Question

the first term and last term of an AP are 1 and 11. sum of its term is 36 . find the number of term ?

Answer 1

given in an AP a=1 and l=11 Sn=36
we know that Sn=n/2(a+l)
substituting the values
36=n/2(1+11)
n=72/12=6
therefore in the. gn AP there are 6 terms

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

$\begin{gathered}\;\;\;\;{\underline{\frak{ \color{navy}{Substituting\;the\:given\;values\;in\;formula\;:}}}}\\\end{gathered}</p><p>⠀$

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

тнαηк үσυ

||TheUnknownStar||