Differentiation (Part-38) | S N De

 $\,1.\,$ The third term of an A.P. is $\,\frac 15\,$ and the $\,5\,$-th term is $\,\frac 13.\,$ Show that the sum to $\,15\,$ terms of the A.P. is $\,8.$

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

So, the $\,n\,$-th term $\,t_n=a+(n-1)d.$

Now, $\,\,t_3=\frac 15\,\,\text{(Given)} \\ \Rightarrow a+(3-1)d=\frac 15\\ \Rightarrow a+2d=\frac 15 \rightarrow(1)$

Again, $\,\,t_5=\frac 13 \\ \Rightarrow a+(5-1)d=\frac 13 \\ \Rightarrow  a+4d=\frac 13 \rightarrow(2)$

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

$\,2d=\frac 13-\frac 15 \\ \Rightarrow 2d=\frac{2}{15} \\ \Rightarrow  d=\frac{1}{15}.$

Now, from $\,(1),\,$ we get, $\,a + 2 \times \frac{1}{15}=\frac 15 \\ \Rightarrow a+\frac{2}{15}=\frac 15 \\ \Rightarrow  a=\frac 15-\frac{2}{15} \\ \Rightarrow  a= \frac{3-2}{15} \\ \Rightarrow  a=\frac{1}{15}.$

So, $\, S_n=\frac n2[2a+(n-1)d] \\ \therefore S_{15}=\frac{15}{2}[2 \times \frac{1}{15}+(15-1)\times \frac{1}{15}] \\ \Rightarrow S_{15}=\frac{15}{2} \times \frac{16}{15}=8.$

 $\,2.\,$ The sum of $\,p\,$ terms of a series is $\,3p^2+5p.\,$ Prove that the terms of the series form an A.P.

Sol. Let the sum of $\,p\,$ terms of a series be denoted by $\,S_p.$

So, $\,S_p=3p^2+5p$

Also, the sum of $\,(p-1)\,$ terms of a series be denoted by $\,S_{p-1}.$

Hence, $\,S_{p-1}=3(p-1)^2+5(p-1)$

Now, $\,p\,$-th term of the series $\,\,t_p\,\,$ so that

$\,t_p=S_p-S_{p-1} \\ \Rightarrow t_p=(3p^2+5p)-[3(p-1)^2+5(p-1)] \\ \Rightarrow t_p=(3p^2+5p)-3(p^2-2p+1)\\ ~~~~~~~~~~-5(p-1) \\ \Rightarrow  t_p=6p+2=6p+(8-6) \\ \therefore t_p=8+(p-1)\times 6 \rightarrow(1)$

Hence, from $\,(1),\,$ it is evident that the terms of the series form an A.P. whose first term is $\,8\,$ and common difference is $\,6.$

$\,3.\,$ Find the $\,12\,$-th term of a series in A.P., whose sum of $\,p\,$ terms is $\,4p^2+3p.$

Sol. Let the sum of $\,p\,$ terms of a series be denoted by $\,S_p.$

So, $\,S_p=4p^2+3p$

Also, the sum of $\,(p-1)\,$ terms of a series be denoted by $\,S_{p-1}.$

Hence, $\,S_{p-1}=4(p-1)^2+3(p-1)$

Now, $\,p\,$-th term of the series $\,\,t_p\,\,$ so that

$\,t_p=S_p-S_{p-1} \\ \Rightarrow t_p=(4p^2+3p)-[4(p-1)^2+3(p-1)] \\ \Rightarrow t_p=(4p^2+3p)-4(p^2-2p+1)\\ ~~~~~~~~~~-3(p-1) \\ \Rightarrow  t_p=8p-1=7+(p-1)\times 8 \\ \therefore t_{12}=7+(12-1)\times 8 =7+88=95\,\,\text{(ans.)}$

$\,4.\,$ If $\,\,51 +53 +55+-+t_n= 5151,\,$ then find the value of $\,\,t_n.$

Sol. For the series $\,\,51,\,53,\,55,\cdots ,t_n,t_{n+1},\cdots\,$, it is evident  that the series is in A.P. with first term $\,51,\,\,$ and common difference $\,2.$

Let the sum of $\,n\,$ terms of a series be denoted by $\,S_n.$

So, $\,S_n=51 +53 +55+-+t_n \\ \Rightarrow S_n=\frac n2[2 \times 51+(n-1)\cdot 2] \\ \Rightarrow S_n=\frac n2[102+2n-2]\\ \Rightarrow S_n=\frac n2[100+2n]$

Now, $\,\frac n2(100+2n)=5151 \\ \Rightarrow n(50+n)-5151=0 \\ \Rightarrow n^2+50n-5151=0 \\ \Rightarrow n^2+101n-51n-5151=0 \\ \Rightarrow n(n+101)-51(n+101)=0 \\ \Rightarrow (n+101)(n-51)=0 \\ \Rightarrow n-51=0\,\,\,[\because n+101 \neq 0] \\ \Rightarrow n=51.\\ \therefore t_n=t_{51}=51+(51-1) \times 2 \\ \Rightarrow t_{51}=51+100=151\,\,\text{(ans.)}$

$\,5.\,$ The sum of first $\,n\,$ terms of an A.P. be $\,n^2+3n,\,$ which term of it has the value $\,162\, ?$

Sol. The sum of first $\,n\,$ terms of an A.P. be $\, S_n=n^2+3n. $

So, $\,t_n=S_n-S_{n-1} \\ \Rightarrow t_n=n^2+3n-[(n-1)^2+3(n-1)]\\~~~~~~~~=n^2+3n-[n^2-2n+1+3n-3]\\~~~~~~~~=n^2+3n-n^2+2n-1-3n+3\\~~~~~~~~=2n+2 \\ \Rightarrow 162=2n+2 \\ \therefore n=\frac{162-2}{2}=80.$

So, $\,\,80\,$-th term has a value $\,162.$ 

$\,6.\,$ How many terms of the series $\,\,\{27+24+21+\cdots\}\,\,$ must be added to get the sum $\,132\, ?$ Explain the double answer.

Sol.  Let $\,n\,$ terms must be added to the series $\,\,\{27+24+21+\cdots\}\,\,$  to get the sum $\,132.$

So, $\,\,\frac n2[2 \times 27+(n-1) \times (-3)]=132 \\ \Rightarrow n[54-3n+3]=264 \\ \Rightarrow  n(57-3n)=264 \\ \Rightarrow  -3n^2+57n-264=0 \\ \Rightarrow 3n^2-57n+264=0 \\ \Rightarrow  3n^2-33n-24n+264=0\\ \Rightarrow 3n(n-11)-24(n-11)=0 \\ \Rightarrow (3n-24)(n-11)=0 \\ \Rightarrow  3(n-8)(n-11)=0 \\ \Rightarrow  n=8,11.$

Explanation for double answer :

We notice that , $\,t_9=27+(9-1)\times (-3)=3,\\ t_{10}=27+(10-1)\times (-3)=0,\\ t_{11}=27+(11-1) \times (-3)=-3.$

So,$\,\,\,t_9+t_{10}+t_{11}=3+0-3=0,\,\,$ and hence $\,S_8=S_{11}.$

To download full PDF solution of Arithmetic Progression (SEQUENCE AND SERIES) of S.N.De Math solution, click here.

$\,7.\,$ The first and last terms of an A.P. having a finite number of terms are respectively $\,(-2)\,$ and $\,124\,$ and the sum of the A.P. is $\,6161.\,$ Find the number of terms in the A.P. and also its common difference.

Sol. Let total number of terms of an A.P. be $\,n.$

So, $\,\,\frac n2[-2+124]=6161 \\ \Rightarrow \frac n2 \times 122=6161 \\ \Rightarrow 61n=6161 \\ \Rightarrow n=\frac{6161}{61}=101.$

According to the problem, $\,t_{101}=124 \\ \Rightarrow -2+(101-1)d=124 \\ \Rightarrow 100d=124+2 \\ \Rightarrow d=\frac{126}{100}=1.26 $

Hence, total number of terms of the series is $\,101\,\,$ and the common difference is $\,\,1.26$

$\,8.\,$ The $\,n\,$-th term of an A.P. is $\,p\,$ and the sum of its first $\,n\,$ terms is $\,q.$ Prove that its first term is $\,\,\frac 1n(2q-pn).$

Sol. $\,t_n=p,\,\,\, S_n=q \rightarrow(1)$

From $\,(1),\,\,$ we have , $\,S_n=q \\ \Rightarrow \frac{n}{2}(a+p)=q\,\,[*] \\ \Rightarrow a+p=\frac{2q}{n} \\ \Rightarrow a=\frac{2q}{n}-p=\frac{2q-pn}{n}\,\,\text{(proved)}$

Note[*] : Since the sum of $\,n\,$ numbers of a series which is in A.P. is : 

$\,\,S_n=\frac n2[\text{First term +Last term}]$

$\,9.\,$ Find the sum of the series $\,\,\{2+3+5+9+8+15+11+21+\cdots \}\,\,$ to $\,(2n+1)\,$ terms.

Sol. $\,\,\{2+3+5+9+8+15+11+21+\cdots \}\\=\{2+5+8+11+\cdots \text{ to (n+1)terms}\}+\\ ~~~~~\{3+9+15+\cdots\text{to n terms}\}\\=\frac{n+1}{2}[2 \times 2+n \times 3]+\frac n2[2 \cdot 3+(n-1)\cdot 6]\\=\frac{(n+1)(3n+4)}{2}+3n^2\\=\frac 12(9n^2+7n+4)\,\,\,\text{(ans.)}$

$\,10.(i)\,\,$ How many even numbers are there between $\,91\,$ and $\,259\,?\,\,$ Find also the sum of the even numbers between them.

Sol.  The even numbers  between $\,91\,$ and $\,259\,\,\,$ are $\,92,94,96,\cdots,258.$

Let $\,\,t_n=258 \\ \Rightarrow 92+(n-1) \cdot 2=258 \\ \Rightarrow 2n=258-92+2 \\ \Rightarrow 2n=168 \\ \Rightarrow n=\frac{168}{2}=84$

Hence, the sum of even numbers 

$=\frac{84}{2}[92+258]=42 \times 350=14700$

$\,10(ii)\,$ Find the sum of all multiples of $\,11\,$ between $\,100\,$ and $\,400.$

Sol. The multiples of $\,11\,$ between $\,100\,$ and $\,400\,$ are $\,110,121,\cdots, 396.$

Let $\,t_n=396 \\ \therefore a+(n-1)d=396 \\ \Rightarrow 110+(n-1)\cdot 11=396 \\ \Rightarrow 11(n-1)=396-110  \\ \Rightarrow  n-1=\frac{286}{11}  \\ \Rightarrow  n=26+1=27.$

Hence, the sum of all multiples of $\,11\,$ between $\,100\,$ and $\,400\,$

$=\frac{27}{2} (110+396)\\=\frac{27}{2} \times 506\\=6831\,\,\text{(ans.)}$