Question

The sum of the series \[3.6{\rm{ }} + {\rm{ }}4.7{\rm{ }} + {\rm{ }}5.8{\rm{ }} + {\rm{ }}......upto{\rm{ }}\left( {n{\rm{ }} - {\rm{ }}2} \right)\] terms
A) \[{n^3} + {n^2} + n + 2\]
B) \[\dfrac{1}{6}(2{n^3} + 12{n^2} + 10n - 84)\]
C) \[{n^3} + {n^2} + n\]
D) None of these

Answer 1

Hint: in this question we have to find sum of \[n - 2\] term of given series. First find the first term of series. Apply formula of sum of AP on the given equation and then simply solve the equation to get require value.

Formula Used: In order to calculate sum of n terms of AP formula is given as:
\[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]
Where
\[{S_n}\] is sum of of AP
a is first term of AP
n is number of terms in AP
d is common difference of AP

Complete step by step solution: Given: \[3.6{\rm{ }} + {\rm{ }}4.7{\rm{ }} + {\rm{ }}5.8{\rm{ }} + {\rm{ }}......upto{\rm{ }}\left( {n{\rm{ }} - {\rm{ }}2} \right)\]
Now apply the sum of n terms of AP on each part of the given equation
In order to calculate sum of n terms of AP formula is given as:
\[{S_n} = \dfrac{n}{2}[2a + (n - 1)d]\]
First term is given as \[3.6\]
Common difference is \[d = 4.7 - 3.6 = 1.1\]
\[\begin{array}{l}{S_n} = \dfrac{{n - 2}}{2}[2 \times 3.6 + ((n - 2) - 1) \times 1.1]\\\end{array}\]
On simplification we get
\[{S_n} = \dfrac{{n - 2}}{2}[7.2 + (n - 3) \times 1.1]\]
\[{S_n} = \dfrac{{n - 2}}{2}[7.2 + (1.1)n - 3.3]\]
\[{S_n} = \dfrac{{n - 2}}{2}[3.9 + (1.1)n]\]
\[{S_n} = \dfrac{{n - 2}}{2}[\dfrac{{39}}{{10}} + \dfrac{{11}}{{10}}n]\]
\[{S_n} = \dfrac{{n - 2}}{{20}}[39 + 11n]\]

Option ‘D’ is correct

Note: Here must find common difference first then apply formula for sum of AP. Don’t try any other concept otherwise it become very complicated.
Whenever given series doesn’t follow any pattern then we first try to rearrange the series. If we get any series then follow that series to get required values. Sometimes by using series we are able to find the first term and common difference therefore always try to find first term and common ratio or common difference if required. Then apply the formula to get the required value.
Sometime students get confused in between AP and GP the only difference in between them is in AP we talk about common difference whereas in GP we talk about common ratio.