Question

The first and last terms of an ap are 1 and 11. If the sum of its terms is 36,then it's the number of terms will be​

Answer 1

AnswEr :

$\frak{ Given}\begin{cases} & \sf{First\: term\;of\;AP,\;(a) = \bf{1}} \\ &\sf{Last\;term \;of \;AP,\;(l) = \bf{11}} \\ & \sf{Sum\:of\;terms\;of\;AP,\: (s_{n}) = \bf{36}} \end{cases}$

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¤ For any Arithmetic Progression ( AP ), the sum of nth terms having last term is Given by :

$\bigstar\;\underline{\boxed{\pmb{\sf{S_n = \dfrac{n}{2} \bigg\lgroup a + l \bigg\rgroup}}}}$

where :

• n = no. of terms
• a = First Term
• l = Last Term

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$\;\;\;\;\bf{\dag}\;{\underline{\frak{Substituting\;the\:given\;values\;in\;formula\;:}}}$

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$:\implies\sf S_n = \dfrac{n}{2}\bigg\lgroup a + l\bigg\rgroup = 36$

$:\implies\sf \dfrac{n}{2} \bigg\lgroup 1 + 11\bigg\rgroup = 36$

$:\implies\sf \dfrac{ \: n}{\cancel{\;2}} \times \: \cancel{12}= 36$

$:\implies\sf 6n = 36$

$:\implies\sf n = \cancel\dfrac{36}{6}$

$:\implies\sf\underline{\boxed{\pmb{\frak{\purple{n = 6}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Hence,\; number\:of\;terms\:of\;AP\;are\; {\textsf{\textbf{6}}}.}}}$

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$\qquad\boxed{\underline{\underline{\bigstar \: \bf\:Formula\:Related\:to\:AP\:\bigstar}}}$⠀⠀⠀⠀⠀⠀⠀

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$\sf (i)\;The\; n^{th}\;term\;of\;an\;AP\; = \; a_n + (n - 1)d$⠀⠀⠀⠀⠀⠀⠀

$\sf (ii)\;Sum\;of\;n\;term\;of\;an\;AP\; = \;S_n = \dfrac{n}{2} \bigg\lgroup\sf 2a + (n - 1)d \bigg\rgroup$⠀⠀

Answer 2

$\quad\odot\:\boxed{\sf{\blue{Number\:of\:terms\:(n)\:of\:AP\:=\:\bf{6}}}}$

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Explanation :

$\underline{\bf \dag{\underline{\red{Given}}}:-}$

• First term of AP, a = 1
• Last term of AP, l = 11
• Sum of terms of AP, Sn = 36

$\underline{\bf\dag{\underline{\red{To\:Find}}}:-}$

• Number of terms of AP, n = ?

$\underline{\bf\dag{\underline{\red{Solution}}}:-}$

• Here, we have first term (a) , last term (l) and sum of all terms of AP (Sn). So, to find out number of terms (n) of AP, Using formula :-

$\qquad\qquad\bf\dag\:\bigg\lgroup{\purple{S_{n} = \dfrac{n}{2} (a + l) }}\bigg\rgroup$

★ Putting all known values :-

$\qquad\leadsto\quad\tt 36 = \dfrac{n}{2} (1 + 11)$

$\qquad\leadsto\quad\tt 36 = \dfrac{n}{2} (12)$

$\qquad\leadsto\quad\tt 36 = \dfrac{n}{2}\:\times\:12$

$\qquad\leadsto\quad\tt 36 = \dfrac{n}{\cancel{2}}\:\times\:\cancel{12}$

$\qquad\leadsto\quad\tt 36 = n\:\times\:6$

$\qquad\leadsto\quad\tt \dfrac{36}{6} = n$

$\qquad\leadsto\quad\tt \dfrac{\cancel{36}}{\cancel{6}} = n$

$\qquad\leadsto\quad\bf{n = \pink{6}}$

Hence,

• Number of terms (n) of AP = 6

$\underline{\bf \dag{\underline{\red{More\:to\:know}}}:-}$

• An arithmetic progression (AP) is a list pf numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
• This fixed number is called the common difference of the AP. Remember that it can be positive, negative or zero.

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