Question

The sum of the fourth and fifth term of an AP is 24 and the sum of the sixth and the tenth term of an AP is 44. Find the first three terms of an AP.

Answer 1

Hint: Assume that the first term of the AP is \[a\] and the common difference of the AP is \[d\] . We know the formula for the \[{{n}^{th}}\] term of the AP, \[{{n}^{th}}term=first\,term+common\,difference\left( n-1 \right)\] . Now, use this formula and calculate the summation of the fourth and fifth term of the AP. It is given that the sum of the fourth and fifth term of an AP is 24. Now, form an equation. Again use the formula and calculate the summation of the sixth and the tenth term of the AP. It is also given that the sum of the sixth and the tenth term of the AP is 44. Now, form another equation. Here, we have two equations and we have two unknown values that are \[a\] and \[d\] . Now, solve it further and get the value of \[a\] and \[d\] . Now, using the formula for the \[{{n}^{th}}\] term of the AP and the value of \[a\] and \[d\] , get the first, second, and the third term of the AP.

Complete step by step answer:
According to the question, it is given that,
The sum of the fourth and fifth term of the AP = 24 ………………………………………………..(1)
The sum of the sixth and tenth term of the AP = 44 ………………………………………………..(2)
First of all, let us assume that the first term of the AP is \[a\] and the common difference of the AP is \[d\] .
The first term of the AP = \[a\] …………………………………………………(3)
The common difference of the AP = \[d\] ………………………………………………(4)
We know the formula for the \[{{n}^{th}}\] term of the AP, \[{{n}^{th}}term=first\,term+common\,difference\left( n-1 \right)\] ………………………………………………………..(5)
From equation (3) and equation (4), we have the first term and common difference of the AP.
Now, using the formula shown in equation (5), we get
Putting \[n=4\] , the fourth term of the AP = \[a+\left( 4-1 \right)d=a+3d\] ……………………………………………….(6)
Putting \[n=5\] , the fifth term of the AP = \[a+\left( 5-1 \right)d=a+4d\] ……………………………………………….(7)
Putting \[n=6\] , the sixth term of the AP = \[a+\left( 6-1 \right)d=a+5d\] ……………………………………………….(8)
Putting \[n=10\] , the tenth term of the AP = \[a+\left( 10-1 \right)d=a+9d\] ……………………………………………….(9)
From equation (1), equation (6), and equation (7), we have
The sum of the fourth and fifth term of the AP = 24
\[\Rightarrow a+3d+a+4d=24\]
\[\Rightarrow 2a+7d=24\] …………………………………………(10)
Similarly, from equation (2), equation (8), and equation (9), we get
The sum of the sixth and tenth term of the AP = 44
\[\Rightarrow a+5d+a+9d=44\]
\[\Rightarrow 2a+14d=44\] ………………………………………(11)
On subtracting equation (10) from equation (11), we get
\[\begin{align}
  & \Rightarrow \left( 2a+14d \right)-\left( 2a+7d \right)=44-24 \\
 & \Rightarrow 7d=20 \\
\end{align}\]
\[\Rightarrow d=\dfrac{20}{7}\] ……………………………………..(12)
Now, putting \[d=\dfrac{20}{7}\] in equation (10), we get
\[\begin{align}
  & \Rightarrow 2a+7\times \dfrac{20}{7}=24 \\
 & \Rightarrow 2a+20=24 \\
 & \Rightarrow 2a=24-20 \\
 & \Rightarrow 2a=4 \\
\end{align}\]
\[\Rightarrow a=2\] ………………………………………….(13)
Now, putting \[a=2\] in equation (3), we get
The first term of the AP = 2 …………………………………………(14)
Putting \[d=\dfrac{20}{7}\] in equation (4), we get
The common difference of the AP = \[\dfrac{20}{7}\] ……………………………………..(15)
We have to find the first three terms of the AP that are first term, second term, and third term of the AP.
On putting \[n=2\] in equation (5), we get
\[{{2}^{nd}}term=first\,term+common\,difference\left( 2-1 \right)\] ………………………………………….(16)
Now, from equation (14), equation (15), and equation (16), we get
\[{{2}^{nd}}term=2+\dfrac{20}{7}=\dfrac{34}{7}\] ………………………………………….(17)
On putting \[n=3\] in equation (5), we get
\[{{3}^{rd}}term=first\,term+common\,difference\left( 3-1 \right)\] ………………………………………(18)
Now, from equation (14), equation (15), and equation (18), we have
\[{{3}^{rd}}term=2+\dfrac{20}{7}\times 2=2+\dfrac{40}{7}=\dfrac{54}{7}\] …………………………………………..(19)
From equation (14), equation (17), and equation (19), we have
The first, second, and the third term of the AP are 2, \[\dfrac{34}{7}\] , and \[\dfrac{54}{7}\] .
Therefore, the first three terms of the AP are 2, \[\dfrac{34}{7}\] , and \[\dfrac{54}{7}\] .

Note:
In this question, one might make a silly mistake in the formula for the \[{{n}^{th}}\] term of an Arithmetic Progression. Here, one might use the formula, \[{{n}^{th}}\,term=first\,term{{\left( common\,ratio \right)}^{n-1}}\] for the \[{{n}^{th}}\] term of an Arithmetic Progression. This is wrong because the formula \[{{n}^{th}}\,term=first\,term{{\left( common\,ratio \right)}^{n-1}}\] is for the \[{{n}^{th}}\] term of the Geometric progression.