Question

The number of terms of an A.P. is even; the sum of odd terms is $\;24$ , one of the even terms is $\;30$ , and the last term exceeds the first by $10\dfrac{1}{2}$. Find the number of terms and the series.

Answer 1

Hint: As the number of terms in the series is even we will assume the total number of terms $\;2n$ . From the given information that the difference between the first and last term is $10\dfrac{1}{2}$ we will get a relation between the first and last term. Then we will use the formula of summation in the A.P. series and then we further get the total terms by solving the two equations of summation of even and odd terms.

Formula used: $n$th term of the series is ${T_n} = a + (n - 1)d$ where $a$ is the first term and $d$ is the common difference between two terms.
The sum of $n$ terms is ${S_n} = \dfrac{n}{2} \cdot \{ 2a + (n - 1)d\}$ where $a$ is the first term and $d$ is the common difference between the two terms.

Complete step-by-step solution:
Let us assume that the total number of terms in the series is $2\;n$ and the first term is $a$ and$d$ is the common difference between the two terms.
We know that $n$ th term of the series is ${T_n} = a + (n - 1)d$ .
If we assume the last term as $l$ we will get;
$l = a + (2n - 1)d$
We have given that the last term exceeds the first by $10\dfrac{1}{2}$ or $\;10.5$ .
$\therefore l - a = 10.5$
Then we get;
$10.5 = (2n - 1)d$ .. $\left( 1 \right)$
If we take the odd terms we get an A.P. series of $n$ terms with a common difference $2\;d$ .
The sum of odd terms is $\;24$ .
Using the formula The sum of $n$ terms is ${S_n} = \dfrac{n}{2} \cdot \{ 2a + (n - 1)d\}$ .
$24 = \dfrac{n}{2} \cdot \{ 2a + (n - 1)2d\}$
Simplifying we get;
$\Rightarrow 24 = n \cdot \{ a + (n - 1)d\}$ .. $\left( 2 \right)$
Similarly, if we take the even terms we can form an A.P. series of $n$ terms with a common difference $2\;d$ and the first term of the series is $a + d$ .
The sum of even terms is $\;30$ .
So using the formula of summation we will get;
$30 = \dfrac{n}{2} \cdot \{ 2(a + d) + (n - 1)2d\}$
Simplifying we get;
$\Rightarrow 30 = n \cdot \{ (a + d) + (n - 1)d\}$
After multiplication, addition, and subtraction we get;
$\Rightarrow 30 = n \cdot (a + nd)$ ….. $\left( 3 \right)$
Putting this value of equation $\left( 3 \right)$ in the equation $\left( 2 \right)$ we get;
$24 = 30 - nd$
Simplifying we get;
$\Rightarrow nd = 6$ …. $\left( 4 \right)$
Put this value in $\left( 1 \right)$ we will get;
$10.5 = 12 - d$
Simplifying we get;
$\Rightarrow d = 1.5$
From $\left( 3 \right)$ and $\left( 4 \right)$ we get;
$n = 4$
And $a = 1.5$ .

So the total number of terms is $8$ .
So the series will be like this $\;1.5$ , $3$ , ……, $\;12$.

Note: To find a series the number of terms of the series, the common difference, and the first term is necessary to find out. In the series, all terms are not necessary to write. Instead of that, we can write the first, second, and last term as it is shown.