
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
513.9k+ views
Hint: 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.
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.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

How do I get the molar mass of urea class 11 chemistry CBSE

How do I convert ms to kmh Give an example class 11 physics CBSE

Where can free central placentation be seen class 11 biology CBSE

What is the molecular weight of NaOH class 11 chemistry CBSE

What is 1s 2s 2p 3s 3p class 11 chemistry CBSE
