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

Answer

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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.

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.

Last updated date: 27th Sep 2023

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