Chapter 9
Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry · 307 exercises
Problem 157
Show that the sum of \(n\) terms of a G.P. of common ratio \(r\) beginning with the \(p^{t h}\) term is \(r^{p-q}\) times the sum of an equal number of terms of the same series beginning with \(q^{t h}\) term.
5 step solution
Problem 158
Find the sum of \(2 n\) term of a series of which every even term is \(a\) times the term before it, and every odd term \(c\) times the term before it, the first term being unity.
3 step solution
Problem 159
If \(S\) is the sum to infinity of a GP, whose first term is \(a\), then find the sum of the first \(n\) terms.
6 step solution
Problem 160
If \(S\) be the sum, \(P\) the product and \(R\) the sum of the reciprocals of \(n\) terms of a G.P., prove that \(\left(\frac{S}{R}\right)^{n}=P^{2}\).
4 step solution
Problem 161
If \(S=1+a+a^{2}+\ldots \ldots\) to \(\infty(a<1)\), then show that \(a=\frac{S-1}{S}\).
7 step solution
Problem 162
If the sum of the series \(1+\frac{3}{x}+\frac{9}{x^{2}}+\frac{27}{x^{3}}+\ldots\) to \(\infty\) is a finite number, then show that \(x>3\).
3 step solution
Problem 163
Prove that \(\sum_{r=1}^{n} \log \left(\frac{a^{r}}{b^{r-1}}\right)=\frac{n}{2} \log \left(\frac{a^{n+1}}{b^{n-1}}\right)\)
3 step solution
Problem 164
If \(A=1+r^{a}+r^{2 a}+\ldots \ldots \ldots\) to \(\infty\) and \(B=1+r^{b}+r^{2 b}+\ldots \ldots \ldots\) to \(\infty\), prove that \(r=\left(\frac{A-1}{A}\right)^{1 / a}=\left(\frac{B-1}{B}\right)^{1 / b}\)
4 step solution
Problem 165
If \(S_{1}, S_{2}, S_{3}\) be respectively the sums of \(n, 2 n, 3 n\) terms of a G.P., then prove that i. \(\quad S_{1}\left(S_{3}-S_{2}\right)=\left(S_{2}-S_{1}\right)^{2}\) ii. \(\quad S_{1}^{2}+S_{2}^{2}=S_{1}\left(S_{2}+S_{3}\right)\).
2 step solution
Problem 166
If \(S_{1}, S_{2}, \ldots, S_{n}\) are the sums of infinite geometric series whose first terms are \(1,2,3, \ldots n\) and common ratios are \(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots, \frac{1}{n+1}\) respectively then prove that \(S_{1}+S_{2}+S_{3}+\ldots+S_{n}=\frac{1}{2} n(n+3)\).
6 step solution
Problem 167
If \(S_{p}\) denotes the sum of series \(1+r^{p}+r^{2 p}+\ldots \ldots\) to \(\infty\) and \(s_{p}\) the sum of the series \(1-r^{p}+r^{2 p}-\ldots \ldots \ldots\) to \(\infty\), prove that \(S_{p}+s_{p}=2 S_{2 p}\)
4 step solution
Problem 168
If \(S_{n}\) represents the sum of \(n\) terms of a G.P. whose first term and common ratio are \(a\) and \(r\) respectively, then prove that i. \(\quad S_{1}+S_{2}+S_{3}+\ldots+S_{n}=\frac{n a}{1-r}-\frac{\operatorname{ar}\left(1-r^{n}\right)}{(1-r)^{2}}\); ii. \(S_{1}+S_{3}+S_{5}+\ldots+S_{2 n-1}=\frac{a n}{1-r}-\frac{a r\left(1-r^{2 n}\right)}{(1-r)^{2}(1+r)}\).
6 step solution
Problem 169
The sum of the squares of three distinct real numbers, which are in G.P. is \(S^{2}\). If their sum is \(\alpha S\), show that \(\alpha^{2} \in\left(\frac{1}{3}, 1\right) \cup(1,3)\).
4 step solution
Problem 171
Find the natural number \(a\) for which \(\sum_{k=1}^{n} f(a+k)=16\left(2^{n}-1\right)\), where the function \(f\) satisfies the relation \(f(x+y)=f(x) \cdot f(y)\) for all natural numbers \(x, y\) and further \(f(1)=2 .\)
3 step solution
Problem 172
Sum the series \((a+b)+\left(a^{2}+2 b\right)+\left(a^{3}+3 b\right)+\ldots\) to \(n\) terms.
4 step solution
Problem 173
Sum the series \(\left(x+\frac{1}{x}\right)^{2}+\left(x^{2}+\frac{1}{x^{2}}\right)^{2}+\left(x^{3}+\frac{1}{x^{3}}\right)^{2} \ldots\left(x^{n}+\frac{1}{x^{n}}\right)^{2}\).
5 step solution
Problem 174
Sum the series \(1+(1+x)+\left(1+x+x^{2}\right)+\left(1+x+x^{2}+x^{3}\right)+\cdots \cdots\) to \(n\) terms.
4 step solution
Problem 175
Sum the series \(x(x+y)+x^{2}\left(x^{2}+y^{2}\right)+x^{3}\left(x^{3}+y^{3}\right)+\cdots \cdots\) to \(n\) terms.
5 step solution
Problem 176
If \(x=1+a+a^{2}+a^{3}+\ldots\) to \(\infty(|a|<1)\) and \(y=1+b+b^{2}+b^{3}+\ldots \ldots\) to \(\infty(|b|<1)\). Prove that \(1+a b+a^{2} b^{2}+a^{3} b^{3}+\ldots\) to \(\infty=\frac{x y}{x+y-1} .\)
4 step solution
Problem 177
Sum the series \(1+\frac{\sqrt{2}-1}{2 \sqrt{3}}+\frac{3-2 \sqrt{2}}{12}+\frac{5 \sqrt{2}-7}{24 \sqrt{3}}+\frac{17-12 \sqrt{2}}{144}+\ldots \infty\)
3 step solution
Problem 179
Let \(r\) be the common ratio of the G.P. \(a_{1}, a_{2}, \ldots\), show that \(\frac{1}{a_{1}^{\mathrm{m}}+a_{2}^{m}}+\frac{1}{a_{2}^{m}+a_{3}^{m}}+\ldots+\frac{1}{a_{n-1}^{\mathrm{m}}+a_{n}^{\mathrm{m}}}=\frac{1-r^{m(1-n)}}{a_{1}^{m}\left(r^{m}-r^{-m}\right)}\)
4 step solution
Problem 180
Find the sum of the infinite series \(1+(1+a) r+\left(1+a+a^{2}\right) r^{2}+\left(1+a+a^{2}+a^{3}\right) r^{3}+\ldots \ldots ., r\) and \(a\) being proper fractions.
4 step solution
Problem 181
Find \(\lim _{x \rightarrow 1} \frac{x+x^{2}+\ldots \ldots . .+x^{n}-n}{x-1} .\)
4 step solution
Problem 182
Find \(\lim _{n \rightarrow \infty} n^{-n^{2}}\left((n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^{2}}\right) \cdots\left(n+\frac{1}{2^{n-1}}\right)\right)^{n}\)
5 step solution
Problem 183
Insert five geometric means between 486 and \(\frac{2}{3}\).
4 step solution
Problem 184
If \(A\) and \(G\) be the A.M. and G.M. between two numbers, prove that the numbers are \(A \pm \sqrt{(A+G)(A-G)}\).
6 step solution
Problem 185
Construct a quadratic in \(x\) such that A.M. of its roots is \(A\) and G.M. is \(G\).
3 step solution
Problem 186
If one G.M. \(G\) and two arithmetic means \(p\) and \(q\) be inserted between any two given numbers then show that \(G^{2}=(2 p-q)(2 q-p)\).
4 step solution
Problem 187
If one A.M. \(A\) and two geometric means \(p\) and \(q\) be inserted between any two given numbers then show that \(p^{3}+q^{3}=2\) Apq.
6 step solution
Problem 188
The A.M. between \(m\) and \(n\) and the G.M. between \(a\) and \(b\) are each equal to \((m a+n b)(m+n)\). Find \(m\) and \(n\) in terms of \(a\) and \(b\).
5 step solution
Problem 190
For what value of \(n, \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}}\) is the geometric mean of \(a\) and \(b ?\)
4 step solution
Problem 192
If \(A_{1}, A_{2}\) be two AM's and \(G_{1}, G_{2}\) be two GM's between \(a\) and \(b\), then prove that \(\frac{A_{1}+A_{2}}{G_{1} G_{2}}=\frac{a+b}{a b}\).
5 step solution
Problem 193
If the A.M. between \(a\) and \(b\) is twice as great as their G.M. show that \(a: b=(2+\sqrt{3}):(2-\sqrt{3})\).
4 step solution
Problem 194
The A.M. of \(a\) and \(b\) is to their G.M as \(m\) to \(n\), show \(a: b=m+\sqrt{m^{2}-n^{2}}: m-\sqrt{m^{2}-n^{2}}\).
6 step solution
Problem 195
If \(A\) be the arithmetic mean of \(b\) and \(c\) and \(G_{1}, G_{2}\), be the two geometric means between them, then prove that \(G_{1}^{3}+G_{2}^{3}=2 G_{1} G_{2} A\).
4 step solution
Problem 196
The 7th term of a H.P. is \(\frac{1}{10}\) and 12 th term is \(\frac{1}{25}\), find the 20 th term.
5 step solution
Problem 197
Solve the equation \(6 x^{3}-11 x^{2}+6 x-1=0\) if its roots are in harmonic progression.
5 step solution
Problem 198
The value of \(x+y+z\) is 15 if \(a, x, y, z, b\) are in A.P. while the value of \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) is \(\frac{5}{3}\) if \(a, x, y, z, b\) are in H.P., find \(a\) and \(b\).
8 step solution
Problem 199
If \(x, y, z\) are in H.P., prove that \(\log (x+z)+\log (x+z-2 y)=2 \log (x-z)\).
5 step solution
Problem 200
Show that \(\log _{3} 2, \log _{6} 2, \log _{12} 2\) are in HP.
4 step solution
Problem 201
Prove that \(a, b, c\) are in A.P., G.P or H.P. according as the value of \(\frac{a-b}{b-c}\) is equal to \(\frac{a}{a}, \frac{a}{b}\) or \(\frac{a}{c}\) respectively.
3 step solution
Problem 202
If \(a_{1}, a_{2}, a_{3}, \ldots ., a_{n}\) are in harmonic progression, prove that \(a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{n-1} a_{n}=(n-1) a_{1} a_{n} .\)
7 step solution
Problem 203
If \(a, b, c\) be respectively the \(p\) th, \(q\) th and \(r\) th terms of an H.P., then prove that \(b c(q-r)+c a(r-p)+a b(p-q)=0 .\)
5 step solution
Problem 204
If \(p\) th term of an H.P. is \(q r\) and \(q\) th term is \(r p\), prove that \(r\) th term is \(p q\).
4 step solution
Problem 205
If the roots of the equation \(a(b-c) x^{2}+b(c-a) x+c(a-b)=0\) be equal, then prove that \(a, b, c\) are in H.P.
4 step solution
Problem 206
If the \(m\) th term of an H.P. is \(n\) and \(n\) th term be \(m\), then prove that \((m+n)\) th term is \(\frac{m n}{m+n}\).
5 step solution
Problem 207
If \(a, b, c\) be in H.P., prove that i. \(\frac{a-b}{b-c}=\frac{a}{c}\); ii. \(\frac{1}{b-a}+\frac{1}{b-c}=\frac{2}{b}\); iii. \(\frac{b+a}{b-a}+\frac{b+c}{b-c}=2\).; iv. \(\left(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}\right)\left(\frac{1}{b}+\frac{1}{c}-\frac{1}{a}\right)=\frac{4}{a c}-\frac{3}{b^{2}}\); v. \((b c+c a-a b)(c a+a b-b c)=a c\left(4 b^{2}-3 a c\right)\).
6 step solution
Problem 208
If \(a, b, c\) be in H.P. prove that i. \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) are in H.P.; ii. \(\frac{a}{b+c-a}, \frac{b}{c+a-b}, \frac{c}{a+b-c}\) are in H.P. iii. \(\frac{1}{a}+\frac{1}{b+c}, \frac{1}{b}+\frac{1}{c+a}, \frac{1}{c}+\frac{1}{a+b}\) are in H.P.
3 step solution
Problem 209
If \(\frac{1}{a(b+c)}, \frac{1}{b(c+a)}, \frac{1}{c(a+b)}\) be in H.P. then \(a, b, c\) are also in H.P.
5 step solution
Problem 210
If \(b+c, c+a, a+b\) are in H.P. then prove that i. \(\frac{a}{b+c}, \frac{b}{c+a}, \frac{c}{a+b}\) are in A.P.; ii. \(a^{2}, b^{2}, c^{2}\) are in A.P.
4 step solution