Question

Find the sum of the following Aps:
A) \[2,7,12,...,\] to 10 terms.
B) \[37,33,29,...,\] to 12 terms.
C) \[0.6,1.7,2.8,...,\] to 100 terms.
D) \[\dfrac{1}{{15}},\dfrac{1}{{12}},\dfrac{1}{{10}},...,\] to 11 terms.

Answer 1

Hint: We can use the formula of the sum of n terms in Arithmetic progression that is \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\] where, \[{a_1}\] is the initial term of the AP, n is the number of terms, and d is the common difference of successive numbers. Substitute the value of a, n, and d and then calculate the sum of the AP \[{S_n}\].

Complete step-by-step answer:
Given data:
A) The series of the AP is \[2,7,12,...,\] to 10 terms.
Now, we know about the formula of the sum of n terms in Arithmetic progression that is \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
Now, calculate the value of ${S_n}$ where $n = 10,a = 2,{\rm{ and }}d = 5\left( {7 - 2} \right)$. Substitute the values in the expression \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
\[{S_{10}} = \dfrac{{10}}{2}\left[ {2\left( 2 \right) + \left( {10 - 1} \right)\left( 5 \right)} \right]\\
 = 5\left[ {4 + 45} \right]\\
 = 5\left( {49} \right)\\
 = 245\]
Hence, the sum of the AP is \[{S_{10}} = 245\].

B) \[37,33,29,...,\] to 12 terms.
Now, we know about the formula of the sum of n terms in Arithmetic progression that is \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
Calculate the value of ${S_n}$ where $n = 12,a = 37,{\rm{ and }}d = - 4\left( {33 - 37} \right)$. Substitute the values in the expression.
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\]
\[{S_{12}} = \dfrac{{12}}{2}\left[ {2\left( {37} \right) + \left( {12 - 1} \right)\left( { - 4} \right)} \right]\\
 = 6\left[ {74 - 44} \right]\\
 = 6\left( {30} \right)\\
 = 180\]
Hence, the sum of the AP is \[{S_{12}} = 180\].

C) \[0.6,1.7,2.8,...,\] to 100 terms.
Now, we know about the formula of the sum of n terms in Arithmetic progression that is \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
Calculate the value of ${S_n}$ where $n = 100,a = 0.6,{\rm{ and }}d = 1.1\left( {1.7 - 0.6} \right)$. Substitute the values in the expression.
\[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
\[{S_{100}} = \dfrac{{100}}{2}\left[ {2\left( {0.6} \right) + \left( {100 - 1} \right)\left( {1.1} \right)} \right]\\
 = 50\left[ {1.2 + 108.9} \right]\\
 = 50\left( {110.1} \right)\\
 = 5,505\]
Hence, the sum of the AP is \[{S_{100}} = 5,505\].

D) \[\dfrac{1}{{15}},\dfrac{1}{{12}},\dfrac{1}{{10}},...,\] to 11 terms.
Now, we know about the formula of the sum of n terms in Arithmetic progression that is \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
Now, calculate the value of ${S_n}$ where $n = 11,a = \dfrac{1}{{15}},{\rm{ and }}d = \dfrac{1}{{60}}\left( {\dfrac{1}{{12}} - \dfrac{1}{{15}}} \right)$. Substitute the values in the expression \[{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]\].
\[{S_{11}} = \dfrac{{11}}{2}\left[ {2\left( {\dfrac{1}{{15}}} \right) + \left( {11 - 1} \right)\left( {\dfrac{1}{{60}}} \right)} \right]\\
 = 5.5\left[ {0.1333 + 0.16667} \right]\\
 = 5.5\left( {0.29999} \right)\\
 = 1.6499\]
Hence, the sum of the AP is \[{S_{11}} = 1.6499\].

Note: The behaviour of the progression in arithmetic depends on the common difference d. there are two conditions, If the common difference is:
Positive, then the members will expand to positive infinity (terms);
Negative, then to negative infinity, the members (terms) expand.