Class 11th

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 = $\frac{a+(m-1)2a}{a+(n-1)2a}$

= $\frac{a+2am-2a}{a+2an-2a}$

= $\frac{2am-a}{2an-a}$

= $\frac{a[2m-1]}{a[2n-1]}$

= $\frac{2m-1}{2n-1}$ .

25. Let e and d the first term and common difference of an AP.

Then,

Sum of first p terms, S_p=a.

$\frac{p}{2}[2e+(p-1)d]=a$

$\frac{1}{2}[2e+(p-1)d]=\frac{a}{p}$ .

Similarly, S_q=b

$\frac{q}{2}[2e+(q-1)d]=b$

$\frac{1}{2}[2e+(q-1)d]=\frac{b}{q}$ ----------------(2)

And. S_r=C

$\frac{r}{2}[2e+q(r-1)d]=c$

$\frac{1}{2}[2c+(r-1)d]=\frac{c}{r}$ -----------------(3)

So, L.H.S. $\frac{a}{p}(q-r)+\frac{b}{p}(n-p)$ + $\frac{c}{r}(p-q)$ (? given)

= $\frac{1}{2}[2e+(p-1)d](q-r)$ + $\frac{1}{2}[2e+(q-1)d](r-p)$ + $\frac{1}{2}[2e+(r-1)d](p-q)$

= $\frac{2e}{2}(q-r)+\frac{d}{2}(p-1)(q-r)+\frac{2e}{2}(n-p)$ + $\frac{d}{2}(q-1)(r-p)$ + $\frac{2e(p-q)}{2e}+\frac{d}{2}(r-1)(p-q)$

=c[q – r+r – p+p – q]+ $\frac{d}{2}$ [(p – 1)(q – r)+(q – 1)(r – p)+(r – 1)(p – q)]

=C * 0 + $\frac{d}{2}$ [pq – pr – q+r+qr – pq – r+p+pr=qr – p+q]

=0+ $\frac{d}{2}$ [pq – pq – pr+pr – q+q+r – r+qr+qr+p – p]

= $\frac{d}{2}[0]$

=0.

=R.H.S.

24. Let a and d be the first elements and common different of the A.P.

Then, Sum of first p terms of the AP=Sum of first q terms of A.P

Sp=Sq

$\frac{p}{2}[2a+(p-1)d]=\frac{q}{2}[2a+(q-1)d]$

p [2a+pd – d]=q [2a+qd – d]

2ap+p²d – pd=2aq+q²d – qd

2ap – 2aq+p²d – q²d – pd+qd=0

2a (p – q)+ [p² – q² – p+q]d=0.

2a (p – q)+ [ (p – q) (p+q) – (p – q)]d=0

(p – q) {2a+ [ (p+q) – 1]d}=0

Deviding both sides by p – q,

2a+ [ (p+q) –1]d=0.

And multiplying by P+Q/2 we get,

$\frac{p+q}{2}\{2a+[(p+q)-1]d\}=0$ which in the form $\frac{n}{2}\{2a+(n-1)d\}$ where n=p+q.

S_p+q=0.

23. Let a1, a2 be the first terms and d1, d2 be the common difference of the first term A.P.S

So, $\frac{\mathrm{Sum\; ofntermofIst}{}^{}AP}{\mathrm{Sum\; of\; n\; terms\; of\; 2nd\; AP}}=\frac{5n+4}{9n+6}$

$\frac{\frac{n}{2}\left[2a_1+(n-1)d_1\right]}{\frac{n}{2}\left[2a_2+(n-1)d_2\right]}=\frac{5n+4}{9n+6}$

$\frac{2a_1+(n-1)d}{2a_2+(n-1)d}=\frac{5n+4}{9n+6}$ ---------------(1)

The ratio of their 18thterms.

$\frac{a_{1\mathrm{8offirstA.P}}}{a_{1\mathrm{8ofsecondA.P}}}=\frac{a_1+(18-1)d_1}{a_2+(18-1)d_2}$

= $\frac{a_1+17d_1}{a_2+17d_2}$

= $\frac{a_1+17d_1}{a_2+17d_2}*\frac{2}{2}$

= $\frac{2a_1+34d_1}{2a_2+35d_2}$ --------(2)

Comparing eqn (1) and (2),

$\frac{2a_1+34d_1}{2a_2+35d_2}=\frac{2a_1+(35-1)d_1}{2a_2+(35-1)d_1}=\frac{5*35+4}{9*35+6}$ = $\frac{179}{321}$

So, ratio of their 18th terms = $\frac{179}{321}$

22. Given, sum of n terms of A.P, S_n= (pⁿ+qn²), p&q are constants substituting n=1,

S₁=p* 1+q* 1²=p+q=a₁ [sum upto1^st term only]

And substituting n=2,

S₂=p* 2+q* 2²=2p+4q=a₁+q₂  [sum upto 2^ndtern only]

So,  a₁+a₂=2p+4q

⇒ a₂=2p+4q – a₁=2p+4q – (p+q)=2p – p+4q – q

⇒ a₂=p+3q

So, common difference,  d=a₂ – a₁

= (p+3q) – (p+q)

=p+3q – p – q

=2q.