Question

The number of terms of an A.P. Is even; the sum of the odd terms is '$24$‘, of the even terms ’$30$‘, and the last term exceeds the first by ’$\frac{21}{2}$', to find the number of terms of the series.

Answer 1

Arithmetic Progression: An arithmetic progression or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant.

Step 1: Find the difference between the sum of odd terms and the sum of even terms

Here, the number of terms is even, So let the number of terms in A.P. be '$2\mathrm{n}$'.

Let the terms of the A.P. be ${\mathrm{T}}_{1,}{\mathrm{T}}_2,{\mathrm{T}}_3,{\mathrm{T}}_4,.................{\mathrm{T}}_{2\mathrm{n}}$

Let the first term of the A.P. be '$\mathrm{a}$‘, and the common difference be ’$\mathrm{d}$'.

It is given that the sum of odd terms is '$24$',

$\therefore {\mathrm{T}}_1+{\mathrm{T}}_3+{\mathrm{T}}_5+{\mathrm{T}}_7+...............+{\mathrm{T}}_{2\mathrm{n}-1}=24$ ………..($1$)

It is given that the sum of even terms is '$30$'.

$\therefore {\mathrm{T}}_2+{\mathrm{T}}_4+{\mathrm{T}}_6+{\mathrm{T}}_8+.................+{\mathrm{T}}_{2\mathrm{n}}=30$ …………..($2$)

Subtracting equation ($1$) from equation ($2$):

$\therefore \left({\mathrm{T}}_2-{\mathrm{T}}_1\right)+\left({\mathrm{T}}_4-{\mathrm{T}}_3\right)+\left({\mathrm{T}}_6-{\mathrm{T}}_5\right)+...........+\left({\mathrm{T}}_{2\mathrm{n}}-{\mathrm{T}}_{2\mathrm{n}-1}\right)=30-24$

$\mathrm{d}+\mathrm{d}+\mathrm{d}+........+\mathrm{d}=6$ ($\because$The difference between two consecutive terms in an A.P.$=\mathrm{d}$)

$\mathrm{nd}=6$………………..($3$)

Step 2: Use the given relation between the first and last term

It is given that the last term exceeds the first term by $\frac{21}{2}$

$\therefore {\mathrm{T}}_{2\mathrm{n}}-{\mathrm{T}}_1=\frac{21}{2}$

$\left[\mathrm{a}+\left(2\mathrm{n}-1\right)\mathrm{d}\right]-\mathrm{a}=\frac{21}{2}$ ($\because {\mathrm{a}}_{\mathrm{n}}=\mathrm{a}+\left(\mathrm{n}-1\right)\mathrm{d}$)

$2\mathrm{nd}-\mathrm{d}=\frac{21}{2}$………………………………($4$)

Step 3: Find the value of the common difference $\text{d}$

Putting the value of $\mathrm{nd}$ from equation ($3$) in equation ($4$):

$\begin{array}{rcl}\therefore 2\times 6-\mathrm{d}& =& \frac{21}{2}\\ \mathrm{d}& =& 12-\frac{21}{2}\end{array}$

$\mathrm{d}=\frac{3}{2}$……………………………………………….($5$)

Step 4: Find the value of $\mathrm{n}$

Putting the value of $\mathrm{d}$ from equation ($5$) in the equation ($3$):

$\begin{array}{rcl}\therefore \mathrm{n}\times \frac{3}{2}& =& 6\\ \mathrm{n}& =& 4\\ & & \end{array}$

$\begin{array}{rcl}\therefore 2\mathrm{n}& =& 2\times 4\\ & =& 8\end{array}$

Therefore, there are '$8$' terms in the given series.