Differentiation (Part-38) | S N De

 

$\,1(i)\,\,$ The $\,10\,$-th term of an A.P. is $\,(-15)\,$ and the $\,31\,$-st term is $\,(-57)\,$. Find the first term and common difference of the A.P.

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

So, $\,\,a+(10-1)d=-15 \\ \Rightarrow a+9d=-15 \rightarrow(1) \\ a+(31-1)d=-57 \\ \Rightarrow a+30d=-57 \rightarrow(2)$

From $\,(1),\,(2)\,$ we get, 

$\,(a+30d)-(a+9d)=-57-(-15) \\ \Rightarrow 21d=-57+15 \\ \Rightarrow d=-\frac{42}{21}=-2 $

Hence, putting $\,d=-2,\,$ we get from $\,(1),$

$a+9 \times (-2)=-15 \\ \Rightarrow a=-15+18 \\ \Rightarrow a=3.$ 

So, the first term and common difference of the A.P. are $\,3\,$ and $\,-2\,$ respectively.

$\,1(ii)\,\,$ If the $\,p\,$-th term of an arithmetic progression is $\,q\,$ and the $\,q\,$-th term is $\,p\,$, show that its $\,(p+q)\,$-th term is $\,0.$

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

So, $\,t_p=q \\ \Rightarrow a+(p-1)d=q \rightarrow(1)$

and , $\,t_q=p \\ \Rightarrow a+(q-1)d=p \rightarrow(2)$

From $\,(1)\,$ and $\,(2),\,$ we get, 

$[a+(p-1)d]-[a+(q-1)d]=q-p \\ \Rightarrow (p-q)d=-(p-q) \\ \Rightarrow  d=-1.$

Hence, putting $\,d=-1,\,$ we get from $\,(1),$

$\,a+(p-1)\times (-1)=q \\ \Rightarrow a=q+p-1 \\ \Rightarrow a=p+q-1.$

So, $\,t_{p+q}=a+(p+q-1)d \\ \Rightarrow t_{p+q}=(p+q-1)+(p+q-1)\times (-1)\\ \Rightarrow t_{p+q}=0\,\,\text{(showed)}$

$\,1(iii)\,\,$ Let $\,T_r\,$ be the $\,r\,$th term of an A.P. If $\,\,m \cdot T_m=n \cdot T_n,\,$ then show that $\,T_{m+n}=0.$

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

$\,T_m=a+(m-1)d \rightarrow(1), \\ ~~T_n=a+(n-1)d\rightarrow(2).$

Now, $\,T_{m+n}=a+(m+n-1)d \\~~~~~~~~~~=a-a\,\,[\text{By (3)}]\\~~~~~~~~~~=0\,\,\text{(showed)}$

Now, $\,\,m \cdot T_m=n \cdot T_n\\ \Rightarrow am+m(m-1)d=an+n(n-1)d\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\text{Using (1),(2)}]\\ \Rightarrow a(m-n)=d(n^2-n-m^2+m)\\ \Rightarrow a(m-n)=-d[m^2-n^2-(m-n)] \\ \Rightarrow a(m-n)=-d[(m+n)(m-n)\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~-(m-n)] \\ \Rightarrow a(m-n)=-d(m-n)(m+n-1) \\ \Rightarrow a=-(m+n-1)d \rightarrow(3)$

So, $\,T_{m+n}=a+(m+n-1)d \\~~~~~~~~~~=a-a\,\,[\text{By (3)}]\\~~~~~~~~~~=0\,\,\text{(showed)}$

$\,2(i)\,$ Which term of the A.P. $\,\{7,11,15,19,\cdots\}\,$ is $\,111.$

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

Then, here $\,a=7,\,\,\,d=11-7=4.$

Let $\,n\,$ th term of the A.P. is $\,111.$

So, $\,t_n=a+(n-1)d \\ \Rightarrow 111=7+(n-1) \times 4 \\ \Rightarrow111-7=4(n-1) \\ \Rightarrow 104=4(n-1) \\ \Rightarrow \frac{104}{4}=n-1 \\ \Rightarrow n=26+1=27.$

Hence, $\,27\,$th term of the A.P. $\,\{7,11,15,19,\cdots\}\,$ is $\,111.$

$\,2(ii)\,$ Is $\,600\,$ a term of the A.P. $\,\{2,9,16,23, \cdots\}\,$ ? Give reasons for your answer.

Sol. If possible , let $\,n\,$ th term of the given  A.P. is $\,600\,$ where the first term $(a)=2,\,$ and the common difference $(d)=9-2=7.$

Now, $\,t_n=a+(n-1)d \\ \Rightarrow 600=2+(n-1) \times 7 \\ \Rightarrow n-1=\frac{600-2}{7} \\ \Rightarrow n-1=\frac{598}{7} \\ \Rightarrow n=\frac{598}{7}+1 \rightarrow(1)$

From $\,(1),\,$ we can say $\,n\,$ is not an integer which is not possible. 

So, there is no term which is $\,600\,$ in the given sequence  $\,\{2,9,16,23, \cdots\}\,$.

$\,3.\,$ Fill up the gaps (indicated by -) in each of the following A.P.'s :

$\,(i)\,\,1,-,-, (-50).$

Sol. Here, the sequence is : $\,a=1,\,a+d,a+2d,(-50),\,$ where the $\,4\,$th term is $\,(-50).$

$\,t_4= -50 \\ \Rightarrow 1+(4-1)d=-50 \\ \Rightarrow 1+3d=-50 \\ \Rightarrow  d=\frac{-51}{3}=-17.$

So, $\,a+d=1-17=-16,\\ a+2d=1-2\times 17=1-34=-33.$

Hence, two terms in the middle is : $\,-16,-33.$

$\,3.\,$ Fill up the gaps (indicated by -) in each of the following A.P.'s :

$\,(ii)\,\,-,-,19,-,- ,31.$

Sol. Here, $\,t_3=19 \\ \Rightarrow a+(3-1)d=19 \\ \Rightarrow a+2d=19 \rightarrow (1) \\ t_6=31 \\ \Rightarrow a+(6-1)d=31 \\ \Rightarrow a+5d=31 \rightarrow(2)$

From $\,(1),(2)\,$ we get, 

$\,(a+5d)-(a+2d)=31-19 \\ \Rightarrow 3d=12 \\ \Rightarrow d=\frac{12}{3} \\ \Rightarrow d=4.$ 

Putting the value of $\,d\,$ in $\,(1),\,$ we get,

$\,a+ 2 \times 4=19 \\ \Rightarrow a=19-8 \\ \Rightarrow a= 11.$

$\,t_2=11+d=11+4=15,\\ t_3=19,\\ t_4=11+(4-1)\times 4 \\ \Rightarrow t_4=11+12=23, \\ t_5=11+(5-1) \times 4 \\ \Rightarrow t_5=11+16=27.$

$\,4(i)\,$ Given, $\,a^2+2a+2,\,3a^2+6a+6\,$ and $\,4a^2+5a+4\,$ are in A.P. Find the value or values of $\,a.$

Sol. Since $\,a^2+2a+2,\,3a^2+6a+6\,$ and $\,4a^2+5a+4\,$ are in A.P., so from properties of A.P., we get,

$\,3a^2+6a+6-(a^2+2a+2)\\=4a^2+5a+4-(3a^2+6a+6) \\ \Rightarrow  2a^2+4a+4=a^2-a-2 \\ \Rightarrow a^2+5a+6=0 \\ \Rightarrow a^2+3a+2a+6=0 \\ \Rightarrow a(a+3)+2(a+3)=0 \\ \Rightarrow (a+3)(a+2)=0 \\ \Rightarrow a=-3,-2\,\,\text{(ans.)}$

$\,4(ii)\,$ Prove that, $\,(x^2+2xy-y^2)^2,\,(x^2+y^2)^2\,$ and $\,(x^2-2xy-y^2)^2\,$ are in A.P.

Sol. Let $\,a_1=(x^2+2xy-y^2)^2,\\ a_2=(x^2+y^2)^2,\\ a_3=(x^2-2xy-y^2)^2.$

To prove that $\,a_1,a_2,a_3\,$ are in A.P. , we have to prove that 

$\,a_2-a_1=a_3-a_2.$

Now, $\,a_2-a_1\\=(x^2+y^2)^2-(x^2+2xy-y^2)^2\\=(x^2+y^2+x^2+2xy-y^2)\\ \times (x^2+y^2-x^2-2xy+y^2)\\=2x(x+y)\times [-2y(x-y)]\\=-4xy(x+y)(x-y)\\=-4xy(x^2-y^2)\rightarrow(1)$

Again, $\,a_3-a_2\\=(x^2-2xy-y^2)^2-(x^2+y^2)^2\\=(x^2-2xy-y^2+x^2+y^2)\\ \times (x^2-2xy-y^2-x^2-y^2)\\=2x(x-y)\times [-2y(x+y)]\\=-4xy(x^2-y^2)\rightarrow(2)$

Hence, from $\,(1),(2)\,$ we can say $\,(x^2+2xy-y^2)^2,\,(x^2+y^2)^2\,$ and $\,(x^2-2xy-y^2)^2\,$ are in A.P.