Question

Find the number of the terms of the A.P: -12, -9, -6, ……… 12. If 1 is added to each of the terms of this A.P, then find the sum of all terms of the A.P. thus obtained.

Answer 1

We will first write the formulas related to A.P. and then put in the values as given to us. Then , we will form a new A.P: -11, -8, -5, ….. and so on and find the sum using its formula.

Complete step-by-step answer:
If a is the first term of the A.P, d is the common difference, then $d = {a_n} - {a_{n - 1}}$and ${a_n}$ is the ${n^{th}}$ term of the A.P, then ${a_n} = a + (n - 1)d$ and if ${s_n}$ is the sum of n terms, then ${s_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$.
Now, we have the AP: -12, -9, -6, ………12.
So, ${a_1} = - 12$, ${a_2} = - 9$.
So, $d = {a_2} - {a_1}$.
Hence, $d = - 9 + 12 = 3$.
Now, in this AP, we have: ${a_n} = 12$.
We will put all these values in ${a_n} = a + (n - 1)d$.
We will get:- $12 = - 12 + 3(n - 1)$
Taking 12 to LHS, we will have:-
$\Rightarrow 3(n - 1) = 24$
Taking 3 from LHS to RHS, we will get:-
$\Rightarrow n - 1 = \dfrac{{24}}{3}$
Simplifying it, we will get:-
$\Rightarrow n - 1 = 8$
Taking 1 from LHS to RHS, we will get:-
$\Rightarrow n = 8 + 1 = 9$.
Hence, there are 9 terms in the A.P: -12, -9, -6, ………12.
Now, if 1 is added to each of the terms of this A.P, then we will have the new AP as: -12 + 1, -9 + 1, -6 + 1, ………12 + 1.
So, the new AP will become:- -11, -8, -5, ……..13.
Here, the first term = a = -11, d will remain the same, so d = 3 and n will also remain the same, so n = 9.
Now, putting these values in ${s_n} = \dfrac{n}{2}\left[ {2a + (n - 1)d} \right]$, we will have:-
$\Rightarrow {s_n} = \dfrac{9}{2}\left[ {2 \times ( - 11) + 3 \times (9 - 1)} \right]$
$\Rightarrow {s_n} = \dfrac{9}{2}\left[ { - 22 + 3 \times 8} \right]$
$\Rightarrow {s_n} = \dfrac{9}{2}( - 22 + 24) = \dfrac{9}{2} \times 2 = 9$.
Hence, the sum of terms of the new AP is 9.

Note: The students must notice that we formed the new AP, but the number of terms and d is not changed. n is not changed because we are just adding 1 to every term but the number of terms remain the same.
And for d we subtract two consecutive terms, but if both have 1 added to them, it will be cancelled out and d will remain the same.