Question

$\Large \color{lime} \bigstar \: \pmb{\mathbb{ \color{navy}QU \color{blue}ES \purple{T I \pink{ \ON}}}} \: \bigstar$



Find the sum of the Arithmetic Series
4 + 9 + 14 + . . . . + 54
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Answer 1

▪Given :-

An Arthematic Series :

$\large\mathfrak{4+9+14+ \:  .\: . \:. \:  +54}$

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▪To Find :-

Sum of Given Series

▪Formulas

1》  for Sum :-

$\bf S_n= \dfrac{ n}{2}  \bigg \{a + a_n\bigg \}$

2》For nth term

$\bf a_n = a + (n - 1)d$

Where ,

• a = First term

• $\sf{a_n}$ = nth Term

• n = Number of terms

• d = Common difference

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▪Solution :-

Here

• a = 4
• d = 5
• $\sf{a_n}$ = 54

《Calculating Number of Terms》

Using Formula of nth term :-

$54 = 4  \: +  \: (n - 1)5 \\  \\ 5n - 5 = 54 - 4 \\  \\ 5n = 50 +5\\  \\ n =  \frac{ \cancel{55 {}} }{ \cancel{5}}  \\  \\ \Large \purple{ \implies  \underline {\boxed{{\bf n = 11} }}}$

《Calculating  Sum》

Using Formula For Sum :-

$\sf SUM =  \dfrac{11}{2}  \bigg \{ 4 + 54\bigg \} \\  \\  = \dfrac{11}{2}\times(58) \\  \\\Large \purple{ \implies  \underline {\boxed{{\bf SUM = 319} }}}$

$\Large \red{\mathfrak{  \text{W}hich \:   \: is  \:  \: the  \:  \: required} }\\ \huge \red{\mathfrak{ \text{ A}nswer.}}$

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Answer 2

Question :

Find the sum of the Arithmetic Series

4 + 9 + 14 + . . . . + 54

To find :

The sum in arithmetic series

Solution :

Given series is 4,9,14,..

Clearly, the given sequence is an AP with first term a=4 and the common difference d=5.

We know that

$\sf \small a_{n} =a+(n−1)d$

By putting the values we get

54 = 4 + (n - 1)5

54 = 4 + 5n - 5

54 - 4 + 5 = 5n

55 = 5n

n = 55/5

n = 11

For getting the sum we need to apply

$\sf \rarr \frac{n}{2} (a + a_{n})$

By inserting the value we have

$\sf \rarr \frac{11}{2} (4 + 54)$

$\rarr \frac{11}{2} \times 58$

$\sf \rarr \frac{11}{ \cancel{2}} \times \cancel{58}$

$\rarr11 \times 29$

$\rarr319$

Hence, The sum is 319