Question

The sum of the first 4 terms of an A.P is 56. The sum of the last 4 terms is 112 then finds the number of terms if the first term is 11.

Answer 1

Hint: Here we will find the number of terms by using the concept of Arithmetic Progression. We will add the first four terms and equate it to 56. We will then substitute the value of the first term in the obtained equation to get a common difference. We will then add the last four terms and equate it to 112. We will again substitute the first term and common difference in the equation to find the value of the number of terms.

Formula Used:
We will use the following formulas:
The formula for \[{n^{th}}\] term is \[{a_n} = {a_1} + \left( {n - 1} \right)d\], where, \[{a_1}\] is the initial term, \[d\] is the common difference.
General formula of \[{n^{th}}\] term of AP is \[{a_n} = {a_m} + \left( {n - m} \right)d\], where \[{a_m}\] is the initial term, \[d\] is the common difference.

Complete step-by-step answer:
Let the first term of an AP be \[a\] and its common difference be \[d\].
So, the terms of the AP will be as:
\[a,a + d,a + 2d,a + 3d,............,a + \left( {n - 4} \right)d,a + \left( {n - 3} \right)d,a + \left( {n - 2} \right)d,d + \left( {n - 1} \right)d\]
Now, it is given that the sum of first four terms are 56 so,
 \[a + \left( {a + d} \right) + \left( {a + 2d} \right) + \left( {a + 3d} \right) = 56\]……..\[\left( 1 \right)\]
Next, it is given that the sum of last four term is 112 so,
\[a + \left( {n - 4} \right)d + a + \left( {n - 3} \right)d + a + \left( {n - 2} \right)d + a + \left( {n - 1} \right)d = 112\]……\[\left( 2 \right)\]
The first term is given as 11 so,
\[a = 11\]
Substituting value of \[a\] in equation \[\left( 1 \right)\], we get
\[11 + \left( {11 + d} \right) + \left( {11 + 2d} \right) + \left( {11 + 3d} \right) = 56\]
Adding like terms, we get
\[ \Rightarrow 44 + 6d = 56\]
Subtracting 44 on both the sides, we get
\[\begin{array}{l} \Rightarrow 6d = 56 - 44\\ \Rightarrow 6d = 12\end{array}\]
Dividing both side by 6, we get
\[ \Rightarrow d = \dfrac{{12}}{6} = 2\]
So, the common difference is 2.
Substituting the value of \[a\] and \[d\] in equation \[\left( 2 \right)\], we get
\[11 + \left( {n - 4} \right) \times 2 + 11 + \left( {n - 3} \right) \times 2 + 11 + \left( {n - 2} \right) \times 2 + 11 + \left( {n - 1} \right) \times 2 = 112\]
Taking 2 common, we get
\[ \Rightarrow 44 + 2\left( {n - 4 + n - 3 + n - 2 + n - 1} \right) = 112\]
Adding and subtracting the like terms inside the bracket, we get
\[ \Rightarrow 44 + 2\left( {4n - 10} \right) = 112\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 8n - 20 = 112 - 44\\ \Rightarrow 8n = 68 + 20\end{array}\]
Adding the terms, we get
\[ \Rightarrow 8n = 88\]
Dividing by 8 on both sides, we get
\[\begin{array}{l} \Rightarrow n = \dfrac{{88}}{8}\\ \Rightarrow n = 11\end{array}\]

Therefore, we get the number of terms in AP as 11.

Note: An Arithmetic Progression (AP) is a sequence or series in which the consecutive terms differ by the same common difference. If it is a finite AP it is also known as Arithmetic Series. If each of the terms of an AP is multiplied or divided by the same number the resulting sequence will also be an AP. If the nth term of a sequence is a linear expression then that sequence is an AP.