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Question: Find the sum \[2,3\dfrac{1}{4},4\dfrac{1}{2},\]…..To \[20\]terms....

Find the sum 2,314,412,2,3\dfrac{1}{4},4\dfrac{1}{2},…..To 2020terms.

Explanation

Solution

Hint:-Here, this is a term of APAPso we apply summation of nnterms ofAPAP.

Given series is2,314,412,2,3\dfrac{1}{4},4\dfrac{1}{2},…..Up to 2020terms.
The given series can be written as84,134,184\dfrac{8}{4},\dfrac{{13}}{4},\dfrac{{18}}{4},……to 2020 terms.
Take 14\dfrac{1}{4} common we get
14(8+13+18+......)\Rightarrow \dfrac{1}{4}(8 + 13 + 18 + ......) To 2020 terms.
Observe that 8,13,18{\text{8,13,18}}…… is in Arithmetic Progression with first term 88and55as common difference
So, the sum of first2020 terms in that series will be
=14(202(2×8+(201)×5))= \dfrac{1}{4}\left( {\dfrac{{20}}{2}\left( {2 \times 8 + (20 - 1) \times 5} \right)} \right) Sn=n2(2a+(n1)d)\because {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)
Here n=n = number of terms, a=a = first term and d=d = common difference.
 = 14(1110)=5552{\text{ = }}\dfrac{1}{4}(1110) = \dfrac{{555}}{2} Answer.

Note: - Whenever such type of series is given first convert into simple form and then find which series is this. Then apply the formula of that series to get the answer.