### <Success>Sequence & series lecture 8</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-2.jpg"></p>

### <Success>Recall</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>AP</p> <p>$a, a+d, a+2 d \cdots$,</p> <p>$a_n=n^{\text {th }} \text { term }=a+(n-1) d$</p> <p>$S_n=\frac{n}{2}[I$ term + Last term]</p> <p>$S_n = \frac{n}{2}[2a + (n-1)d]$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-3.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-3a.jpg"></p>

### <Success>Problem 1</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>If $S_1$ is the sum of first " $n$ " terms of an A.P, where $n$ is odd.</p> <p>and $S_2$ is the sum of terms of this series in odd places. Find the ratio $\frac{S_1}{S_2}$.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-4.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-4a.jpg">

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>Let $a_1, a_2\cdots$ be the A.P.</p> <p>Let us denote its common difference by “d”</p> <p>$S_1=a_1+a_2+\ldots+a_n $</p> <p>$=\frac{n}{2}\left[2 a_1+(n-1) \cdot d\right] $</p> <p>$S_2=a_1+a_3+a_5+\cdots+a_n$</p> <p>$a_3-a_1 =a_3-a_2+a_2-a_1 $</p> <p>$=d+d =2 d $</p> <p>$a_5-a_3 =a_5-a_4+a_4-a_3 =2 d$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-5.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-5a.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-5b.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$S_1=a_1+a_2+\ldots+a_n =\frac{n}{2}\left[2 a_1+(n-1) \cdot d\right] $</p> <p>$S_2=a_1+a_3+a_5+\cdots+a_n \\ \ =\frac{n+1}{4}\left[2 a_1+(n-1) \cdot d\right]$</p> <p>Thus,</p> <p>$\frac{S_1}{S_2} =\frac{\frac{n}{2}\left[2 a_1+(n-1) d\right]}{\frac{n+1}{4}\left[2 a_1+(n-1) d\right]} $</p> <p>$=\frac{2 n}{n+1} $</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-6.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-6a.jpg"></p>

### <Success>Recall</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>GP</p> <p>$a, a r, a r^2\cdots$,</p> <p>$a_n=n^{th} \text { term }=a r^{n-1} $</p> <p>$S_n=\frac{a\left(r^n-1\right)}{r-1} \text { if } r \neq 1$</p> <p>$a+a r+a r^2+\cdots $</p> <p>$=\frac{a}{1-r} \text { if }|r|<1 $</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-7.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-7a.jpg"></p>

### <Success>Problem 2</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$P^{\text {th }}$ term of an AP is $\frac{1}{q}$ and $q^{\text {th }}$ term of the same $A P$ is $\frac{1}{p}$.</p> <p>Prove that sum of first “$pq$” terms is $\frac{1}{2}(p q+1) , \ p \neq q$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-8.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-8a.jpg">

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>Let “a” be the first term and “d” be the common difference.<br> $n^{\text {th }} \text { term }=a+(n-1) d$</p> <p>Sum of “n “terms</p> <p>$S_n =\frac{n}{2}[2 a+(n-1) d] $</p> <p>$=\frac{n}{2} [I^{\text{st}} \text { term }+ \text { Last term}] $</p> <p>$a+(p-1) d=\frac{1}{q} \longrightarrow (1) $</p> <p>$a+(q-1) d=\frac{1}{p} \longrightarrow (2)$</p> <p>${[(p-1)-(q-1)] d=\frac{1}{q}-\frac{1}{p}} $</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-9.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$\Rightarrow(p-q) d=\frac{p-q}{q p}$<br> $\Rightarrow d=\frac{1}{q p} $</p> <p>$a =\frac{1}{q}-(p-1) \cdot d $</p> <p>$=\frac{1}{q}-(p-1) \cdot \frac{1}{q p} $</p> <p>$=\frac{1}{q}-\frac{1}{q}+\frac{1}{p q}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-11.jpg">

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$a =\frac{1}{p q} $</p> <p>$S_{p q} =\frac{p q}{2}\left[2 \cdot \frac{1}{p q}+(p q-1) \cdot \frac{1}{p q}\right] $</p> <p>$=\frac{p q}{2}\left[\frac{2}{p q}+1-\frac{1}{p q}\right]$</p> <p>$=\frac{p q}{2}\left[\frac{1}{p q}+1\right] $</p> <p>$=\frac{p q}{2}\left[\frac{1+p q}{p q}\right] $</p> <p>$=\frac{1+p q}{2}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-12.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-12a.jpg">

### <Success>Problem 3</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$a, b$ and $c$ are three consecutive terms of a GP.</p> <p>Further $a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}$</p> <p>Prove that $x, y, z$ are in A.P</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-13.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$a^{\frac{1}{x}}=b^{\frac{1}{y}}=c^{\frac{1}{z}}=k$</p> <p>Thus $\quad a^{\frac{1}{x}}=k$</p> <p>$\Rightarrow a=k^x$</p> <p>$b=k^y $</p> <p>$c=k^z$</p> <p>Since $a, b, c$ are in GP</p> <p>$b=\sqrt{a c} $</p> <p>$b^2=a c$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-14.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-14a.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$i.e.\quad\left(k^y\right)^2=k^x \cdot k^z $</p> <p>$\Rightarrow k^{2 y}=k^{x+z} $</p> <p>$\Rightarrow 2 y=x+z $</p> <p>$\Rightarrow y=\frac{x+z}{2}$<br> Thus $y$ is AM of $x$ and $z$.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-15.jpg"></p>

### <Success>Problem 4</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>Let m, n be positive reals.</p> <p>Assume that arithmetic mean of m, n is A and geometric mean of m, n is G.</p> <p>Then show that the quadratic whose roots are m, n is $x^2 - 2Ax +G^2 = 0.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-16.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-16a.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$A M(m, n)=A $</p> <p>$\text { i.e. } \quad\frac{m+n}{2}=A $</p> <p>$\Rightarrow m+n=2 A .$</p> <p>$G M(m, n)=G $</p> <p>$\sqrt{m n}=G $</p> <p>$\Rightarrow m n=G^2$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-17.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-17a.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$m+n=2 A$</p> <p>$m n=G^2$</p> <p>A quadratic with roots $m, n$ is given by</p> <p>$(x-m)(x-n)=0.$</p> <p>$x^2 - (m + n)x + mn = 0.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-18.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-18a.jpg"></p>

### <Success>Problem 5</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>If $A$ is Arithmetic mean and $G$ is Geometric mean of two positive numbers then show that numbers are</p> <p>$A \pm \sqrt{A^2 - G^2}.$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-19.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-19a.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>Let the numbers be m and n.<br> $\frac{m+n}{2} = A \ \Rightarrow m + n = 2A$</p> <p>$\sqrt{mn} = G$</p> <p>$\Rightarrow mn = G^2$</p> <p>$(m-n)^2 = m^2 - 2mn + n^2 = (m+n)^2 -4mn$</p> <p>$= 4A - 4 G^2$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-20.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-20a.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$(m-n)^2 =4\left(A^2-G^2\right) $</p> <p>$m-n = \pm \sqrt{4\left(A^2-G^2\right)} $</p> <p>$= \pm 2 \sqrt{A^2-G^2}$</p> <p>Thus</p> <p>$m+n =2 A $</p> <p>$m-n = \pm 2 \sqrt{A^2-G^2} $</p> <p>$2 m=2 A \pm 2 \sqrt{A^2-G^2} $</p> <p>$\Rightarrow m=A \pm \sqrt{A^2-G^2}$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-21.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-21a.jpg"></p>

### <Success>Problem 6</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>Given $f$ is a function satisfying</p> <p>$f(x+y)=f(x) \cdot f(y) \ \ \forall x, y \in \mathbb{N} .$</p> <p>Let $f(1)=3$</p> <p>If $\sum_{x=1}^n f(x)=120$, find value of $n$.</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-22.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-22a.jpg">

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$f(x+y)=f(x) \cdot f(y) $</p> <p>$\forall x, y \in \mathbb{N} $</p> <p>$f(2)=f(1+1)$</p> <p>$=f(1) \cdot f(1) $</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-23.jpg">

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$\text { i.e.. } \quad f(2)=[f(1)]^2$</p> <p>Similarly</p> <p>$f(3) =f(2+1) $</p> <p>$=f(2) \cdot f(1) =[f(1)]^3$</p> <p>continuing like this</p> <p>$f(n)=[f(1)]^n$</p> <p>Given</p> <p>$\sum_{x=1}^n f(x)=120$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-24.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-24a.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$\text { i.e. } $</p> <p>$f(1)+f(2)+\cdots+f(n) =120. $</p> <p>$\text {i.e.} $</p> <p>$f(1)+f(1)^2+\cdots+[f(1)]^n =120 . $</p> <p>i.e.</p> <p>$3+3^2+\cdots+3^n =120 $</p> <p>The LHS represents Sum of terms of a GP with $a=3, \text{and} \ r=3$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-25.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-25a.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$\text{using} \quad S_n=\frac{a\left(r^n-1\right)}{r-1} $</p> <p>$\frac{3\left(3^n-1\right)}{3-1}=120 .$</p> <p>$3\left(3^n-1\right)=120 \times 2 =240 $</p> <p>$3^n-1=\frac{240}{3}=80 $</p> <p>$3^n=81$<br> Thus</p> <p>$3^n = 81 = 3^4$</p> <p>$\Rightarrow n = 4$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-26.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-26a.jpg"></p>

### <Success>Problem 7</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>Find sum of the terms of the following sequence</p> <p>$ 7, 77, 777, 7777, \cdots (n$ terms).</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-28.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:30px;text-align:left'> <p>$S_n = 7 + 77 + 777 + \cdots (n$ terms).</p> <p>$ = 7 (1 + 11 + 111 + \cdots )$</p> <p>$ = \frac{7}{9} [ 9 + 99 + 999 + \cdots ]$</p> <p>$ = \frac{7}{9} [ (10 - 1) + (100 - 1) + \cdots ]$</p> <p>$ = \frac{7}{9} [ (10 + 100 + 1000 + \cdots -n]$</p> <p>$ =\frac{7}{9}\left[\frac{10\left(10^n-1\right)}{10-1}-n\right]$</p> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-29.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-29a.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-10-chapter-08-sequence-and-series-l-8-10-ws8zmsebxnw-29b.jpg"></p> </div> <p>======</p> <h3 id="successthank-yousuccess"><Success>Thank you</Success></h3> <div style='float: left; width:100%;font-size:30px;text-align:left'> </div>