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

Last updated date: 17th Apr 2024
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Hint:-Here, this is a term of $AP$so we apply summation of $n$terms of$AP$.

Given series is\[2,3\dfrac{1}{4},4\dfrac{1}{2},\]…..Up to \[20\]terms.
The given series can be written as$\dfrac{8}{4},\dfrac{{13}}{4},\dfrac{{18}}{4}$,……to \[20\] terms.
Take $\dfrac{1}{4}$ common we get
$ \Rightarrow \dfrac{1}{4}(8 + 13 + 18 + ......)$ To \[20\] terms.
Observe that ${\text{8,13,18}}$…… is in Arithmetic Progression with first term $8$and$5$as common difference
So, the sum of first\[20\] terms in that series will be
$ = \dfrac{1}{4}\left( {\dfrac{{20}}{2}\left( {2 \times 8 + (20 - 1) \times 5} \right)} \right)$ $\because {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$
Here $n = $ number of terms, $a = $ first term and $d = $ common difference.
${\text{ = }}\dfrac{1}{4}(1110) = \dfrac{{555}}{2}$ 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.