Sequences and Series

84. Let a, ar and $a{r}^2$ be the three nos. which is in G.P.

Then, a + ar + ar² = 56

a ( 1 + r + r²) =56  -I

Given, that a1, ar 7, ar² - 21 from an AP we have,

$(ar-7)-(a-1)=\left(a{r}^2-21\right)-(ar-7)$

$\Rightarrow ar-7-a+1=a{r}^2-21-ar+7$

$\Rightarrow ar-a-6=a{r}^2-ar-14$

$\Rightarrow a{r}^2-ar-ar+a-a-14-6$

$\Rightarrow a{r}^2-2ar+a=8$

$\Rightarrow a\left(r^2-2r+1\right)=8$ ………………. II

Now, dividing equation I by II we get,

$\Rightarrow \frac{a\left(1+n+n^2\right)}{a\left(r^2-2r+1\right)}=\frac{56}{8}$

$\Rightarrow 1+r+r^2=7\left(n^2-2n+1\right)$

$\Rightarrow 1+r+r^2=7r^2-14n+7$

$\Rightarrow 7r^2-14n+7-1-x-r^2=0$

$\Rightarrow 6r^2-15r+6=0$

$\Rightarrow 2r^2-5r+2=0$ (dividing by 3 throughout)

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

$\Rightarrow 2r(r-2)-(r-2)=0$

$\Rightarrow (r-2)(2r-1)=0$

$\Rightarrow r-2=02r-1=0$

$\Rightarrow r=2r=\frac{1}{2}$

So, when r = 2, putting in equation I,

$\Rightarrow a\left(1+2+2^2\right)=56$

$\Rightarrow a(1+2+4)=56$

$\Rightarrow a(7)=56$

$\Rightarrow a=\frac{56}{7}=8$

The numbers are 8, 8* 2, 8* 2² = 8, 16, 32.

And When $r=\frac{1}{2}$ putting in equation I,

$a\left(1+\frac{1}{2}+\frac{1}{2^2}\right)=56$

$\Rightarrow a\left(1+\frac{1}{2}+\frac{1}{4}\right)=56$

$\Rightarrow a\left(\frac{4+2+1}{4}\right)=56$

$\Rightarrow a*\frac{7}{4}=56$

$\Rightarrow a=56*\frac{4}{7}=32$

So, the numbers are $32,32*\frac{1}{2},32*{\left(\frac{1}{2}\right)}^2\Rightarrow 32,16,8$

83. Given, a = 1

$a_3+a_5=90$

Let r be the common ratio of the G.P.

So,

Let r be the common ratio of the G.P.

So,

$a_3+a_5=a{r}^{3?1}+a{r}^{5?1}=90$

$?a\left[r^2+r^4\right]=90$

$?1\left[r^2+r^4\right]=90\text{?}\{?a=1\}$

$?r^4+r^2?90=0$

Let $x=r^2$ so we can write above equation as

$x^2+x?90=0x=r^2$

82. Given, a = 5

$r=2>1$

$s_n=315$

So,  $\frac{a\left(r^n-1\right)}{r-1}=315$

$\Rightarrow \frac{5\left(2^n-1\right)}{2-1}=315$

$\Rightarrow 2^n-1=\frac{315}{5}$

$\Rightarrow 2^n=63+1$

$\Rightarrow 2^n=64=26$

$\therefore n=6$

Hence, last term $=a_6=a{r}^{6-1}=a{r}^5=5*2^5=5*32$

= 160

81. Given, $f(x+y)=f(x)\cdot f(y).Ux,y\in N$ and $f\left(1\right)=3$

Putting (x, y) = (1+1) we get

Putting (x, y) = (1,1) we get,

$f(1+1)=f(1)\cdot f(1)=3\cdot 3=9$

$\Rightarrow f(2)=9$

And putting $(x,y)=(1,2)$ we get,

$f(1+2)=f(1)\cdot f(2)=3*9=27$

$f(3)=27$

$f(1)+f(2)+f(3)+?+f(x)={\displaystyle \sum _{x=1}^{n}f}(x)=120$ (Given)

As, With a = 3

$r=\frac{9}{3}=3>1$

We can write equation I as ,

$\frac{a\left(r^n-1\right)}{r-1}=120$

$\Rightarrow \frac{3\left(3^n-1\right)}{3-1}=120$

$\frac{3}{2}\left(3^n-1\right)=120$

$\Rightarrow \left(3^n-1\right)=120*\frac{2}{3}$

3ⁿ 1 = 80

3ⁿ = 81 +1

3ⁿ = 81

3ⁿ = 3⁴

n = 4

80. Two digits no. when divided by 4 yields 1 as remainder are, 12+1, 16+1, 20+1 …., 96+1

13, 17, 21, ………97 which forms an A.P.

So, a = 13

$d=17-13=4$

$l=97$

$\Rightarrow a+(x-1)d=97$

$\Rightarrow 13+(x-1)4=97$

$\Rightarrow \left(x-1\right)4=97-13=84$

$\Rightarrow x-1=\frac{84}{4}=21$

$\Rightarrow x=21+1=22$

Sum of numbers in A.P. = $\frac{x}{2}(a+l)$

$=\frac{22}{2}(13+97)$

= 11* 110

= 1210

79. An A.P. of numbers from 1 to 100 divisible by 2 is

2, 4, 6, ……….98, 100.

So, a = 2 and d = 4-2 = 2

$l=$ 100

$\Rightarrow a+(n-1)d=100$

$\Rightarrow 2+(n-1)2=100$

$\Rightarrow \frac{2}{2}+(n-1)\frac{2}{2}=\frac{100}{2}$

$\Rightarrow 1+(n-1)=50$

$\Rightarrow n=50$

Let, $s_1=2+4+6?+9+100=\frac{50}{2}[2+100]\left\{?S_n=\frac{n}{2}(a+1)\right\}$

$=\frac{50*102}{2}$

= 2550

Similarly, an A.P. of numbers from 1 to 100 divisible by 5 is

5,10,15, ……. 95,100

So, a = 5 and d = 100

$l=$ 100

$\Rightarrow a+(n-1)d=100$

$\Rightarrow 5+(n-1)5=100$

$\Rightarrow 1+(n-1)=20$

$\Rightarrow n=20$

$\therefore S_2=5+10+15\dots +95+100=\frac{20}{2}[5+100]$

$=\frac{20*105}{2}$

= 1050

As there are also no divisible by both 2 and 5 , i.e., LCM of 2 and 5 = 10 An A.P. of no. from 1 to 100 divisible by 10 is 10, 20, …………100

So, a = 10, d = 10

$l=100$

$\Rightarrow a+(x-1)l=100$

$\Rightarrow 10+(x-1)10=100$

$\Rightarrow 1+(x-1)=10$

$\Rightarrow x=10$

So, $S_3=10+20+30+\dots +100=\frac{10}{2}[10+100]$

= 550

The required sum of number $=S_1+S_2{S}_3=2550+1050-550=3050$

= 3050

78. $\begin{array}{ccccc}\hfill 7\frac{28}{200}\hfill & \hfill \frac{57}{400}\hfill & \hfill \frac{14}{\frac{60}{\frac{56}{4}}}\hfill & \hfill \frac{35}{\begin{array}{l}50\\ \frac{49}{1}\\ \end{array}}\hfill & \hfill \hfill \\ \hfill \hfill & \hfill Smallest =200-4+7=203\hfill & \hfill \hfill & \hfill Largest =400-1=399\hfill & \hfill \hfill \end{array}$

So we can form an A.P. 203,203+7, …., 399-7, 399

So, i.e. a = 203 and d = 7 $$

l = 399

$\Rightarrow a+(x-1)d=399$

$\Rightarrow 203+(x-1)7=399$

$\Rightarrow (x-1)7=399-203$

$\Rightarrow x-1=\frac{196}{7}$

$\Rightarrow x=28+1$

$\Rightarrow x=29$

Sum of the 29 number of the AP = Required sum = $\frac{\eta }{2}[a+l)$

$=\frac{29}{2}[203+399]$

$=\frac{29}{2}*602$

= 8729.

77. Let a and d be the first term and common difference of an A.P.

Then, 

$n_{}=\frac{n}{2}[2a+(n-1)d]=S_1$

$S_{2n}=\frac{2n}{2}[2a+(2n-1)d]=S_2$

$S_{3n}=\frac{3n}{2}[2a+(3n-1)d]=S_3$

Now, R.H.S $=3\left(S_2-S_1\right)$

$=3\left\{\frac{2n}{2}[2a+(2n-1)d]-\frac{x}{2}[2a+(n-1)d]\right\}$

$=\frac{3n}{2}\{2[2a+(2n-1)d-[2a+(x-1)d]\}$

$=\frac{3n}{2}\{4a+(4x-2)d-2a-(x-1)d\}$

$=\frac{3n}{2}\{2a+\left[4x-2\left(n-1\right)d\right]\}$

$=\frac{3n}{2}\{2a+[4n-2-n+1]d\}$

$=\frac{3n}{2}\{2a+[3n-1]d\}=S_3$

$\therefore S_3=3\left(S_2-s_1\right)$

76. Let the three numbers a $-$ d, a, a + d be in A.P.

Then, $\left(a-d\right)+a+\left(a+d\right)=24$

$\Rightarrow$ 3a = 24

$\Rightarrow$ a = 8

and, $(a-d)*a*(a+d)=440$

$\Rightarrow (a-d)\left(a^2+ad\right)=440$

$\Rightarrow a^3+a^{2}d-a^{2}d-a{d}^2=440$

$\Rightarrow a^3-a{d}^2=440.$

$\Rightarrow a\left(a^2-d^2\right)=440$

Put, $a=8,\Rightarrow 8\left(a^2-d^2\right)=440$

$\Rightarrow 64-d^2=\frac{440}{8}$

$\Rightarrow 64-d^2=55$

$d^2=64-55$

$\Rightarrow d^2=9$

$\Rightarrow d=\pm 3$

When d = 3, a = 8 the three number are.

8 $-$ 3, 8, 8 + 3 5,8,11

When d = $-$ 3, a = 8 the three numbers are.

$8-(-3),8,8+(-3)\Rightarrow 8+3,8,8-3\Rightarrow 11,8,5$

75. Let a and d be the first term and common difference of the A.P.

So,  $a(m+n)+a(m-n)=\{a+[(m+n)-1]d\}+\{a+[[m-n)-1]d\}$

$=a+\left(m+n-1\right)d+a+\left(m-n-1\right)d$

$=2a+\left(m+n-1+m-n-1\right)d$

$=2a+\left(2m+0-2\right)d$

$=2[a+(m-1)d]$

= 2 am