Question

if the sum of an Arithemetic progression whose first term and the last term are 1 and 11 respectively is 36 then the number of its term is ______ A)5 B)6 C)7 D)8​

Answer 1

Answer:

the correct answer is:

B) 6

Answer 2

Answer:

• The correct option is B)6

Given:

• $\sf{First\: term\;of\;AP,\;(a) = \textsf{ \textbf{1}}}$
• $\sf{Last\;term \;of \;AP,\;(l) = \textsf{ \textbf{11}}}$
• $\sf{Sum\:of\;terms\;of\;AP,\: (s_{n}) ={ \textsf{ \textbf{36}}}}$

⠀⠀⠀

To find:

• The number of term

Solution:

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

$\begin{gathered}\begin{gathered} \blue\bigstar \underline{\boxed{{\sf{S_n = \dfrac{n}{2} \bigg\lgroup a + l \bigg\rgroup}}}}\\ \\\end{gathered}\end{gathered}$

Where,

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

$\begin{gathered}\begin{gathered}\;\;\;\;{\underline{\frak{ \color{navy}{Substituting\;the\:given\;values\;in\;formula\;:}}}}\\\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}:\implies\sf S_n = \dfrac{n}{2}\bigg\lgroup a + l\bigg\rgroup = 36\\ \\\end{gathered} \\ \\ \begin{gathered}:\implies\sf \dfrac{n}{2} \bigg\lgroup 1 + 11\bigg\rgroup = 36 \\ \\\end{gathered} \\ \\ \begin{gathered}:\implies\sf \dfrac{ \: n}{\cancel{\;2}} \times \: \cancel{12}= 36\\ \\\end{gathered} \\ \\ \begin{gathered}:\implies\sf 6n = 36\\ \\\end{gathered} \\ \\ \begin{gathered}:\implies\sf n = \cancel\dfrac{36}{6}\\ \\\end{gathered} \\ \\ \begin{gathered}:\implies\sf\underline{\boxed{{\frak{{n = 6}}}}}\pink\bigstar\\ \\\end{gathered} \\ \\ \begin{gathered}\therefore\:{\underline{\sf{Hence,\; number\:of\;terms\:of\;AP\;are\; {\textsf{\textbf{6}}}.}}}\\\end{gathered}\end{gathered}$

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