
The sum of n term of the series is
\[{\text{1}}{\text{.4 + 3}}{\text{.04 + 5}}{\text{.004 + 7}}{\text{.0004}}\]
A.\[{{\text{n}}^{\text{2}}}{\text{ + }}\dfrac{{\text{4}}}{{\text{9}}}{\text{(1 + }}\dfrac{{\text{1}}}{{{\text{1}}{{\text{0}}^{\text{n}}}}}{\text{)}}\]
B.\[{{\text{n}}^{\text{2}}}{\text{ + }}\dfrac{{\text{4}}}{{\text{9}}}{\text{(1 - }}\dfrac{{\text{1}}}{{{\text{1}}{{\text{0}}^{\text{n}}}}}{\text{)}}\]
C.\[{{\text{n}}^{\text{2}}}{\text{ + }}\dfrac{{\text{4}}}{{\text{9}}}{\text{(1 + }}\dfrac{{\text{1}}}{{{\text{1}}{{\text{0}}^{\text{n}}}}}{\text{)}}\]
D.None of these
Answer
594.3k+ views
Hint: The above given series can be rearranged and written as the sum of A.P and G.P as
\[{\text{1 + 3 + 5}}...{\text{and 0}}{\text{.4 + 0}}{\text{.04 + 0}}{\text{.004}}...\] And so now proceed with sum of terms of A.P and G.P as \[\dfrac{{\text{n}}}{{\text{2}}}{\text{(2a + (n - 1)d)}}\] and \[\dfrac{{{\text{a(}}{{\text{r}}^{\text{n}}}{\text{ - 1)}}}}{{{\text{r - 1}}}}\]
Complete step-by-step answer:
The given series is given as \[{\text{1}}{\text{.4 + 3}}{\text{.04 + 5}}{\text{.004 + 7}}{\text{.0004}}\]
As the above series can be written as
\[{\text{1 + 3 + 5 + }}...{\text{ + 0}}{\text{.4 + 0}}{\text{.04 + 0}}{\text{.004 + }}..\]
And thus we can see in the above equation that they are summation of both A.P and G.P as
In\[{\text{A}}{\text{.P,a = 1,d = 2}}\]
In \[{\text{G}}{\text{.P,a = 0}}{\text{.4,r = 0}}{\text{.1}}\]
Substituting all the values in the above general equation of A.P and G.P
So we obtain,
\[
\Rightarrow \dfrac{n}{2}[2(1) + (n - 1)2] + \dfrac{{{\text{0}}{\text{.4((0}}{\text{.1}}{{\text{)}}^{\text{n}}}{\text{ - 1)}}}}{{{\text{(0}}{\text{.1) - 1}}}} \\
\Rightarrow \dfrac{n}{2}(2n) + \dfrac{{{\text{0}}{\text{.4((1 - }}\dfrac{1}{{{{10}^n}}}{\text{)}}}}{{0.9}} \\
\Rightarrow {n^2} + \dfrac{4}{9}({\text{1 - }}\dfrac{1}{{{{10}^n}}}) \\
\]
Hence , option (a) is our required correct answer.
Note: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1
A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3.
\[{\text{1 + 3 + 5}}...{\text{and 0}}{\text{.4 + 0}}{\text{.04 + 0}}{\text{.004}}...\] And so now proceed with sum of terms of A.P and G.P as \[\dfrac{{\text{n}}}{{\text{2}}}{\text{(2a + (n - 1)d)}}\] and \[\dfrac{{{\text{a(}}{{\text{r}}^{\text{n}}}{\text{ - 1)}}}}{{{\text{r - 1}}}}\]
Complete step-by-step answer:
The given series is given as \[{\text{1}}{\text{.4 + 3}}{\text{.04 + 5}}{\text{.004 + 7}}{\text{.0004}}\]
As the above series can be written as
\[{\text{1 + 3 + 5 + }}...{\text{ + 0}}{\text{.4 + 0}}{\text{.04 + 0}}{\text{.004 + }}..\]
And thus we can see in the above equation that they are summation of both A.P and G.P as
In\[{\text{A}}{\text{.P,a = 1,d = 2}}\]
In \[{\text{G}}{\text{.P,a = 0}}{\text{.4,r = 0}}{\text{.1}}\]
Substituting all the values in the above general equation of A.P and G.P
So we obtain,
\[
\Rightarrow \dfrac{n}{2}[2(1) + (n - 1)2] + \dfrac{{{\text{0}}{\text{.4((0}}{\text{.1}}{{\text{)}}^{\text{n}}}{\text{ - 1)}}}}{{{\text{(0}}{\text{.1) - 1}}}} \\
\Rightarrow \dfrac{n}{2}(2n) + \dfrac{{{\text{0}}{\text{.4((1 - }}\dfrac{1}{{{{10}^n}}}{\text{)}}}}{{0.9}} \\
\Rightarrow {n^2} + \dfrac{4}{9}({\text{1 - }}\dfrac{1}{{{{10}^n}}}) \\
\]
Hence , option (a) is our required correct answer.
Note: An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant. For example, the sequence 1, 2, 3, 4, ... is an arithmetic progression with common difference 1
A geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54, ... is a geometric progression with common ratio 3.
Recently Updated Pages
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

