Question

if first and last term of AP is 1 and 11 respectively and sum of its term is 36 then find the number of terms​

Answer 1

Step-by-step explanation:

Given:

$a = 1 \\ t_n = 11 \\ S_n = 36 \\ n = ? \\ \because \: S_n = \frac{n}{2} (a + t_n) \\ \therefore \: 36 = \frac{n}{2} (1 +11) \\ \therefore \: 36 = \frac{n}{2} \times 12\\ \therefore \: 36 = n \times 6\\ \therefore \: n = \frac{36}{6} \\ \therefore \: n = 6$

Answer 2

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 nth terms having last term is Given by :

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

Where,

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

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

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

тнαηк үσυ

||TheUnknownStar||