 & \Rightarrow 2a=2d \\
 & \Rightarrow a=d \\
\end{align}$
Substituting the above value of “a” in eq. (1) we will get the value of “d”.
$\begin{align}
  & 2a+5d=14 \\
 & \Rightarrow 2\left( d \right)+5d=14 \\
 & \Rightarrow 7d=14 \\
 & \Rightarrow d=2 \\
\end{align}$
Now, we have shown above that $a=d$ so the value of “a” is also equal to 2.
We are asked in the question to find the first and the thirteenth term of an A.P.
First term of an A.P. is “a” which is equal to 2.
Thirteenth term of an A.P. we are going to calculate by using the general term of an A.P.
$\begin{align}
  & {{T}_{n}}=a+\left( n-1 \right)d \\
 & \Rightarrow {{T}_{13}}=2+\left( 13-1 \right)2 \\
\end{align}$
$\begin{align}
  & \Rightarrow {{T}_{13}}=2+26-2 \\
 & \Rightarrow {{T}_{13}}=26 \\
\end{align}$
From the above solution, we have got the values of first and the thirteenth term as:
First term is equal to 2 and the thirteenth term is equal to 26.

Note: The question is pretty simple but the mistake that could happen in writing the formula for summation of n terms in A.P. and the calculation mistake could happen.
In the formula of summation of first n terms of an A.P.
${{S}_{n}}=\dfrac{n}{2}\left( 2a+\left( n-1 \right)d \right)$
You might write n in place of $\dfrac{n}{2}$ mistakenly and in this formula you might write “a” in place of “2a” because the general term of an A.P. contains only “a” is $a+\left( n-1 \right)d$ so might mingle the general term formula with the summation one. So, be careful while writing the formulae and try to avoid calculation mistakes.