Question

In an AP the first term is 8, nth term is 33 and the sum to n terms is 123. Find the number of terms and common differences.

Answer 1

Hint: In this question, we will use the formula of the last term as:
$${T_n} = a + (n - 1)d$$
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$

Complete step-by-step answer:
It is given in the question that a= 8 and $T_n$= 33 and $S_n$=123
We have to find the value of d and n.
We will use the formula of the nth term which is mentioned above.
$${T_n} = a + (n - 1)d$$
We will put the values of a and $T_n$, we get
$$33 = 8 + (n - 1)d$$
From here we will find the value of d
 $$33 - 8 = (n - 1)d$$
On solving and taking n-1 to the left side, we get
$$\dfrac{{25}}{{n - 1}} = d$$………………….. (1)
Now, we will use the formula of sum to n terms.
${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$
Put the values of a and $S_n$, we get
$\eqalign{
  & 123 = \dfrac{n}{2}[2 \times 8 + (n - 1)(\dfrac{{25}}{{n - 1}})] \cr
  & 123 = \dfrac{n}{2}[16 + 25] \cr
  & 246 = n[41] \cr
  & \dfrac{{246}}{{41}} = n \cr
  & n = 6 \cr} $
Put the value of n in (1)
$$\eqalign{
  & \dfrac{{25}}{{6 - 1}} = d \cr
  & d = \dfrac{{25}}{5} = 5 \cr} $$

Note: Using the formula of Tn, find the value of d in terms of n and then use the formula of sum to n terms and put the value of d in that, you will get the value of n.
After that put the value of n in d to get the result