Differentiation (Part-38) | S N De

 

$\,1.\,$ If the sum of first $\,2n\,$ terms of the A.P. $\,2,5,8,\cdots$ is equal to the sum of the first $\,n\,$ terms of the A.P. $\,\,57, 59,61,\cdots$ then find $\,n.$

Sol. For the first series , the sum $\,S_1=\frac{2n}{2}[2\times 2+(2n-1).3]$

For the second series, the sum $\,S_2=\frac{n}{2}[2 \times 57+(n-1).2]$

By question, we have $\,\,S_1=S_2 \\ \Rightarrow \frac{2n}{2}[2\times 2+(2n-1).3] \\ =\frac{n}{2}[2 \times 57+(n-1).2] \\ \Rightarrow n[4+6n-3]=n[57+n-1] \\ \Rightarrow 6n+1=56+n \\ \Rightarrow 6n-n=56-1 \\ \Rightarrow 5n=55 \\ \Rightarrow n=\frac{55}{5}=11\,\,\text{(ans.)}$

$\,2.\,$ Let $\,S_n\,$ be the sum of first $\,n\,$ terms of an A.P.  If $\,S_{2n}=5S_n\,$, then find the value of $\,S_{3n}:S_{2n}.$

Sol. We have, $\,S_{2n}=5S_n\, \\ \Rightarrow \frac{2n}{2}[2a+(2n-1)d]=5.\frac n2[2a+(n-1)d]\\ \Rightarrow n[2a+(2n-1)d]=\frac{5n}{2}[2a+(n-1)d]\\ \Rightarrow 4an+(4n-2)nd=10an+(5n-5)nd \\ \Rightarrow nd(4n-2-5n+5)=10an-4an \\ \Rightarrow nd(-n+3)=6an \\ \Rightarrow 6a=(3-n)d \rightarrow(1)$

Now, $\,\frac{S_{3n}}{S_{2n}}\\=\frac{(3n/2)[2a+(3n-1)d]}{(2n/2)[2a+(2n-1)d]}\\=\frac{3[2a+(3n-1)d]}{2[2a+(2n-1)d]}\\=\frac{3[6a+(9n-3)d]}{2[6a+(6n-3)d]}\\=\frac{3[(3-n)d+(9n-3)d]}{2[(3-n)d+(6n-3)d]},\,\,\text{[By (1)]}\\=\frac{3 \times 8nd}{2 \times 5nd}\\=\frac{12}{5}\,\,\text{(ans.)}$

$\,3.\,$ The sum of first $\,m\,$ terms of an A.P. is $\,n\,$ and the sum of first $\,n\,$ terms of the same A.P. is $\,m.\,$ Find the sum of first $\,(m+n)\,$ terms of the A.P.

Sol. Let the sum of first $\,m\,$ terms of an A.P. be denoted by $\,S_m\,$ and the sum of first $\,n\,$ terms of the same A.P. is $\,S_n.\,$

Then, by question, $\, S_m=n \\ \Rightarrow \frac{m}{2}[2a+(m-1)d]=n \\ \Rightarrow 2a+(m-1)d=\frac{2n}{m}\rightarrow(1)$

Similarly, $\,S_n=m \\ \Rightarrow \frac n2[2a+(n-1)d]=m \\ \Rightarrow 2a+(n-1)d=\frac{2m}{n} \rightarrow(2)$

Subtracting $\,(2),\,$ from $\,(1),\,\,$ we get,

$\,[(m-1)-(n-1)]d=\frac{2n}{m}-\frac{2m}{n} \\ \Rightarrow (m-n)d=-2\left[\frac{m^2-n^2}{mn}\right]\\ \Rightarrow d=-2\times \frac{(m+n)(m-n)}{mn} \times \frac{1}{(m-n)}\\ \Rightarrow d=\frac{-2(m+n)}{mn}$

Now, $\,S_{m+n}=\frac{m+n}{2}[2a+(m+n-1)d] \\ \Rightarrow S_{m+n}=\frac{m+n}{2}[2a+(m-1)d+nd] \\ \Rightarrow S_{m+n}=\frac{m+n}{2}[\frac{2n}{m}-n\times \frac{2(m+n)}{mn}] \\ \Rightarrow S_{m+n}=\frac{m+n}{2} \times 2\left[\frac{n}{m}-\frac{(m+n)}{m}\right]\\~~~~~~~~~~~~~~=(m+n)\times \frac{n-m-n}{m}\\~~~~~~~~~~~~~~=-(m+n)\,\,\text{(ans.)}$

$\,4.\,$ If the $\,p\,$-th and $\,q\,$-th term of an A.P. be $\,a\,$ and $\,b\,$ respectively, show that the sum of the first $\,(p+q)\,$terms of the A.P. is $\,\frac{p+q}{2}\left(a+b+\frac{a-b}{p-q}\right).$

Sol. Let the first term of the A.P. be $\,x\,$ and the common difference be $\,d.$

By question, we have, $\,x+(p-1)d=a \rightarrow(1) \\ x+(q-1)d=b\rightarrow(2)$

Now, by $\, (1)-(2),\,$ we get, 

$\,[(p-1)-(q-1)]d=a-b \\ \Rightarrow (p-q)d=a-b \\ \Rightarrow d=\frac{a-b}{p-q}. \\ \therefore S_{p+q}\\ =\frac{p+q}{2}[2x+(p+q-1)d]\\=\frac{p+q}{2}[x+(p-1)d+x+(q-1)d+d]\\=\frac{p+q}{2}\left[a+b+\frac{a-b}{p-q}\right]\,\,\text{(showed)}$

$\,5.\,$ If $\,S_1,\,S_2,\,S_3\,$ be the sums of $\,n\,$terms of three A.P.'s , the first term of each A.P. being $\,1\,$ and the respective common differences are $\,1,2,3\,$then show that , $\,S_1+S_3=2S_2.$

Sol. By question, $\,S_1=\frac n2[2\times 1+(n-1).1]=\frac n2(n+1)\rightarrow(1) \\ S_3=\frac n2[2  \times 1+(n-1).3]=\frac n2[3n-1]\rightarrow(2)\\ \therefore S_1+S_3\\=\frac n2[n+1+3n-1]\,\,[\text{By (1),(2)}]\\=\frac n2[4n]\\=2n^2\rightarrow(3)$

Again, we see $\,\,2S_2\\=2.\frac n2[2 \times 1+(n-1).2]\\=n \times 2n\\=2n^2\rightarrow(4)$

From $\,(3),\,(4)\,\,$ we can conclude that $\,S_1+S_3=2S_2.$

$\,6.\,$ If the sums of $\,n,\,2n\,$ and $\,3n\,$ terms of an A.P. be $\,S_1,S_2,S_3\,$ respectively, then show that, $\,\,S_3=3(S_2-S_1).$

Sol. By question, we have $\,S_2=\frac{2n}{2}[2a+(2n-1)d],\\ S_1=\frac{n}{2}[2a+(n-1)d] \\ \therefore S_2-S_1\\=\frac n2[4a+(4n-2)d-2a-(n-1)d]\\=\frac n2[2a+(4n-2-n+1)d]\\=\frac n2[2a+(3n-1)d] \\ \Rightarrow 3(S_2-S_1)=\frac{3n}{2}[2a+(3n-1)d]=S_3\\ ~~~~~~~~~~~~~~~~~~~~\text{(showed)}$

$\,7.\,$ The sums of $\,n\,$ terms of two A.P.'s are in the ratio $\,(4n-13):(3n+10).\,\,$ Find the ratio of their $\,9\,$-th terms.

Sol.  By question, $\,\,\frac{\frac n2[2a_1+(n-1)d_1]}{\frac n2[2a_2+(n-1)d_2]}=\frac{4n-13}{3n+10}\,[*]\\ \Rightarrow \frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{4n-13}{3n+10}\rightarrow(1)$

Now, putting $\,\,n=17,\,\,$we get from $\,(1),\,$

$\,\frac{2a_1+(17-1)a_1}{2a_2+(17-1)d_2}=\frac{4\times 17-13}{3\times 17+10} \\ \Rightarrow \frac{a_1+8d_1}{a_2+8d_2}=\frac{55}{61} \\ \Rightarrow\frac{a_1+(9-1)d_1}{a_2+(9-1)d_2}=\frac{55}{61}\rightarrow(2)$

From $\,(2),\,$we can say that the ratio of their $\,9\,$-th terms are : $\,55:61.$

Note[*] : $\,\,a_1,a_2\,$ indicate the first terms of two A.P.s, $\,d_1,d_2\,$ denote the common differences of two different A.P.s.