Question

For an A.P, if \[{T_1} = 22\] , \[{T_n} = - 11\;\] and \[{S_{n}} = 66\] , then find n.

Answer 1

Hint: We have general formulas for both nth term and submission or addition up to the nth term. The nth term in an AP is given by \[{t_n} = {t_1} + (n - 1)d\] where \[{t_1}\] is the first term of the A.P. and \[d\] is the common difference between the consecutive terms. Addition up to the nth term is given \[{S_n} = \dfrac{n}{2} \times [2 \times {t_1} + (n - 1)d)]\] . By substituting the given values we can get the required answer.

Complete step-by-step answer:
We are given that the first term of AP is 22 and the nth term is -11. We are also given the sum that the sum of n terms is equal to 66.
We have,
\[{t_n} = {t_1} + (n - 1)d = - 11\]….. (1)
\[{t_1} = 22\]…… (2)
\[{S_n} = \dfrac{n}{2} \times [2 \times {t_1} + (n - 1)d)] = 66\]….. (3)
Substituting equation (2) in equation (1) we get,
\[ \Rightarrow {t_{n}} = {t_1} + (n - 1)d\]
\[ \Rightarrow - 11 = 22 + (n - 1)d\]
\[ \Rightarrow (n - 1)d = - 33\]….. (4)
Now, we substitute equation (2) in equation (2):
\[{S_n} = \dfrac{n}{2} \times [2 \times {t_1} + (n - 1)d)] = 66\]
\[{S_n} = \dfrac{n}{2} \times [2 \times 22 + (n - 1)d)] = 66\]
Multiplying both equations by 2 we get,
\[ \Rightarrow 132 = n[44 + (n - 1)d)]\]…… (5)
From equation (4) we have \[(n - 1)d = - 33\] , substitute this value in equation (5) we get,
\[ \Rightarrow 132 = n(44 - 33) \\
   \Rightarrow 132 = n(11) \\
   \Rightarrow n = 12 \\
 \]

Therefore, the total number of terms (n) in this AP are 12.

Note: We can directly use the $n^{th}$ term of AP in the formula of submission of the AP series. The formula can be written as –
\[{S_n} = \dfrac{n}{2} \times [2 \times {t_1} + (n - 1)d)]\]
\[{S_n} = \dfrac{n}{2} \times \{ {t_1} + [{t_1} + (n - 1)d]\} \]
From Equation 1 we can substitute the value of \[{t_n} = {t_1} + (n - 1)d\] in the equation and we get,
\[{S_n} = \dfrac{n}{2} \times ({t_1} + {t_n})\]
From the above derived formula the question can be solved in a simpler and shorter way. It goes as –
\[{S_n} = \dfrac{n}{2} \times ({t_1} + {t_n}) = 66\]
Now, substituting the values from equation (1) and equation (2) we get,
\[ \Rightarrow \dfrac{n}{2} \times [22 + ( - 11)] = 66\]
\[
   \Rightarrow 66 \times 2 = n(11) \\
   \Rightarrow 132 = n(11) \\
   \Rightarrow n = 12 \\
\]
Since, the derived formula is very used to find the value of a number of terms and sometimes even the nth term it can be memorized to solve the questions in lesser steps.