Question

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

Answer 1

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(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.