NCERT Solutions · Class 11 Mathematics Sequences and Series

Turn the functional equation into a G.P.Put \(\displaystyle y=1\) in \(\displaystyle f(x+y)=f(x)f(y)\): \[f(x+1) = f(x)\,f(1) = 3f(x). \]So each value is \(\displaystyle 3\) times the previous one: \(\displaystyle f(1), f(2), f(3),\dots\) is a G.P. with first term \(\displaystyle f(1)=3\) and common ratio \(\displaystyle 3\), i.e. \(\displaystyle f(x)=3^{x}\).Summing \(\displaystyle n\) terms of this G.P., \[\sum_{x=1}^{n} f(x) = \frac{3\left(3^{n}-1\right)}{3-1} = \frac{3\left(3^{n}-1\right)}{2}. \]Set this equal to \(\displaystyle 120\): \[\frac{3\left(3^{n}-1\right)}{2} = 120 \;\Longrightarrow\; 3^{n}-1 = 80 \;\Longrightarrow\; 3^{n} = 81 = 3^{4}. \]Check: \(\displaystyle 3+9+27+81 = 120\).Hence \(\displaystyle n = 4\).

Sum of a G.P.Here \(\displaystyle a=5\), \(\displaystyle r=2\) and \(\displaystyle \mathrm{S}_{n}=315\). Since \(\displaystyle r\neq 1\), \[\mathrm{S}_{n} = \frac{a\left(r^{n}-1\right)}{r-1} = \frac{5\left(2^{n}-1\right)}{2-1} = 5\left(2^{n}-1\right). \]So \[5\left(2^{n}-1\right) = 315 \;\Longrightarrow\; 2^{n}-1 = 63 \;\Longrightarrow\; 2^{n} = 64 = 2^{6} \;\Longrightarrow\; n = 6. \]The last term is the \(\displaystyle 6^{\text{th}}\) term: \[a_{6} = ar^{5} = 5 \times 2^{5} = 5 \times 32 = 160. \]Check: \(\displaystyle 5+10+20+40+80+160 = 315\).Number of terms \(\displaystyle =6\), last term \(\displaystyle =160\).

Quadratic in \(\displaystyle r^{2}\).With first term \(\displaystyle 1\) and common ratio \(\displaystyle r\), the third and fifth terms are \(\displaystyle r^{2}\) and \(\displaystyle r^{4}\). Given \[r^{2}+r^{4} = 90. \]Write \(\displaystyle t=r^{2}\): \[t^{2}+t-90 = 0 \;\Longrightarrow\; (t+10)(t-9)=0 \;\Longrightarrow\; t = -10 \text{ or } t = 9. \]Since \(\displaystyle t=r^{2}\ge 0\), we reject \(\displaystyle t=-10\) and take \(\displaystyle r^{2}=9\), so \(\displaystyle r=\pm 3\).Check for \(\displaystyle r=3\): third term \(\displaystyle 9\), fifth term \(\displaystyle 81\), and \(\displaystyle 9+81=90\). The same holds for \(\displaystyle r=-3\).Hence \(\displaystyle r = 3\) or \(\displaystyle r = -3\).

Two conditions in the two unknowns \(\displaystyle a\) and \(\displaystyle r\).Let the three numbers in G.P. be \(\displaystyle a,\; ar,\; ar^{2}\).Sum condition — call it ($\displaystyle 1$): \[a\left(1+r+r^{2}\right) = 56. \]After subtracting \(\displaystyle 1, 7, 21\) the numbers \(\displaystyle a-1,\; ar-7,\; ar^{2}-21\) are in A.P., and in an A.P. twice the middle term equals the sum of the outer two: \[2(ar-7) = (a-1)+\left(ar^{2}-21\right) \;\Longrightarrow\; 2ar-14 = a+ar^{2}-22, \] \[ar^{2}-2ar+a = 8 \;\Longrightarrow\; a(r-1)^{2} = 8. \]Call this last one ($\displaystyle 2$).Divide ($\displaystyle 1$) by ($\displaystyle 2$): \[\frac{1+r+r^{2}}{(r-1)^{2}} = \frac{56}{8} = 7 \;\Longrightarrow\; 1+r+r^{2} = 7\left(r^{2}-2r+1\right), \] \[6r^{2}-15r+6 = 0 \;\Longrightarrow\; 2r^{2}-5r+2 = 0 \;\Longrightarrow\; (2r-1)(r-2)=0, \] so \(\displaystyle r=2\) or \(\displaystyle r=\tfrac{1}{2}\).For \(\displaystyle r=2\): equation ($\displaystyle 2$) gives \(\displaystyle a(2-1)^{2}=8\), so \(\displaystyle a=8\) and the numbers are \(\displaystyle 8, 16, 32\).For \(\displaystyle r=\tfrac{1}{2}\): equation ($\displaystyle 2$) gives \(\displaystyle a\left(\tfrac{1}{2}-1\right)^{2}=8\), i.e. \(\displaystyle \tfrac{a}{4}=8\), so \(\displaystyle a=32\) and the numbers are \(\displaystyle 32, 16, 8\) — the same three numbers in reverse order.Check: \(\displaystyle 8+16+32=56\), and \(\displaystyle 8-1,\;16-7,\;32-21\) give \(\displaystyle 7, 9, 11\), an A.P. with common difference \(\displaystyle 2\).The numbers are \(\displaystyle 8, 16, 32\) (or \(\displaystyle 32, 16, 8\)).

Sum of all terms versus the sum of the odd-placed terms.Let the G.P. have \(\displaystyle 2n\) terms, first term \(\displaystyle a\) and common ratio \(\displaystyle r\) (\(\displaystyle r\neq 1\)). Then \[\mathrm{S}_{\text{all}} = \frac{a\left(r^{2n}-1\right)}{r-1}. \]The terms in odd places are \[a,\; ar^{2},\; ar^{4},\dots,\; ar^{2n-2}, \] which are \(\displaystyle n\) terms of a G.P. with first term \(\displaystyle a\) and common ratio \(\displaystyle r^{2}\): \[\mathrm{S}_{\text{odd}} = \frac{a\left(\left(r^{2}\right)^{n}-1\right)}{r^{2}-1} = \frac{a\left(r^{2n}-1\right)}{r^{2}-1}. \]The condition \(\displaystyle \mathrm{S}_{\text{all}} = 5\,\mathrm{S}_{\text{odd}}\) gives \[\frac{a\left(r^{2n}-1\right)}{r-1} = \frac{5a\left(r^{2n}-1\right)}{r^{2}-1}. \]Cancelling \(\displaystyle a\left(r^{2n}-1\right)\neq 0\) and using \(\displaystyle r^{2}-1=(r-1)(r+1)\), \[1 = \frac{5}{r+1} \;\Longrightarrow\; r+1 = 5 \;\Longrightarrow\; r = 4. \]Check with \(\displaystyle a=2\) and \(\displaystyle 6\) terms: the sum is \(\displaystyle 2730\), and the odd-placed terms \(\displaystyle 2, 32, 512\) sum to \(\displaystyle 546\), with \(\displaystyle 5\times 546 = 2730\).Hence the common ratio is \(\displaystyle 4\).

Componendo and dividendo on each equality.Take the first two ratios: \[\frac{a+bx}{a-bx} = \frac{b+cx}{b-cx}. \]Applying componendo and dividendo (add and subtract the numerator and denominator on each side, then divide), \[\frac{(a+bx)+(a-bx)}{(a+bx)-(a-bx)} = \frac{(b+cx)+(b-cx)}{(b+cx)-(b-cx)} \;\Longrightarrow\; \frac{2a}{2bx} = \frac{2b}{2cx}. \]Since \(\displaystyle x \neq 0\), cancelling \(\displaystyle x\) gives \[\frac{a}{b} = \frac{b}{c} \;\Longrightarrow\; b^{2} = ac. \]Doing exactly the same with the second and third ratios, \[\frac{2b}{2cx} = \frac{2c}{2dx} \;\Longrightarrow\; \frac{b}{c} = \frac{c}{d} \;\Longrightarrow\; c^{2} = bd. \]Putting the two results together, \(\displaystyle \dfrac{b}{a} = \dfrac{c}{b} = \dfrac{d}{c}\): the ratio of each term to its predecessor is the same.Hence \(\displaystyle a, b, c, d\) are in G.P.

Write \(\displaystyle \mathrm{S}\), \(\displaystyle \mathrm{P}\) and \(\displaystyle \mathrm{R}\) in terms of \(\displaystyle a\) and \(\displaystyle r\).Let the \(\displaystyle n\) terms be \(\displaystyle a, ar, ar^{2},\dots, ar^{\,n-1}\), with \(\displaystyle r \neq 1\).Sum: \[\mathrm{S} = \frac{a\left(r^{n}-1\right)}{r-1}. \]Product: \[\mathrm{P} = a^{n}\,r^{\,1+2+\cdots+(n-1)} = a^{n}\,r^{\frac{n(n-1)}{2}}. \]Sum of reciprocals — itself a G.P. with first term \(\displaystyle \dfrac{1}{a}\) and ratio \(\displaystyle \dfrac{1}{r}\): \[\mathrm{R} = \frac{1}{a}\cdot\frac{1-\left(\dfrac{1}{r}\right)^{n}}{1-\dfrac{1}{r}} = \frac{1}{a}\cdot\frac{r^{n}-1}{r^{\,n-1}(r-1)}. \]Divide \(\displaystyle \mathrm{S}\) by \(\displaystyle \mathrm{R}\); the factor \(\displaystyle \dfrac{r^{n}-1}{r-1}\) cancels: \[\frac{\mathrm{S}}{\mathrm{R}} = a \cdot a\, r^{\,n-1} = a^{2} r^{\,n-1}. \]Raise to the \(\displaystyle n^{\text{th}}\) power: \[\frac{\mathrm{S}^{n}}{\mathrm{R}^{n}} = \left(a^{2} r^{\,n-1}\right)^{n} = a^{2n} r^{\,n(n-1)} = \left(a^{n} r^{\frac{n(n-1)}{2}}\right)^{2} = \mathrm{P}^{2}. \]Hence \(\displaystyle \mathrm{P}^{2}\mathrm{R}^{n} = \mathrm{S}^{n}\).

Check that the middle term squared equals the product of the outer two.Since \(\displaystyle a, b, c, d\) are in G.P., write \(\displaystyle b=ar\), \(\displaystyle c=ar^{2}\), \(\displaystyle d=ar^{3}\) for the common ratio \(\displaystyle r\).Three numbers are in G.P. exactly when the square of the middle one equals the product of the other two. Test that for \(\displaystyle \left(a^{n}+b^{n}\right), \left(b^{n}+c^{n}\right), \left(c^{n}+d^{n}\right)\).Middle term squared: \[\left(b^{n}+c^{n}\right)^{2} = \left(a^{n}r^{n}+a^{n}r^{2n}\right)^{2} = \left(a^{n}r^{n}\right)^{2}\left(1+r^{n}\right)^{2} = a^{2n}r^{2n}\left(1+r^{n}\right)^{2}. \]Product of the outer terms: \[\left(a^{n}+b^{n}\right)\left(c^{n}+d^{n}\right) = a^{n}\left(1+r^{n}\right)\cdot a^{n}r^{2n}\left(1+r^{n}\right) = a^{2n}r^{2n}\left(1+r^{n}\right)^{2}. \]The two agree, so \[\left(b^{n}+c^{n}\right)^{2} = \left(a^{n}+b^{n}\right)\left(c^{n}+d^{n}\right), \] and the common ratio of the new sequence is \(\displaystyle \dfrac{b^{n}+c^{n}}{a^{n}+b^{n}} = r^{n}\).Hence \(\displaystyle \left(a^{n}+b^{n}\right), \left(b^{n}+c^{n}\right), \left(c^{n}+d^{n}\right)\) are in G.P.

Relations between roots, then the G.P.For \(\displaystyle x^{2}-3x+p=0\) with roots \(\displaystyle a,b\): \[a+b=3,\qquad ab=p.\] For \(\displaystyle x^{2}-12x+q=0\) with roots \(\displaystyle c,d\): \[c+d=12,\qquad cd=q.\]Since \(\displaystyle a,b,c,d\) are in G.P., write them as \(\displaystyle a,\;ar,\;ar^{2},\;ar^{3}\) with \(\displaystyle a\neq 0,\;r\neq 0\). Then \(\displaystyle b=ar,\;c=ar^{2},\;d=ar^{3}\), and the two sum conditions become \[a(1+r)=3,\qquad ar^{2}(1+r)=12.\]Dividing the second by the first (the first is non‑zero, being \(\displaystyle 3\)): \[r^{2}=4 .\]Now express \(\displaystyle p\) and \(\displaystyle q\) through \(\displaystyle a\) and \(\displaystyle r\): \[p=ab=a\cdot ar=a^{2}r,\qquad q=cd=ar^{2}\cdot ar^{3}=a^{2}r^{5}.\]Hence \[\frac{q+p}{q-p}=\frac{a^{2}r^{5}+a^{2}r}{a^{2}r^{5}-a^{2}r}=\frac{r^{4}+1}{r^{4}-1},\] after cancelling \(\displaystyle a^{2}r\;(\neq 0)\). With \(\displaystyle r^{2}=4\) we get \(\displaystyle r^{4}=16\), so \[\frac{q+p}{q-p}=\frac{16+1}{16-1}=\frac{17}{15}.\]The single condition \(\displaystyle r^{2}=4\) covers both admissible G.P.s: \(\displaystyle r=2\) gives \(\displaystyle 1,2,4,8\) (so \(\displaystyle p=2,\;q=32\)), and \(\displaystyle r=-2\) gives \(\displaystyle -3,6,-12,24\) (so \(\displaystyle p=-18,\;q=-288\)); both satisfy \(\displaystyle a+b=3,\;c+d=12\) and both give \(\displaystyle \tfrac{q+p}{q-p}=\tfrac{34}{30}=\tfrac{-306}{-270}=\tfrac{17}{15}\).Therefore \(\displaystyle (q+p):(q-p)=17:15\).

Build \(\displaystyle a+b\) and \(\displaystyle ab\) from the given ratio, then use \(\displaystyle (a-b)^{2}=(a+b)^{2}-4ab\).For two positive numbers \(\displaystyle a,b\), \[\text{A.M.}=\frac{a+b}{2},\qquad \text{G.M.}=\sqrt{ab}.\]Given \(\displaystyle \dfrac{a+b}{2}:\sqrt{ab}=m:n\), the two quantities are proportional to \(\displaystyle m\) and \(\displaystyle n\), so there is a \(\displaystyle k>0\) with \[\frac{a+b}{2}=mk,\qquad \sqrt{ab}=nk,\] that is \[a+b=2mk,\qquad ab=n^{2}k^{2}.\]Then \[(a-b)^{2}=(a+b)^{2}-4ab=4m^{2}k^{2}-4n^{2}k^{2}=4k^{2}\left(m^{2}-n^{2}\right),\] so, taking \(\displaystyle a\ge b\), \[a-b=2k\sqrt{m^{2}-n^{2}}.\](The A.M.–G.M. inequality guarantees \(\displaystyle m\ge n\), so the square root is real.)Adding and subtracting, \[a=\tfrac12\left[(a+b)+(a-b)\right]=k\left(m+\sqrt{m^{2}-n^{2}}\right),\] \[b=\tfrac12\left[(a+b)-(a-b)\right]=k\left(m-\sqrt{m^{2}-n^{2}}\right).\]The common factor \(\displaystyle k\) cancels in the ratio.Hence \(\displaystyle a:b=\left(m+\sqrt{m^{2}-n^{2}}\right):\left(m-\sqrt{m^{2}-n^{2}}\right)\).