Sequences and Series

Given log?/? x + log?/? x + log?/? x + . 20 times = 460
⇒ (2+3+4+.+21)log?x = 460
⇒ (20/2)(2+21)log?x = 460
⇒ log?x = 2
⇒ x = 49

a? + a? = 4 ⇒ a? + a?r = 4
a? + a? = 16 ⇒ a?r² + a?r³ = 16
⇒ r = ±2
r = 2 ⇒ a? = 4/3
r = -2 ⇒ a? = -4
Σ(i=1 to 9) a? = (a(r?-1))/(r-1) = (-4)((-2)? - 1)/(-2-1)
= 4/3 (-513) = 4λ
⇒ λ = -171

Let a, ar, ar² . G.P.
T? + T? + T? = 3 ⇒ ar (1+r+r²) = 3
T? + T? + T? = 243 ⇒ ar? (1+r+r²) = 243
by (i) and (ii)
r? = 81 ⇒ r=3
∴ a = 1/13
S? = a (r? -1)/ (r-1) = (3? -1)/26

$3+4+8+9+13+14+\dots .$ . upto 40 terms

$\Rightarrow 7+17+27+\dots \dots ..20$ terms

$\mathrm{S}=\frac{20}{2}[2*7+19*10]$

$=102*20=102\mathrm{m}$

$\therefore \mathrm{m}=20$

2log? (2? -5) = log?2 + log? (2? -7/2).
(2? -5)² = 2 (2? -7/2) = 2*2? -7.
Let t=2? (t-5)² = 2t-7.
t²-10t+25=2t-7 ⇒ t²-12t+32=0 ⇒ (t-4) (t-8)=0.
t=4 or t=8.
2? =4 ⇒ x=2. log? (4-5) undefined.
2? =8 ⇒ x=3. log? (8-5)=log?3=1. log? (8-3.5)=log?4.5.
2 (1) = log?2+log?4.5 = log?9=2. Correct.
x=3.

${\displaystyle \sum _{n=1}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}$

put 2n + 1 = r, r = 3, 5, 7,.

so n = $\frac{r-1}{2}$

$Now{\displaystyle \sum _{r=\left(3,5,7,.....\right)}^{\infty }\frac{n^2+6n+10}{\left(2n+1\right)!}}={\displaystyle \sum _{r=\left(3,5,7.....\right)}^{\infty }\left(\frac{\frac{{\left(r-1\right)}^2}{4}+3r-3+10}{r!}\right)}={\displaystyle \sum _{r=\left(3,5,7,......\right)}^{\infty }\left(\frac{r^2+10r+29}{4.r!}\right)}$

$=\frac{1}{8}\{41e-\frac{19}{e}-80)=\frac{41}{8}e-\frac{19}{8}{e}^{-1}-10$

Given $P_n={\alpha }^n+{\beta }^n,P_{n-1}=11\&P_{n+1}=29$

    $P_n={\alpha }^{n-2}.{\alpha }^2+{\beta }^{n-2}.{\beta }^2.........\left(i\right)$                          

Now quadratic equation having roots α & β will be x² – (α + β)x + αβ = 0

i.e.         x² – x – 1 = 0,     put x = α and put x = β

So          α² = α + 1           & β² = β + 1

(i)     $P_n={\alpha }^{n-2}\left(\alpha +1\right)+{\beta }^{n-2}\left(\beta +1\right)$

$P_n=P_{n-1}+P_{n-2}$          

= > ${P_n}^2=234$  

Given sequence is -16, 8, -4, 2, .

are in G.P. with first term a = -16 & common ratio r =  $\frac{-1}{2}$

Now $t_p=a{r}^{P-1}=-16{\left(-\frac{1}{2}\right)}^{p-1}\&t_q=a{r}^{q-1}=-16{\left(-\frac{1}{2}\right)}^{q-1}$

So A.M. = -8   $\left[{\left(-\frac{1}{2}\right)}^{p-1}+{\left(\frac{-1}{2}\right)}^{q-1}\right]=\alpha \left(let\right)$

Given equation is 4x² – 9x + 5 = 0 gives $x=1,\frac{5}{4}$

From roots we get possible value of b = 1 so

$16{\left(-\frac{1}{2}\right)}^{\frac{P+q-2}{2}}=1OR{\left(-\frac{1}{2}\right)}^{\frac{p+q-2}{2}}=\frac{1}{16}-{\left(-\frac{1}{2}\right)}^4$

$\Rightarrow \frac{p+q-2}{2}=4\Rightarrow p+q=10$

Let a, ar, ar², . an increasing G.P then r > 1 & a > 0

given : ar + ar5 = $\frac{25}{2}$ .(i)

and $a{r}^{2}a{r}^4=25\Rightarrow a^2{r}^6=25\Rightarrow {\left(a{r}^3\right)}^2=25$

$\Rightarrow a{r}^3=5.......\left(ii\right)$

From (i) and (ii), $\frac{ar\left(1+r^4\right)}{a{r}^3}=\frac{25}{2*5}$

$\Rightarrow 2+2r^4=5r^2\Rightarrow 2r^4-5r^2+2=0$

$\Rightarrow \left(2r^2-1\right)\left(r^2-2\right)=0$

$\Rightarrow r^2=\frac{1}{2},2\therefore r^2=2\Rightarrow r=\sqrt[]{2}$

$\therefore t_4+t_6+t_8=5+10+20=35$

Put x = $\frac{1}{3}$

S = 1 + 2x + 7x² + 12x³ + 17x⁴ + 22x⁵ + .

$\underset{\_}{xS=x+2x^2+7x^3+12x^4+17x^5+.........}$

Subtracting,

$\left(1-x\right)S=1+x+5x^2+5x^3+5x^4+5x^5+......$

= $1-4x+5x+5x^2+5x^3+.........$

$=\frac{1-x-4x+4x^2+5x}{1-x}=\frac{4x^2+1}{1-x}$

$S=\frac{4x^2+1}{{\left(1-x\right)}^2}putx=\frac{1}{3}wegets=\frac{\frac{4}{9}+1}{\frac{4}{9}}=\frac{13}{4}$