SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Sequences and Series

64 questions · 64 still being checked

EXERCISE 8.2 21–32 (part 5 of 7)

  1. Exercise 21

    Find four numbers forming a geometric progression in which the third term is greater than the first term by 9\displaystyle 9 , and the second term is greater than the 4th \displaystyle 4{ }^{\text {th }} by 18\displaystyle 18 .

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle 3$, -$\displaystyle 6$, $\displaystyle 12$, -$\displaystyle 24$
    Write the four terms and translate each sentence. Let them be \(\displaystyle a,\ ar,\ ar^{2},\ ar^{3}\)."The third term is greater than the first by $\displaystyle 9$": \[ar^{2} = a + 9 \;\Longrightarrow\; a\left(r^{2} - 1\right) = 9. \tag{1}\]"The second term is greater than the fourth by $\displaystyle 18$": \[ar = ar^{3} + 18 \;\Longrightarrow\; -ar\left(r^{2} - 1\right) = 18. \tag{2}\]Divide ($\displaystyle 2$) by ($\displaystyle 1$) — allowed because the left side of ($\displaystyle 1$) equals \(\displaystyle 9 \neq 0\): \[\frac{-ar\left(r^{2} - 1\right)}{a\left(r^{2} - 1\right)} = \frac{18}{9} \;\Longrightarrow\; -r = 2 \;\Longrightarrow\; r = -2 .\]Substituting \(\displaystyle r = -2\) in ($\displaystyle 1$): \[a(4 - 1) = 9 \;\Longrightarrow\; 3a = 9 \;\Longrightarrow\; a = 3 .\]So the numbers are \[3,\ -6,\ 12,\ -24 .\]Check: third \(\displaystyle -\) first \(\displaystyle = 12 - 3 = 9\) ✓; second \(\displaystyle -\) fourth \(\displaystyle = -6 - (-24) = 18\) ✓.The four numbers are \(\displaystyle 3,\ -6,\ 12,\ -24\).

  2. Exercise 22

    If the pth ,qth \displaystyle p^{\text {th }}, q^{\text {th }} and rth \displaystyle r^{\text {th }} terms of a G.P. are a,b\displaystyle a, b and c\displaystyle c, respectively. Prove that aq−rbr−pcP−q=1.a^{q-r} b^{r-p} c^{P-q}=1 .

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Write every given term in the form \(\displaystyle \mathrm{A}\mathrm{R}^{\,k-1}\). Let the G.P. have first term \(\displaystyle \mathrm{A}\) and common ratio \(\displaystyle \mathrm{R}\) (capitals, so as not to clash with the index \(\displaystyle r\) in the question). Then \[a = \mathrm{A}\mathrm{R}^{\,p-1}, \qquad b = \mathrm{A}\mathrm{R}^{\,q-1}, \qquad c = \mathrm{A}\mathrm{R}^{\,r-1}.\]Substituting into the left-hand side and collecting powers: \[a^{q-r}\,b^{r-p}\,c^{p-q} = \mathrm{A}^{\,(q-r)+(r-p)+(p-q)}\;\mathrm{R}^{\,(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}.\]Exponent of \(\displaystyle \mathrm{A}\): \[(q - r) + (r - p) + (p - q) = 0 .\]Exponent of \(\displaystyle \mathrm{R}\): expand each product, \[(p-1)(q-r) = pq - pr - q + r,\] \[(q-1)(r-p) = qr - pq - r + p,\] \[(r-1)(p-q) = rp - rq - p + q .\] Adding, the products cancel in pairs (\(\displaystyle pq\) with \(\displaystyle -pq\), \(\displaystyle qr\) with \(\displaystyle -rq\), \(\displaystyle -pr\) with \(\displaystyle rp\)) and so do the single letters (\(\displaystyle -q\) with \(\displaystyle +q\), \(\displaystyle +r\) with \(\displaystyle -r\), \(\displaystyle +p\) with \(\displaystyle -p\)), leaving \(\displaystyle 0\).Therefore \[a^{q-r}\,b^{r-p}\,c^{p-q} = \mathrm{A}^{0}\,\mathrm{R}^{0} = 1 .\]Hence \(\displaystyle a^{q-r}\,b^{r-p}\,c^{p-q} = 1\), as required.

  3. Exercise 23

    If the first and the nth \displaystyle n^{\text {th }} term of a G.P. are a\displaystyle a and b\displaystyle b, respectively, and if P is the product of n\displaystyle n terms, prove that P2=(ab)n\displaystyle \mathrm{P}^{2}=(a b)^{n}.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Product of the terms of a G.P.Let the G.P. have first term \(\displaystyle a\) and common ratio \(\displaystyle r\). Then its \(\displaystyle n^{\text{th}}\) term is \[b = ar^{\,n-1}. \]The product of the \(\displaystyle n\) terms is \[\mathrm{P} = a \cdot ar \cdot ar^{2} \cdots ar^{\,n-1} = a^{n}\, r^{\,1+2+\cdots+(n-1)} = a^{n}\, r^{\frac{n(n-1)}{2}}, \] using \(\displaystyle 1+2+\cdots+(n-1)=\dfrac{n(n-1)}{2}\).Squaring, \[\mathrm{P}^{2} = a^{2n}\, r^{\,n(n-1)} = \left(a^{2} r^{\,n-1}\right)^{n} = \left(a \cdot ar^{\,n-1}\right)^{n}. \]Since \(\displaystyle ar^{\,n-1}=b\), \[\mathrm{P}^{2} = (ab)^{n}. \]Hence \(\displaystyle \mathrm{P}^{2}=(ab)^{n}\).

  4. Exercise 24

    Show that the ratio of the sum of first n\displaystyle n terms of a G.P. to the sum of terms from (n+1)th  to (2n)th  term is 1rn.(n+1)^{\text {th }} \text { to }(2 n)^{\text {th }} \text { term is } \frac{1}{r^{n}} .

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Sum of a G.P., applied to two blocks of \(\displaystyle n\) terms.Let the G.P. be \(\displaystyle a, ar, ar^{2},\dots\) with \(\displaystyle r \neq 1\). The sum of the first \(\displaystyle n\) terms is \[\mathrm{S}_{1} = \frac{a\left(r^{n}-1\right)}{r-1}. \]The terms from the \(\displaystyle (n+1)^{\text{th}}\) to the \(\displaystyle (2n)^{\text{th}}\) are \[ar^{n},\; ar^{n+1},\; \dots,\; ar^{2n-1}, \] which are again \(\displaystyle n\) terms of a G.P., with first term \(\displaystyle ar^{n}\) and the same ratio \(\displaystyle r\). So \[\mathrm{S}_{2} = \frac{ar^{n}\left(r^{n}-1\right)}{r-1} = r^{n}\,\mathrm{S}_{1}. \]Therefore \[\frac{\mathrm{S}_{1}}{\mathrm{S}_{2}} = \frac{\mathrm{S}_{1}}{r^{n}\,\mathrm{S}_{1}} = \frac{1}{r^{n}}. \]Hence the required ratio is \(\displaystyle \dfrac{1}{r^{n}}\).

  5. Exercise 25

    If a,b,c\displaystyle a, b, c and d\displaystyle d are in G.P. show that (a2+b2+c2)(b2+c2+d2)=(ab+bc+cd)2.\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(a b+b c+c d)^{2} .

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Grouping the difference as a sum of squares.Since \(\displaystyle a, b, c, d\) are in G.P., consecutive ratios are equal: \[\frac{b}{a}=\frac{c}{b}=\frac{d}{c} \quad\Longrightarrow\quad b^{2}=ac,\qquad c^{2}=bd,\qquad ad=bc. \]Expand the two sides. On the left, \[\left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right) = a^{2}b^{2}+a^{2}c^{2}+a^{2}d^{2}+b^{4}+2b^{2}c^{2}+b^{2}d^{2}+c^{4}+c^{2}d^{2}, \] and on the right, \[(ab+bc+cd)^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}d^{2}+2ab^{2}c+2bc^{2}d+2abcd. \]Subtracting and regrouping the surviving terms into three perfect squares, \[\text{LHS}-\text{RHS} = \left(b^{4}-2ab^{2}c+a^{2}c^{2}\right)+\left(c^{4}-2bc^{2}d+b^{2}d^{2}\right)+\left(a^{2}d^{2}-2abcd+b^{2}c^{2}\right) \] \[= \left(b^{2}-ac\right)^{2}+\left(c^{2}-bd\right)^{2}+\left(ad-bc\right)^{2}. \]Each bracket is \(\displaystyle 0\) by the G.P. relations, so \(\displaystyle \text{LHS}-\text{RHS}=0\).Hence \(\displaystyle \left(a^{2}+b^{2}+c^{2}\right)\left(b^{2}+c^{2}+d^{2}\right)=(ab+bc+cd)^{2}\).

  6. Exercise 26

    Insert two numbers between 3\displaystyle 3 and 81\displaystyle 81 so that the resulting sequence is G.P.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle 9$ and $\displaystyle 27$
    Inserting geometric means.Let the two numbers be \(\displaystyle \mathrm{G}_{1}\) and \(\displaystyle \mathrm{G}_{2}\), so that \[3,\; \mathrm{G}_{1},\; \mathrm{G}_{2},\; 81 \] is a G.P. of four terms with first term \(\displaystyle 3\) and common ratio \(\displaystyle r\). Then \(\displaystyle 81\) is the \(\displaystyle 4^{\text{th}}\) term: \[3r^{3} = 81 \quad\Longrightarrow\quad r^{3}=27 \quad\Longrightarrow\quad r=3. \]Hence \[\mathrm{G}_{1}=3\cdot 3=9, \qquad \mathrm{G}_{2}=3\cdot 3^{2}=27. \]Check: \(\displaystyle 3, 9, 27, 81\) has constant ratio \(\displaystyle 3\).The required numbers are \(\displaystyle 9\) and \(\displaystyle 27\).

  7. Exercise 27

    Find the value of n\displaystyle n so that an+1+bn+1an+bn\displaystyle \frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} may be the geometric mean between a\displaystyle a and b\displaystyle b.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle n=\frac{-1}{2}$
    Set the expression equal to \(\displaystyle \sqrt{ab}\).The geometric mean between \(\displaystyle a\) and \(\displaystyle b\) is \(\displaystyle \sqrt{ab}\), so the condition is \[\frac{a^{n+1}+b^{n+1}}{a^{n}+b^{n}} = \sqrt{ab} = a^{1/2}b^{1/2}. \]Cross-multiplying, \[a^{n+1}+b^{n+1} = a^{n+\frac{1}{2}}b^{\frac{1}{2}} + a^{\frac{1}{2}}b^{\,n+\frac{1}{2}}. \]Collect the \(\displaystyle a\)-terms on one side and the \(\displaystyle b\)-terms on the other: \[a^{n+1}-a^{n+\frac{1}{2}}b^{\frac{1}{2}} = a^{\frac{1}{2}}b^{\,n+\frac{1}{2}}-b^{n+1}, \] \[a^{n+\frac{1}{2}}\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right) = b^{\,n+\frac{1}{2}}\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right). \]For \(\displaystyle a \neq b\) we may cancel \(\displaystyle \left(a^{1/2}-b^{1/2}\right)\), which gives \[a^{\,n+\frac{1}{2}} = b^{\,n+\frac{1}{2}} \quad\Longrightarrow\quad \left(\frac{a}{b}\right)^{n+\frac{1}{2}} = 1 \quad\Longrightarrow\quad n+\frac{1}{2}=0. \]Hence \(\displaystyle n=-\dfrac{1}{2}\).

  8. Exercise 28

    The sum of two numbers is 6\displaystyle 6 times their geometric mean, show that numbers are in the ratio (3+22):(3−22)\displaystyle (3+2 \sqrt{2}):(3-2 \sqrt{2}).

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Componendo and dividendo.Let the two positive numbers be \(\displaystyle a\) and \(\displaystyle b\). Their geometric mean is \(\displaystyle \sqrt{ab}\), so the given condition is \[a+b = 6\sqrt{ab}. \]Apply componendo and dividendo with \(\displaystyle 2\sqrt{ab}\): \[\frac{a+b+2\sqrt{ab}}{a+b-2\sqrt{ab}} = \frac{6\sqrt{ab}+2\sqrt{ab}}{6\sqrt{ab}-2\sqrt{ab}} = \frac{8\sqrt{ab}}{4\sqrt{ab}} = 2. \]The left side is a ratio of perfect squares: \[\frac{\left(\sqrt{a}+\sqrt{b}\right)^{2}}{\left(\sqrt{a}-\sqrt{b}\right)^{2}} = 2 \quad\Longrightarrow\quad \frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}} = \sqrt{2}. \]Apply componendo and dividendo once more: \[\frac{2\sqrt{a}}{2\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1} \quad\Longrightarrow\quad \frac{\sqrt{a}}{\sqrt{b}} = \frac{\sqrt{2}+1}{\sqrt{2}-1}. \]Squaring both sides, \[\frac{a}{b} = \frac{\left(\sqrt{2}+1\right)^{2}}{\left(\sqrt{2}-1\right)^{2}} = \frac{3+2\sqrt{2}}{3-2\sqrt{2}}. \]Hence \(\displaystyle a:b = \left(3+2\sqrt{2}\right):\left(3-2\sqrt{2}\right)\).

  9. Exercise 29

    If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A±(A+G)(A−G)\displaystyle \mathrm{A} \pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    Recover the numbers as roots of a quadratic.Let the two positive numbers be \(\displaystyle a\) and \(\displaystyle b\). By definition of A.M. and G.M., \[\mathrm{A} = \frac{a+b}{2} \;\Longrightarrow\; a+b = 2\mathrm{A}, \qquad \mathrm{G} = \sqrt{ab} \;\Longrightarrow\; ab = \mathrm{G}^{2}. \]A pair of numbers with a known sum and product are the roots of \[x^{2}-(a+b)x+ab = 0, \qquad\text{i.e.}\qquad x^{2}-2\mathrm{A}x+\mathrm{G}^{2}=0. \]By the quadratic formula, \[x = \frac{2\mathrm{A} \pm \sqrt{4\mathrm{A}^{2}-4\mathrm{G}^{2}}}{2} = \mathrm{A} \pm \sqrt{\mathrm{A}^{2}-\mathrm{G}^{2}} = \mathrm{A} \pm \sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}. \](Since \(\displaystyle \mathrm{A}\ge \mathrm{G}\) for positive numbers, the square root is real.)Hence the two numbers are \(\displaystyle \mathrm{A}+\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\) and \(\displaystyle \mathrm{A}-\sqrt{(\mathrm{A}+\mathrm{G})(\mathrm{A}-\mathrm{G})}\).

  10. Exercise 30

    The number of bacteria in a certain culture doubles every hour. If there were 30\displaystyle 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd \displaystyle 2^{\text {nd }} hour, 4th \displaystyle 4^{\text {th }} hour and nth \displaystyle n^{\text {th }} hour ?

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle 120$, $\displaystyle 480$, $\displaystyle 30 ~\left(2^{\mathrm{n}}\right)$
    Doubling every hour gives a G.P. with \(\displaystyle r=2\).The counts at the start, and at the end of the \(\displaystyle 1^{\text{st}}, 2^{\text{nd}}, \dots\) hour form a G.P. with first term \(\displaystyle 30\) and common ratio \(\displaystyle 2\): \[30,\; 30(2),\; 30\left(2^{2}\right),\; 30\left(2^{3}\right),\dots \] so the count at the end of the \(\displaystyle n^{\text{th}}\) hour is \(\displaystyle 30 \cdot 2^{n}\).At the end of the \(\displaystyle 2^{\text{nd}}\) hour: \[30 \cdot 2^{2} = 30 \times 4 = 120. \]At the end of the \(\displaystyle 4^{\text{th}}\) hour: \[30 \cdot 2^{4} = 30 \times 16 = 480. \]End of \(\displaystyle 2^{\text{nd}}\) hour: \(\displaystyle 120\) bacteria; end of \(\displaystyle 4^{\text{th}}\) hour: \(\displaystyle 480\) bacteria; end of \(\displaystyle n^{\text{th}}\) hour: \(\displaystyle 30 \cdot 2^{n}\) bacteria.

  11. Exercise 31

    What will Rs 500\displaystyle 500 amounts to in 10\displaystyle 10 years after its deposit in a bank which pays annual interest rate of 10\displaystyle 10% compounded annually?

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    Rs $\displaystyle 500(1.1)^{10}$
    Compound interest as a G.P.At \(\displaystyle 10\%\) per annum compounded annually, each year's amount is \(\displaystyle 1.1\) times the previous year's, so the amounts at the ends of successive years form a G.P. with common ratio \(\displaystyle 1.1\): \[500(1.1),\; 500(1.1)^{2},\; 500(1.1)^{3},\; \dots \]The amount at the end of \(\displaystyle 10\) years is the \(\displaystyle 10^{\text{th}}\) term: \[\text{Amount} = 500\left(1.1\right)^{10}. \]Numerically, \(\displaystyle (1.1)^{10} = 2.5937424601\), so \[500 \times 2.5937424601 = 1296.87123\ldots \]The amount is Rs \(\displaystyle 500(1.1)^{10} \approx\) Rs \(\displaystyle 1296.87\).

  12. Exercise 32

    If A.M. and G.M. of roots of a quadratic equation are 8\displaystyle 8 and 5\displaystyle 5, respectively, then obtain the quadratic equation.

    Not cross-checked

    This solution has not been cross-checked against the answer printed in NCERT.

    NCERT’s answer
    $\displaystyle x^{2}-16 x+25=0$
    Sum and product of the roots.Let the roots be \(\displaystyle \alpha\) and \(\displaystyle \beta\). Given \[\frac{\alpha+\beta}{2} = 8 \;\Longrightarrow\; \alpha+\beta = 16, \qquad \sqrt{\alpha\beta} = 5 \;\Longrightarrow\; \alpha\beta = 25. \]A quadratic equation with these roots is \[x^{2}-(\alpha+\beta)x+\alpha\beta = 0. \]Substituting, \[x^{2}-16x+25 = 0. \]The required equation is \(\displaystyle x^{2}-16x+25=0\).