Answer :
Let's work through the problem step-by-step to find the arithmetic progression (AP) and the sum of the last 10 terms.
1. Identify the given values:
- The first term ([tex]\[a\][/tex]) is 3.
- The last term is 253.
- The 20th term ([tex]\[a_{20}\][/tex]) is 98.
2. Find the common difference ([tex]\[d\][/tex]):
The formula for the n-th term of an AP is:
[tex]\[
a_n = a + (n-1)d
\][/tex]
Plugging in the values for the 20th term:
[tex]\[
98 = 3 + (20-1)d
\][/tex]
[tex]\[
98 = 3 + 19d
\][/tex]
Solving for [tex]\[d\][/tex]:
[tex]\[
98 - 3 = 19d
\][/tex]
[tex]\[
95 = 19d
\][/tex]
[tex]\[
d = \frac{95}{19} = 5
\][/tex]
3. Determine the number of terms ([tex]\[n\][/tex]) in the AP:
Using the last term (253) formula:
[tex]\[
a_n = a + (n-1)d
\][/tex]
Plugging in the values:
[tex]\[
253 = 3 + (n-1)5
\][/tex]
[tex]\[
253 = 3 + 5n - 5
\][/tex]
Simplify:
[tex]\[
253 = 5n - 2
\][/tex]
Solving for [tex]\[n\][/tex]:
[tex]\[
253 + 2 = 5n
\][/tex]
[tex]\[
255 = 5n
\][/tex]
[tex]\[
n = \frac{255}{5} = 51
\][/tex]
There are 51 terms in the AP.
4. Find the first term of the last 10 terms in the AP:
We need the term from which the last 10 terms start:
[tex]\[
a_{42} = a + (42-1)d
\][/tex]
Since [tex]\[42 = 51 - 10 + 1\][/tex] (where 41 terms from 1 to 41 and 10 starting from 42):
[tex]\[
a_{42} = 3 + 41 \cdot 5
\][/tex]
[tex]\[
a_{42} = 3 + 205
\][/tex]
[tex]\[
a_{42} = 208
\][/tex]
5. Calculate the sum of the last 10 terms ([tex]\[S\][/tex]):
The sum [tex]\[S_{last10}\][/tex] of the last 10 terms can be calculated using the sum formula:
[tex]\[
S_m = \frac{m}{2} \cdot (2a + (m-1)d)
\][/tex]
Here, [tex]\[m = 10\][/tex], and the first term of the last 10 terms [tex]\[a_{last10} = 208\][/tex]:
[tex]\[
S_{last10} = \frac{10}{2} \cdot (2 \cdot 208 + (10-1) \cdot 5)
\][/tex]
Simplify the inner expression:
[tex]\[
S_{last10} = 5 \cdot (416 + 45)
\][/tex]
[tex]\[
S_{last10} = 5 \cdot 461
\][/tex]
[tex]\[
S_{last10} = 2305
\][/tex]
So, the common difference is 5, the number of terms in the AP is 51, the first term of the last 10 terms is 208, and the sum of the last 10 terms in this arithmetic progression is 2305.
1. Identify the given values:
- The first term ([tex]\[a\][/tex]) is 3.
- The last term is 253.
- The 20th term ([tex]\[a_{20}\][/tex]) is 98.
2. Find the common difference ([tex]\[d\][/tex]):
The formula for the n-th term of an AP is:
[tex]\[
a_n = a + (n-1)d
\][/tex]
Plugging in the values for the 20th term:
[tex]\[
98 = 3 + (20-1)d
\][/tex]
[tex]\[
98 = 3 + 19d
\][/tex]
Solving for [tex]\[d\][/tex]:
[tex]\[
98 - 3 = 19d
\][/tex]
[tex]\[
95 = 19d
\][/tex]
[tex]\[
d = \frac{95}{19} = 5
\][/tex]
3. Determine the number of terms ([tex]\[n\][/tex]) in the AP:
Using the last term (253) formula:
[tex]\[
a_n = a + (n-1)d
\][/tex]
Plugging in the values:
[tex]\[
253 = 3 + (n-1)5
\][/tex]
[tex]\[
253 = 3 + 5n - 5
\][/tex]
Simplify:
[tex]\[
253 = 5n - 2
\][/tex]
Solving for [tex]\[n\][/tex]:
[tex]\[
253 + 2 = 5n
\][/tex]
[tex]\[
255 = 5n
\][/tex]
[tex]\[
n = \frac{255}{5} = 51
\][/tex]
There are 51 terms in the AP.
4. Find the first term of the last 10 terms in the AP:
We need the term from which the last 10 terms start:
[tex]\[
a_{42} = a + (42-1)d
\][/tex]
Since [tex]\[42 = 51 - 10 + 1\][/tex] (where 41 terms from 1 to 41 and 10 starting from 42):
[tex]\[
a_{42} = 3 + 41 \cdot 5
\][/tex]
[tex]\[
a_{42} = 3 + 205
\][/tex]
[tex]\[
a_{42} = 208
\][/tex]
5. Calculate the sum of the last 10 terms ([tex]\[S\][/tex]):
The sum [tex]\[S_{last10}\][/tex] of the last 10 terms can be calculated using the sum formula:
[tex]\[
S_m = \frac{m}{2} \cdot (2a + (m-1)d)
\][/tex]
Here, [tex]\[m = 10\][/tex], and the first term of the last 10 terms [tex]\[a_{last10} = 208\][/tex]:
[tex]\[
S_{last10} = \frac{10}{2} \cdot (2 \cdot 208 + (10-1) \cdot 5)
\][/tex]
Simplify the inner expression:
[tex]\[
S_{last10} = 5 \cdot (416 + 45)
\][/tex]
[tex]\[
S_{last10} = 5 \cdot 461
\][/tex]
[tex]\[
S_{last10} = 2305
\][/tex]
So, the common difference is 5, the number of terms in the AP is 51, the first term of the last 10 terms is 208, and the sum of the last 10 terms in this arithmetic progression is 2305.
