Question

The first and last terms of an
a.P. Is 1 and 11. If the sum of its terms is 36, then the number of terms will be

Answer 1

$\sf \bf Given:\ \sf The\ first\ and\ last\ terms\ of\ an\ A.P.\ are\ 1\ and\ 11\ and\ that\ the\ sum\ of\ the\ terms\ is\ 36.\\\\\bf To\ find: \sf The\ number\ of\ terms\ in\ the\ A.P.\\\\\bf Answer:\\\\\sf \bf First\ term\ (a)\ : \sf\ 1\\\\\bf Last\ term\ (l)\ : \sf\ 11\\\\\bf Sum\ of\ all\ the\ terms\ (S_{n})\ :\sf\ 36\\\\\\\sf \bf S_{n}\ =\ \sf \dfrac{n}{2} \bigg[a\ +\ l \bigg]\\\\\\\\\sf 36\ =\ \dfrac{n}{2} \bigg[1\ +\ 11 \bigg]$

$\sf 72\ =\ n \bigg[12 \bigg]\\\\\\ \dfrac{72}{12}\ =\ n \\\\\\6\ =\ n\\\\\\Therefore,\ there\ are\ 6\ terms\ in\ the\ A.P.$

Answer 2

$\Huge{\underline{\underline{\mathfrak{Answer \colon }}}}$

Given,

• The first term is 1 and the last term is 11

• The sum of n terms is 36

Let a and l be the first term and last term respectively

Implies,

• a = 1

• l = 11

Sum of n terms is given by :

$\huge{ \boxed{ \boxed{ \sf{ {s}_{n} = \frac{n}{2}(a + l) }}}}$

Putting the values,we get :

$\sf{36 = \frac{n}{2}(1 + 11) } \\ \\ \leadsto \ \sf{72 = 12n} \\ \\ \leadsto \: \sf{n = \frac{72}{12} } \\ \\ \huge{ \leadsto \: \sf{n = 6}}$