Solveeit Logo
Login

Question

Question: Let \({a_n}\), \(n \geqslant 1\), be an arithmetic progression with first term 2 and common differen...

Let an{a_n}, n⩾1n \geqslant 1, be an arithmetic progression with first term 2 and common difference 4. Let Mn{M_n} be the average of the first nn terms, Then the sum ∑n=110Mn\sum\limits_{n = 1}^{10} {{M_n}} is
A) 110
B) 335
C) 770
D) 1100

Explanation

Solution

Here, we will find the value of Mn{M_n} using the formula of average and sum of nn terms of an AP. Then, substituting the value of Mn{M_n} in ∑n=110Mn\sum\limits_{n = 1}^{10} {{M_n}} and using the general formula, we will be able to find the required answer.

Formula Used:
We will use the following formulas:

  1. Average == Sum of observations ÷\div Total number of observations
  2. In an arithmetic progression, the sum of nn terms, Sn=n2[2a+(n−1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right], where aa and dd are the first term and the common difference respectively.
  3. The general formula of sum of nn terms is n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2}

Complete step by step solution:
According to the question,
The first term of this AP, a=2a = 2
And, the common difference, d=4d = 4
Hence, this Arithmetic progression can be written as: 2,6,10,...an2,6,10,...{a_n}
Now, it is given that Mn{M_n} is the average of the first nn terms.
We know that,
Average == Sum of observations ÷\div Total number of observations
Now, in an arithmetic progression, the sum of nn terms, Sn=n2[2a+(n−1)d]{S_n} = \dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]
And, the total number of observations =n = n
Therefore we can write the average Mn{M_n} of the first nn terms as:
Mn=Snn=n2[2a+(n−1)d]n{M_n} = \dfrac{{{S_n}}}{n} = \dfrac{{\dfrac{n}{2}\left[ {2a + \left( {n - 1} \right)d} \right]}}{n}
Here, substituting the known values, we get,
⇒Mn=n[2(2)+(n−1)(4)]2n\Rightarrow {M_n} = \dfrac{{n\left[ {2\left( 2 \right) + \left( {n - 1} \right)\left( 4 \right)} \right]}}{{2n}}
Cancelling out the same terms from the numerator and the denominator and solving further, we get,
⇒Mn=4+4n−42=4n2=2n\Rightarrow {M_n} = \dfrac{{4 + 4n - 4}}{2} = \dfrac{{4n}}{2} = 2n
Now,
∑n=110Mn=∑n=110(2n)\sum\limits_{n = 1}^{10} {{M_n}} = \sum\limits_{n = 1}^{10} {\left( {2n} \right)}
Here, we will use the general formula of the sum of nn terms which is n(n+1)2\dfrac{{n\left( {n + 1} \right)}}{2}.
But, here are 2n2n terms multiplying the numerator by 2 and substituting n=10n = 10, we get,
∑n=110Mn=∑n=110(2n)=2(10)(11)2=11×10=110\sum\limits_{n = 1}^{10} {{M_n}} = \sum\limits_{n = 1}^{10} {\left( {2n} \right)} = \dfrac{{2\left( {10} \right)\left( {11} \right)}}{2} = 11 \times 10 = 110
Therefore, the sum ∑n=110Mn\sum\limits_{n = 1}^{10} {{M_n}} is 110.

Hence, option A is the correct answer.

Note:
An Arithmetic Progression is a sequence of numbers such that the difference between any term and its preceding term is constant. This difference is known as the common difference of an Arithmetic Progression (AP). A real life example of AP is when we add a fixed amount in our money bank every week. Similarly, when we ride a taxi, we pay an amount for the initial kilometer and pay a fixed amount for all the further kilometers, this also turns out to be an AP.