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

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 ${S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]$.

= 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.