The first term of an Arithmetic Progression is 22 and the last term is -11. If the sum is 66, the number of terms in the sequence are:

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Number of terms = n (l e t) \\ First term (a) = 22 \\ Last term (l) = - 11 \\ Sum = 66 \\ Sum of an AP is given by : \\ = Number of terms \times {\frac{(First term + Last term)}{2}} \\ 66 = n \times {\frac{(a + l)}{2}} \\ 66 = n \times \frac{(22 - 11)}{2} \\ 66 = n \times (\frac{11}{2}) \\ n = \frac{(66 \times 2)}{11} \\ n = 12 \\ No . of terms = 12 \end{aligned}

$\eqalign{ & {\text{Number}}\,{\text{of}}\,{\text{terms}} = n\,\left( {let} \right) \cr & {\text{First}}\,{\text{term}}\,\left( a \right) = 22 \cr & {\text{Last}}\,{\text{term}}\left( l \right) = - 11 \cr & {\text{Sum}} = 66 \cr & {\text{Sum}}\,{\text{of}}\,{\text{an}}\,{\text{AP}}\,{\text{is}}\,{\text{given}}\,{\text{by}}: \cr & = {\text{Number}}\,{\text{of}}\,{\text{terms}} \times \left\{ {\frac{{\left( {{\text{First}}\,{\text{term}} + {\text{Last}}\,{\text{term}}} \right)}}{2}} \right\} \cr & 66 = n \times \left\{ {\frac{{\left( {a + l} \right)}}{2}} \right\} \cr & 66 = n \times \frac{{\left( {22 - 11} \right)}}{2} \cr & 66 = n \times \left( {\frac{{11}}{2}} \right) \cr & n = \frac{{\left( {66 \times 2} \right)}}{{11}} \cr & n = 12 \cr & {\text{No}}{\text{.}}\,{\text{of}}\,{\text{terms}} = 12 \cr}$