Question

The sum of first 55 terms is an A.P is 3300
Find its 28th term​

Answer 1

❍ Given, sum of first 55 terms of Arithmetic Progression (AP) is 3300.

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For any Arithmetic Progression (AP), the sum of n terms is Given by :

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$\bf{\dag}\quad\large\boxed{\sf S_n = \dfrac{n}{2}\bigg(2a + (n - 1)d \bigg)}$

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Where :

• n = No. of terms
• a = First Term
• d = Common difference

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$\underline{\bf{\dag} \:\mathfrak{Substituting \: values \: now \: :}}$

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$:\implies\sf S_{55} = \dfrac{55}{2} \bigg(2a + (n - 1) d\bigg) \\\\\\:\implies\sf S_{55} = \dfrac{55}{2} \Big(2a + 54d \Big) \\\\\\:\implies\sf 3300 = \dfrac{55}{\cancel{\:2}} \:\cancel{\:2} \Big(a + 27 \Big)\\\\\\:\implies\sf 3300 = 55 \Big(a + 27 d \Big)\\\\\\:\implies\sf \cancel\dfrac{3300}{55} = a + 27d\\\\\\:\implies\sf a + 27d = 60\qquad\quad\bigg\lgroup\bf Equation \:(I) \bigg\rgroup$

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◗ We've to find out the 28th term.

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$\bf{\dag}\quad\large\boxed{\sf t_n = \bigg(a + (n - 1)d \bigg)}$

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Therefore,

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$:\implies\sf t_{28} = a + (28 - 1)d \\\\\\:\implies\sf t_{28} = a + 27d\qquad\quad\bigg\lgroup\bf From \: eq. \: (I) \:\bigg\rgroup\\\\\\:\implies{\underline{\boxed{\frak{\purple{t_{28} = 60}}}}}\:\bigstar$

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$\therefore\:{\underline{\sf{Hence, \ 28th \: term \: of \: AP \ is \ \bf{60}.}}}$⠀⠀⠀

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• The sum of first 55 terms is an A.P is 3300. Find its 28th term.

$\huge {AηsωeR} ✍$

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${ \boxed {\bf{Given : - }}}$

• The sum of first 55 terms is an A.P is 3300.

${ \boxed {\bf{To \: Find : - }}}$

• Find its 28th term

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$\large\underline\blue{\bold{Formula \: used:- }}$

Let us consider an AP series whose first term is a and common difference is d, then

Sum of first 'n' terms is given by

$\bf \:S_n = \dfrac{n}{2} (2a + (n - 1)d)$

and nth term is given by

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

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$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{a \: be \: the \: first \: term \: of \: an \: AP} \\ &\sf{d \: be \: the \: common \: difference \: of \: an \: AP} \\ &\sf{n \: be \: the \: number \: of \: terms \: of \: AP} \end{cases}\end{gathered}\end{gathered}$

☆Since, sum of 55 terms = 3300

☆It means n = 55.

☆Using formula of sum of n terms,

$\bf \:S_n = \dfrac{n}{2} (2a + (n - 1)d)$

☆On substituting the values, we get

$\sf \: ⟼3300 = \dfrac{55}{2} (2a + (55 - 1) \times d)$

$\sf \:  ⟼3300 = \dfrac{55}{2} (2a + 54d)$

$\sf \:  ⟼3300 = \dfrac{55}{2} \times 2(a + 27d)$

$\sf \:  ⟼3300 = 55 \times (a + 27d)$

$\bf\implies \:a + 27d = 60$

$\bf\implies \:a + (28 - 1)d = 60$

$\bf\implies \:t_{28} = 60$

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${ \boxed {\bf{Hence, {28}^{th} \: term \:of \: an \: AP \: is \: 60}}}$