
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 \[10\dfrac{1}{2}\] , find the number of terms of the series.
Answer
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Hint: Here we will first assume the total number of terms as 2n, first term as a and common difference as d. Then we will form the linear equations in terms of a and d and n and then finally solve them to get the value of n.
The nth term of an AP is given by:-
\[{T_n} = a + \left( {n - 1} \right)d\]
Also, the sum of n terms of an AP is given by:-
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Complete step-by-step answer:
Let the total no. of terms be 2n, & first term, a and common difference be d.
The last term of an AP having 2n terms is given by:-
\[{T_{2n}} = a + \left( {2n - 1} \right)d\]
Now it is given that, the last term exceeds the first by \[10\dfrac{1}{2}\]
Hence,
\[{T_{2n}} - a = 10\dfrac{1}{2}\]
Putting in the value we get:-
\[a + \left( {2n - 1} \right)d - a = 10\dfrac{1}{2}\]
Simplifying it we get:-
\[\left( {2n - 1} \right)d = \dfrac{{21}}{2}\]……………………… (1)
Now it is given that, the sum of the odd terms is 24, of the even terms 30
If we take odd terms only it will also be a A.P. of n terms and if we take even terms only it will also be a A.P of n terms with common diff. of 2d.
Therefore, sum of odd terms is given by:-
\[{S_{odd}} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)2d} \right]\]
Putting in the value we get:-
\[24 = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)2d} \right]\]
Simplifying it further we get:-
\[24 = n\left[ {a + \left( {n - 1} \right)d} \right]\]…………………………. (2)
Similarly, the sum of even terms is given by:-
\[{S_{even}} = \dfrac{n}{2}\{ 2(a + d) + (n - 1)2d\} \]
Putting in the value we get:-
\[30 = \dfrac{n}{2}\{ 2(a + d) + (n - 1)2d\} \]
Simplifying it further we get:-
\[30 = n\left[ {a + nd} \right]\]………………………………….. (3)
Putting the value from equation 3 in equation 2 we get:
\[24 = 30 - {n^2}d + {n^2}d - nd\]
Solving it further we get:-
\[nd = 30 - 24\]
\[ \Rightarrow nd = 6\]………………………… (4)
Putting this value in equation 1 we get:-
\[2\left( 6 \right) - d = \dfrac{{21}}{2}\]
Solving for d we get:-
\[d = 12 - \dfrac{{21}}{2}\]
Simplifying it we get:-
\[d = \dfrac{3}{2}\]
Now putting this value in equation 4 we get:-
\[n\left( {\dfrac{3}{2}} \right) = 6\]
Solving for n we get:-
\[n = \dfrac{{6 \times 2}}{3}\]
\[ \Rightarrow n = 4\]
Now since total number of terms are 2n therefore,
Total number of terms \[ = 2\left( 4 \right)\] \[ = 8\]
Therefore, there are 8 terms in the AP.
Note: Students might forget to multiply the value of n with 2 at the last which may lead to wrong answers.
Also, in such questions we have to form various equations according to the given information in the question and then solve for the unknown variables.
The alternate formula of sum of n terms in AP is given by:-
\[{S_n} = \dfrac{n}{2}\left[ {a + l} \right]\] where a is the first term, l is the last term and n is the total number of terms.
The nth term of an AP is given by:-
\[{T_n} = a + \left( {n - 1} \right)d\]
Also, the sum of n terms of an AP is given by:-
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
Complete step-by-step answer:
Let the total no. of terms be 2n, & first term, a and common difference be d.
The last term of an AP having 2n terms is given by:-
\[{T_{2n}} = a + \left( {2n - 1} \right)d\]
Now it is given that, the last term exceeds the first by \[10\dfrac{1}{2}\]
Hence,
\[{T_{2n}} - a = 10\dfrac{1}{2}\]
Putting in the value we get:-
\[a + \left( {2n - 1} \right)d - a = 10\dfrac{1}{2}\]
Simplifying it we get:-
\[\left( {2n - 1} \right)d = \dfrac{{21}}{2}\]……………………… (1)
Now it is given that, the sum of the odd terms is 24, of the even terms 30
If we take odd terms only it will also be a A.P. of n terms and if we take even terms only it will also be a A.P of n terms with common diff. of 2d.
Therefore, sum of odd terms is given by:-
\[{S_{odd}} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)2d} \right]\]
Putting in the value we get:-
\[24 = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)2d} \right]\]
Simplifying it further we get:-
\[24 = n\left[ {a + \left( {n - 1} \right)d} \right]\]…………………………. (2)
Similarly, the sum of even terms is given by:-
\[{S_{even}} = \dfrac{n}{2}\{ 2(a + d) + (n - 1)2d\} \]
Putting in the value we get:-
\[30 = \dfrac{n}{2}\{ 2(a + d) + (n - 1)2d\} \]
Simplifying it further we get:-
\[30 = n\left[ {a + nd} \right]\]………………………………….. (3)
Putting the value from equation 3 in equation 2 we get:
\[24 = 30 - {n^2}d + {n^2}d - nd\]
Solving it further we get:-
\[nd = 30 - 24\]
\[ \Rightarrow nd = 6\]………………………… (4)
Putting this value in equation 1 we get:-
\[2\left( 6 \right) - d = \dfrac{{21}}{2}\]
Solving for d we get:-
\[d = 12 - \dfrac{{21}}{2}\]
Simplifying it we get:-
\[d = \dfrac{3}{2}\]
Now putting this value in equation 4 we get:-
\[n\left( {\dfrac{3}{2}} \right) = 6\]
Solving for n we get:-
\[n = \dfrac{{6 \times 2}}{3}\]
\[ \Rightarrow n = 4\]
Now since total number of terms are 2n therefore,
Total number of terms \[ = 2\left( 4 \right)\] \[ = 8\]
Therefore, there are 8 terms in the AP.
Note: Students might forget to multiply the value of n with 2 at the last which may lead to wrong answers.
Also, in such questions we have to form various equations according to the given information in the question and then solve for the unknown variables.
The alternate formula of sum of n terms in AP is given by:-
\[{S_n} = \dfrac{n}{2}\left[ {a + l} \right]\] where a is the first term, l is the last term and n is the total number of terms.
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