Ncert Solutions Maths class 11th

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.

21. Given, a_k =5k+1

Putting k =1,

a₁ =5 * 1+1=5+1=6.

a_n =5n+1=l

So, sum of unto n teams of the AP, S_n = $\frac{n}{2}[2a+l]$

S_n = $\frac{n}{2}[6+5n+1]$

= $\frac{n}{2}[7+5n]$ .

20. The given A.P. is 25,22,19, …

So, a=25

d=22 – 25= –3

Given that, sum of first n terms of the AP=116.

n [2 * 25+ (n – 1) (–3)]=116 * 2

n [50 – 3n+3]=232

n [53 – 3n]=232 .

53n – 3n²=232.

3n² – 53n+232.

Using quadratic formula, a=3, b= –53, c=232.

n = $\frac{53+5}{6}$ or $\frac{53-5}{6}$

= $\frac{58}{6}$ or $\frac{48}{6}$

= $\frac{29}{3}$ or 8.

As n $\in$ N, n=8.

? Last term=a+ (n – 1)d

=a+ (8 – 1)d

=25+7* (–3)

=25 – 21

=4.

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

Then, given a_p = $\frac{1}{q}$

a+(p – 1)d = $\frac{1}{q}$ --------(1)

and a_q = $\frac{1}{p}$

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

Subtracting eqn (2) from (1) we get,

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

a+(p – 1)d– a–(q– 1)d= $\frac{p-q}{pq}$

[(p – 1)–(q – 1)]d= $\frac{p-q}{pq}$

[p – 1 – q+1]d= $\frac{p-q}{pq}$

[p – q]d= $\frac{p-q}{pq}$ .

d= $\frac{1}{pq}$ .

Putting d= $\frac{1}{pq}$ in eqn (1) we get,

$a+(p-1)\frac{1}{pq}=\frac{1}{q}$ .

$a+\frac{1}{pq}-\frac{1}{pq}=\frac{1}{q}$

$a+\frac{1}{q}-\frac{1}{pq}=\frac{1}{q}$

$a=\frac{1}{q}-\frac{1}{q}+\frac{1}{pq}$ $a=\frac{1}{pq}$ .

So, sum of first pq terms,

$S_{pq}=\frac{pq}{2}\left[2*\frac{1}{pq}+(pq-1)\frac{1}{pq}\right]$

= $\frac{pq}{2}*\frac{1}{pq}[2+pq-1]$ .

= $\frac{1}{2}[pq+1]$

18. Let the sum of n farms of the A.P –6, $-\frac{11}{2}$ , –5, …. gives –25.

Then, 

From the given A.P.

a= –6

d= – $\frac{11}{2}-(-6)=-\frac{11}{2}+6=-\frac{11+12}{2}=\frac{1}{2}$

So, –25= $\frac{n}{2}\left[2*(-6)+(n-1)\left(\frac{1}{2}\right)\right]$

–25 * 2=n $\left[-12+\frac{n-1}{2}\right]$

–50=n $\left[\frac{-12*2+(n-1)}{2}\right]$

–50=n $\left[\frac{-24+n-1}{2}\right]$

–50 * 2 =n[n – 25]=n² – 25n

n² – 25n+100=0.

n² – 5n – 20n+100=0

n(n – 5) – 20(n – 5)=0

(n – 5)(n – 20)=0

So, n=5 and n=20.

When n=5

$S_5=\frac{5}{2}\left[-12+(5-1)\frac{1}{2}\right]=\frac{5}{2}\left[-12+4*\frac{1}{2}\right]=\frac{5}{2}[-12+2]=\frac{5}{2}*(-10)$ = –25.

When n=20

$S_{20}=\frac{20}{2}\left[-12+(20-1)\frac{1}{2}\right]=10\left[-12+\frac{19}{2}\right]$ = $-120+\frac{19*10}{2}$ = –120 + 19 * 5 = –120 + 95 = –25.

17. Given, a=2

Let A1, A2, A3, A4, A5, A6, A1, A6, A7, A10 be the first ten terms. So, A1=a.

Given A1+A2+A3+A4+A5= $\frac{1}{4}$  (A6+A1+A8+A1+A10)

a+ (a+d)+ (a+2d)+ (a+3d)+ (a+4d)

= $\frac{1}{4}$  [ (a+5d)+ (a+6d)+ (a+7d)+ (a+8d)+ (a+9d)]

5a+10d= $\frac{1}{4}$  [5a+35d].

4 [5a+10d]=9a+35d.

20a+40d=5a+35d.

40d – 35d=5a – 20a

5d= –15a

d= –3a

d= –3 * 2 [as a=2]

d= –6

So, A₂₀=a+ (20 – 1)d

=2+19 * (–6)

=2 – 114

= –112.

16. Sum of all natural number between 100 and 1000 which are multiple of 5.

=105+110+115+ … +995.

So, a=105.

a=110 – 105=5.

As the last term is 995 which is the nthterm,

a+ (n – 1)d =995.

105+ (n – 1) * 5 =995.

(n – 1) * 5=995 – 105.

(n – 1)5 =890

n – 1 = $\frac{890}{5}=178$

n =178+1

n =179

So, required sum S_n $\frac{n}{2}$ (a+l); l=last term.

$=\frac{179}{2}(105+995)$

$=\frac{179}{2}*1100$

= 179 * 550

=98450.