Q-1) Find the sum of \(\sum\limits_{r=1}^n\frac{r}{(r+1)!}\), where \(n!=n\times (n-1)\times (n-2)…3\times 2\times 1\).
Solution:
We have \(T_r=\frac{r}{(r+1)!}=\frac{r+1-1}{(r+1)!}=\frac{1}{r!}-\frac{1}{(r+1)!}\Rightarrow \sum\limits_{r=1}^nT_r=\sum\limits_{r=1}^n\left(\frac{1}{r!}-\frac{1}{(r+1)!}\right)=1-\frac{1}{(n+1)!}\)
Q-2) Find the sum of
Solution:
We have the factors in the denominator in AP. In this case, what we normally do is to multiply and divide \(T_r\) with the difference of the first factor and the last. Thus
If we define \(f(r)=\frac{1}{(r+1)(r+2)(r+3)}\), we have \(T_r=-\frac{1}{3}[f(r)-f(r-1)]\). Thus we have
Q-3) Find the sum \(\sum\limits_{r=1}^nr(r+1)(r+2)(r+3)\)
Solution:
Here the factors are in AP, thus for solving this type of problems we multiply and divide \(T_r\) by difference of factors after the last term and before the first term.
where \(f(r)=r(r+1)(r+2)(r+3)(r+4)\).
Thus the sum upto \(n\) terms equals \(\frac{1}{5}[f(n)-f(0)]=\frac{1}{5}n(n+1)(n+2)(n+3)(n+4)\).
Q-4) Find the value of the sum of the first \(n\)terms of the sequence given by \(a_k=(k^2+1)k!\)
Solution:
We have \(a_k=(k^2+1)k!=(k[k+1]-[k-1])k!=k(k+1)!-(k-1)k!\). Thus as before we have \(\sum\limits_{k=1}^na_k=n(n+1)!\)
Q-5) If the ratio of the sum of \(n\) terms of 2 APs be \(\frac{5n+3}{3n+4}\), then the ratio of their 17th terms is
a) \(172:99\) b) \(168:103\) c) \(175:99\) d) \(171:103\)
Solution:
We have :
Note that we need to find the ratio of the 17th terms, and since 7 is an odd number, we replace \(n\) by \(2n-1\). Thus this ratio becomes:
Putting \(n=17\), we have the answer as \(\frac{168}{103}\).
Q-6) If the sum of \(n\) terms of an AP is \(cn(n-1)\), then find the sum of the squares of these \(n\) terms.
Solution:
We have the rth term of this AP as \(T_r=S_r-S_{r-1}=2c(r-1)\). This gives us \(\sum\limits_{r=1}^nT_r^2=\sum\limits_{r=1}^n4c^2(r-1)^2=4c^2\frac{(n-1)n(2n-1)}{6}=\frac{2}{3}c^2n(n-1)(2n-1)\).
Q-7) If \(\alpha,\beta^2\) be the roots of the equation \(x^2-px+1=0\) and \(\alpha^2,\beta\) be the roots of the quadratic equation \(x^2-qx+8=0\), with reals \(\alpha, \beta\), and \(\frac{r}{8}\) be the AM of \(p,q\), then find the value of \(r\).
Solution:
We have \(\alpha\beta^2=1\) and \(\alpha^2\beta=8\), which implies \(\alpha^3\beta^3=8\Rightarrow\alpha\beta=2\).
Also \(\alpha^2\beta=2\alpha=8\Rightarrow \alpha=4\Rightarrow \beta=\frac{1}{2}\).
We thus have \(p=\alpha+\beta^2=4+\frac{1}{4}=\frac{17}{4}\).
And similarly \(q=\alpha^2+\beta=16+\frac{1}{2}=\frac{33}{2}\).
AM of \(p,q\) is \(\frac{r}{8}=\frac{p+q}{2}\Rightarrow r=4(p+q)=83\).
Q-8) Find the sum of the following infinite series for \(|x|<1\)
Solution:
We have the general term of this sequence as:
Thus the sum eventually becomes:
The last term, for \(|x|<1\) tends to \(1\) as \(n\rightarrow \infty\). Thus finally:
Q-9) Find the value of the following expression:
Solution:
As before, we consider the product of the first \(n\) terms:
Now as \(n\rightarrow \infty\), \(\left(\frac{1}{3} \right )^{2^{n+1}}\rightarrow 0\), thus the value of the infinite product is \(\frac{3}{2}\).
Passage for Questions – 10,11,12
Let \(A_1,A_2,…A_m\) be the Arithmetic means between \(-2\) and \(1027\). Also let \(G_1,G_2,…G_n\) be the Geometric means between \(1\) and \(1024\). The product of the geometric means is \(2^{45}\) and the sum of Arithmetic means is \(1025\times 171\).
Q-10) The value of \(\sum\limits_{r=1}^n G_r\) is:
a) \(512\) b) \(2046\) c) \(1022\) d) None of these
Solution:
We have \(G_1G_2G_3..G_n=\left(\sqrt{1\times 1024} \right)^n=2^{5n}=2^{45}\Rightarrow n=9\).
This gives \(r=(1024)^{\frac{1}{1+9}}=2\).
Thus \(\sum\limits_{r=1}^nG_r=\frac{2\times (2^9-1)}{2-1}=1022\)
Q-11) The number of Arithmetic Means is
a) \(442\) b) \(342\) c) \(378\) d) None of these
Solution:
\(A_1+A_2+A_3+..+A_m=1025\times 171\Rightarrow \left( \frac{-2+1027}{2} \right)m=1025\times 171\Rightarrow m=342\).
Q-12) The numbers \(2A_{171},G_5^2+1,2A_{172}\) are in
a) A.P b) G.P c) H.P d) None of these
Solution:
We have \(A_{171}+A_{172}=-2+1027=1025\Rightarrow \frac{2A_{171}+2A_{172}}{2}=1025\).
Also \(G_5=1\times 2^5=32\Rightarrow G_5^2+1=1025\).
Thus the numbers are in A.P.
Q-13) \(ABC\) is a right angled triangle with \(\angle ABC=90^\circ\) and \(BC=a\). \(AB\) is divided in \(n+1\) equal parts with points \(L_1,L_2,L_3,…L_n\) on \(AB\). Given that the line segments \(L_iM_i\) are parallel to \(BC\) with \(M_i\) on \(AC\) for \(i=1,2,..n\). Find the value of \(\sum\limits_{i=1}^nL_iM_i\).
Solution:
We apply the concepts of similarity in Eucleidian Geometry:
\(\frac{AL_i}{AB}=\frac{i}{n+1}=\frac{L_iM_i}{BC}\Rightarrow L_iM_i=\frac{ai}{n+1}\).
Thus \(\sum\limits_{i=1}^nL_iM_i=\frac{a}{n+1}\sum\limits_{i=1}^ni=\frac{an}{2}\).
Q-14) If \(1+2x+3x^2+4x^3+…\infty \ge 4\), then which of the following is/are true:
a) \(x\ge\frac{1}{2}\) b) \(x\le \frac{4}{3}\) c) \(x \ge \frac{2}{3}\) d) We cannot find the maximum value of \(x\).
Solution:
Note that the last option is correct, and it is trivial to prove that, since, if we have \(S=1+2x+3x^2+…\infty\), then we have \(S\rightarrow \infty\) as \(x\rightarrow\infty\).
Assuming that \(|x|<1\), we have
\(S=1+2x+3x^2+…\infty\Rightarrow xS=x+2x^2+3x^3+…\infty\)Subtracting the 2 equations we have
\((1-x)S=1+x+x^2+…\infty=\frac{1}{1-x}\Rightarrow S=\frac{1}{(1-x)^2}\).
Thus if we have \(S\ge 4\), we must have \((x-1)^2\le \frac{1}{4}\Rightarrow -\frac{1}{2}\le x-1\le \frac{1}{2}\Rightarrow x\ge \frac{1}{2}\).
Q-15) Let \(X\) be the set consisting of the first \(2018\) terms of the arithmetic progression \(1, 6, 11, ….,\) and \(Y\) be the set consisting of the first \(2018\) terms of the arithmetic progression \(9, 16, 23, …..\) . Find the number of elements in the set \(X\bigcup Y\).
Solution:
Note that \(n(X)=n(Y)=2018\). Also \(n\left( X\bigcup Y \right)=n(X)+n(Y)-n\left(X\bigcap Y \right)\).
Now let’s see the common terms of this sequence :
If the xth term of \(X\) is the yth term of \(Y\), then we have:
\(1+(x-1)\times 5= 9+(y-1)\times 7\Rightarrow 5x-4=7y+2\Rightarrow y=\frac{5x-6}{7}\).
Also note that both \(x,y \le 2018\). We’ll see when we cover the topic of congruences modulo primes later on, that this equation has a solution if and only if \(x\) is a number of the form \(7k+4\) with \(k\in \mathbb{Z^+}\bigcup \left \{ 0 \right \}\). Since \(7k+4\le 2018 \Rightarrow k\le 287\). Thus we have \(n\left(X\bigcap Y \right)=288\).
We have \(n\left(X\bigcup Y\right)=2018+2018-288=3748\).
<<In case there is any confusion over any example (solved or not), problem or concept, please feel free to create a thread in the appropriate forum. Our memebrs/experts will try their best to help solve the problem for you>>