Solveeit Logo
Login

Question

Question: Find the sum: \(\dfrac{a+b}{2}, a, \dfrac{3a-b}{2},.......\) upto \(21\) terms....

Find the sum: a+b2,a,3a−b2,.......\dfrac{a+b}{2}, a, \dfrac{3a-b}{2},....... upto 2121 terms.

Explanation

Solution

Hint: First check whether the series is in A.P or not by finding the difference between them. After this use formula to find the sum of first 2121 terms. You will get the answer.

Complete step-by-step answer:
Arithmetic Progression (A.P) is a sequence of numbers in a particular order. If we observe in our regular lives, we come across progression quite often. For example, roll numbers of a class, days in a week, or months in a year. This pattern of series and sequences has been generalized in maths as progressions. Let us learn here AP definition, important terms such as common difference, the first term of the series, nth term and sum of nth term formulas along with solved questions based on them.
It is a mathematical sequence in which the difference between two consecutive terms is always a constant and it is abbreviated as A.P.
The fixed number that must be added to any term of an A.P to get the next term is known as the common difference of the A.P.
An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.
nth{{n}^{th}}term of A.P,
an=a+(n−1)d{{a}_{n}}=a+(n-1)d
Where,
a=a=First-term
d=d= Common difference
n=n= number of terms
an=nth{{a}_{n}}={{n}^{th}}term
Now let us check if the above equation is A.P or not.
So for that common difference should be the same.
So =a−(a+b)2=a−b2=a-\dfrac{(a+b)}{2}=\dfrac{a-b}{2}
and=3a−b2−a=a−b2=\dfrac{3a-b}{2}-a=\dfrac{a-b}{2}
Since the difference is common the series is A.P.
Now we know for the A.P sum ofnth{{n}^{th}}the term is,
S=n2(2a+(n−1)d)S=\dfrac{n}{2}\left( 2a+(n-1)d \right)
So we want to find the sum of the first 2121terms.
So we get,
a=a+b2,d=a−b2,n=21a=\dfrac{a+b}{2},d=\dfrac{a-b}{2},n=21.
So now substituting above values we get,

& {{S}_{21}}=\dfrac{21}{2}\left( 2\left( \dfrac{a+b}{2} \right)+(21-1)\left( \dfrac{a-b}{2} \right) \right) \\\ & {{S}_{21}}=\dfrac{21}{2}\left( \left( a+b \right)+(20)\left( \dfrac{a-b}{2} \right) \right) \\\ & {{S}_{21}}=\dfrac{21}{2}\left( \left( a+b \right)+(10)\left( a-b \right) \right) \\\ & {{S}_{21}}=\dfrac{21}{2}\left( \left( a+b \right)+10a-10b \right) \\\ & {{S}_{21}}=\dfrac{21}{2}\left( 11a-9b \right) \\\ \end{aligned}$$ So the sum of $\dfrac{a+b}{2},a,\dfrac{3a-b}{2},.......$ upto $21$terms is $$\dfrac{21}{2}\left( 11a-9b \right)$$. Note: Read the question carefully. If it is mentioned in the question that the series is in A.P then use the sum of first $n$ terms. If not then it is your work to check whether it is in A.P or not. Students make many mistakes in finding the sum of terms. Sometimes the terms are missing so take care of that. You should know the formula of the first $n$terms of A.P $S=\dfrac{n}{2}\left( 2a+(n-1)d \right)$.