Ncert Solutions Maths class 11th

35. Let a and r be the first term and the common ratio between G.P.

So, a₅= p $\Rightarrow$ a r^t-1 $$ $$ = p $\Rightarrow$ ar⁴ = p

a₈ = q $\Rightarrow$ a r^8-1 =q $\Rightarrow$ ar⁷= q

a₁₁= r $\Rightarrow$ ar^11-1 = r $\Rightarrow$ ar¹⁰= 5

So, L.H.S. q^{2 =}  (ar⁷)² = a²r¹⁴

R.H. S. = p.s= (ar⁴) (ar)¹⁰ = a¹⁺¹ r⁴⁺¹⁰  = a²r¹⁴

$\therefore$ L.H.S. = R.H.S.

34. Given, r=2

Let a be the first term,

Then, a₈= 92

$\Rightarrow$ ar^8-1=192

$\Rightarrow$a (2)⁷ = 192

$\Rightarrow$ a = $\frac{192}{2^7}=\frac{192}{128}=\frac{3}{2}$

so, a = $\frac{3}{\begin{array}{l}2\\ \end{array}}$

$\therefore$ a₁₂ = ar^12-1= $\begin{array}{l}\left(\frac{3}{2}\right)\\ \end{array}$ (2)¹¹ = 3 2^11-1

= 3*2¹⁰

= 31024

= 3072.

33. Here, a= $\frac{5}{2}$

=  $r=\frac{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{2}$

so, a_n =  ar^n-1

i.e  a₂₀=ar^20-1= $\left(\frac{5}{2}\right)*{\left(\frac{1}{2}\right)}^{19}=\frac{5}{2}*\frac{1}{2^{19}}=\frac{5}{2^{19+1}}=\frac{5}{2^{20}}$

and an= ar^n-1= $\left(\frac{5}{2}\right)*{\left(\frac{1}{2}\right)}^{n-1}=\frac{5}{2}*\frac{1}{2^{n-1}}=\frac{5}{2^{n-1+1}}=\frac{5}{2^n}$

31. Since the man starts paying installmentas? 100 and increases? 5 every month.

a=100

d=5.

So, the A.P is 100,105,110,115, ….

? Amount of 30thinstallment=30th term of A.P =a₃₀

=a+ (30 – 1)d

=100+29 * 5.

=100+145

= ?245

i.e., He will pay? 245 in his 30th instalment.

29. Given,

A.M between a and b= $\frac{a+b}{2}$

And $\frac{a^n+b^n}{a^{n-1}+b^{n-1}}=\frac{a+b}{2}$

2 [aⁿ+bⁿ]= (a+b) (a^{n – 1}+b^{n – 1})

2aⁿ+2bⁿ=a^{n – 1}.a+ab^{n – 1}+ba^{n – 1} + b^{n – 1}.b

2aⁿ+2bⁿ=a^{n – 1 + 1}+ ab^{n – 1}+ba^{n – 1} + b^{n – 1 + 1}

2aⁿ+2bⁿ=aⁿ+ ab^{n – 1}+b^{n – 1} + bⁿ

2aⁿ – aⁿ – ban ^{– 1} = ab^{n – 1} + bⁿ– 2bⁿ

aⁿ – ba^{n – 1}=ab^{n – 1} – bn

a^{n – 1} [a – b]=b^{n – 1} [a – b] [?   aⁿ=a^{n – 1+1}=a^{n – 1} .a¹]

a^{n – 1}=b^{n – 1}.

$\frac{a^{n-1}}{b^{n-1}}=1$ .

${\left(\frac{a}{b}\right)}^{n-1}={\left(\frac{a}{b}\right)}^0$ [? a⁰=1].

n – 1=0

n=1.

28. Let A₁, A₂, A₃, A₄and A₅ be five numbers between 8 and 26.

So, the A.P is 8, A₁, A₂, A₃, A₄, A₅, 26. with n=7 terms.

Also, a = 8.

l=26, last term.

a+ (7 – 1)d=26

8+6d=26

6d=26 – 8

d= $\frac{18}{6}$

d=3

So,

A₁=a+d=8+3=11

A₂=a+2d=8+2 * 3=8+6=14

A₃=a+3d=8+3 * 3=8+9=17.

A₄=a+9d=8+4 * 3=8+12=20

A₅=a+5d=8+5 * 3=8+15=23.

Hence the reqd. five nos are 11,14,17,20 and 23.

27. Given,

Sum of n terms of AP, S_n=3n²+5n

Put n=1,

S₁=3 * 1²+5 * 1=3+5=8=a₁ (?   S₁=sum of 1* terms of AP)

Put n=2,

S₂=3 * 2²+5 * 2=3 * 4+10=12+10 =22

=a₁+a₂ (?   Sum of first two term of A.P)

So,  a₁+a₂=22

8 +a₂=22

a₂=22 – 8

a₂=14

? First term,  a=a₁=8

Common difference,  d=a₂ – a₁=14 – 8=6

Now, given that, a_m = 164.

a+ (m – 1)d=164.

8+ (m – 1)6=164.

(m – 1)6=164 – 8.

m – 1= $\frac{156}{6}$

m=26+1

m=27.

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

Then, $\frac{\mathrm{Sum\; of\; M\; terms\; of\; A.P.}}{\mathrm{SumofntermsofA.P.}}=\frac{m^2}{n^2}$

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

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

Dividing both sides by m/n we get,

$\frac{2a+(m-1)d}{2a+(n-1)d}=\frac{m}{n}$

[2a+(n – 1)d]n=m[2a+(n – 1)d].

2an+dmn – dn=2am+dmn – dm.

2an – 2am=dn – dm+2mn – dmn

2a(n – m)=d(n – m)

d= $\frac{2a(n-m)}{(n-m)}$

d=2a.

Now, Ratio of m^th  and n^thterm

= $\frac{a+(m-1)d}{a+(n-1)d}$

Putting d=2a in the above we get,

ratio of mth and nth term = $\frac{a+(m-1)2a}{a+(n-1)2a}$

= $\frac{a+2am-2a}{a+2an-2a}$

= $\frac{2am-a}{2an-a}$

= $\frac{a[2m-1]}{a[2n-1]}$

= $\frac{2m-1}{2n-1}$ .