Answer :
Sure, let's solve the problem step-by-step!
### Step 1: Identify the Common Difference
We start with the arithmetic progression (AP): [tex]\(-12, -9, -6, \ldots, 21\)[/tex].
The common difference [tex]\(d\)[/tex] is found by subtracting the first term from the second term:
[tex]\[
d = -9 - (-12) = -9 + 12 = 3
\][/tex]
### Step 2: Find the Number of Terms
The formula for the [tex]\(n\)[/tex]-th term of an arithmetic progression is:
[tex]\[
a_n = a + (n-1) \times d
\][/tex]
where [tex]\(a\)[/tex] is the first term, and [tex]\(a_n\)[/tex] is the [tex]\(n\)[/tex]-th term.
Here, the first term [tex]\(a\)[/tex] is [tex]\(-12\)[/tex] and the [tex]\(n\)[/tex]-th term is [tex]\(21\)[/tex]. Plugging in these values:
[tex]\[
21 = -12 + (n-1) \times 3
\][/tex]
Solve for [tex]\(n\)[/tex]:
[tex]\[
21 + 12 = 3(n-1)
\][/tex]
[tex]\[
33 = 3(n-1)
\][/tex]
[tex]\[
n-1 = \frac{33}{3} = 11
\][/tex]
[tex]\[
n = 11 + 1 = 12
\][/tex]
So, there are 12 terms in the sequence.
### Step 3: Calculate the Sum of Terms if 1 is Added to Each Term
If we add 1 to each term, the new sequence becomes:
[tex]\[
(-12 + 1), (-9 + 1), (-6 + 1), \ldots, (21 + 1)
\][/tex]
which simplifies to:
[tex]\[
-11, -8, -5, \ldots, 22
\][/tex]
The first term of this new AP is [tex]\(-11\)[/tex], and the last term is [tex]\(22\)[/tex].
The sum [tex]\(S_n\)[/tex] of an arithmetic progression is given by the formula:
[tex]\[
S_n = \frac{n}{2} \times (\text{first term} + \text{last term})
\][/tex]
Here, [tex]\(n = 12\)[/tex], first term = [tex]\(-11\)[/tex], and last term = [tex]\(22\)[/tex]:
[tex]\[
S_n = \frac{12}{2} \times (-11 + 22)
\][/tex]
[tex]\[
S_n = 6 \times 11 = 66
\][/tex]
Thus, the sum of all terms of the new AP is 66.
### Step 1: Identify the Common Difference
We start with the arithmetic progression (AP): [tex]\(-12, -9, -6, \ldots, 21\)[/tex].
The common difference [tex]\(d\)[/tex] is found by subtracting the first term from the second term:
[tex]\[
d = -9 - (-12) = -9 + 12 = 3
\][/tex]
### Step 2: Find the Number of Terms
The formula for the [tex]\(n\)[/tex]-th term of an arithmetic progression is:
[tex]\[
a_n = a + (n-1) \times d
\][/tex]
where [tex]\(a\)[/tex] is the first term, and [tex]\(a_n\)[/tex] is the [tex]\(n\)[/tex]-th term.
Here, the first term [tex]\(a\)[/tex] is [tex]\(-12\)[/tex] and the [tex]\(n\)[/tex]-th term is [tex]\(21\)[/tex]. Plugging in these values:
[tex]\[
21 = -12 + (n-1) \times 3
\][/tex]
Solve for [tex]\(n\)[/tex]:
[tex]\[
21 + 12 = 3(n-1)
\][/tex]
[tex]\[
33 = 3(n-1)
\][/tex]
[tex]\[
n-1 = \frac{33}{3} = 11
\][/tex]
[tex]\[
n = 11 + 1 = 12
\][/tex]
So, there are 12 terms in the sequence.
### Step 3: Calculate the Sum of Terms if 1 is Added to Each Term
If we add 1 to each term, the new sequence becomes:
[tex]\[
(-12 + 1), (-9 + 1), (-6 + 1), \ldots, (21 + 1)
\][/tex]
which simplifies to:
[tex]\[
-11, -8, -5, \ldots, 22
\][/tex]
The first term of this new AP is [tex]\(-11\)[/tex], and the last term is [tex]\(22\)[/tex].
The sum [tex]\(S_n\)[/tex] of an arithmetic progression is given by the formula:
[tex]\[
S_n = \frac{n}{2} \times (\text{first term} + \text{last term})
\][/tex]
Here, [tex]\(n = 12\)[/tex], first term = [tex]\(-11\)[/tex], and last term = [tex]\(22\)[/tex]:
[tex]\[
S_n = \frac{12}{2} \times (-11 + 22)
\][/tex]
[tex]\[
S_n = 6 \times 11 = 66
\][/tex]
Thus, the sum of all terms of the new AP is 66.
