### Question 1. Find the sum of odd integers from 1 to 2001.

**Solution:**

The odd integers forms an A.P. with

Common difference(d)=2 and first term (a)=1

Here, a+(n-1)d=2001

⇒1+(n-1)2=2001

⇒2n-2=2000

⇒n=1001

S

_{n }= n[2a + (n – 1)d]/2S

_{1001}=1001[2(1)+(1001-1)2]/2=1001[2+2000]/2

=1001[2002]/2

=1001[1001]

=1002001

Therefore, the sum of odd integers from 1 to 2001 is 1002001.

### Question 2. In an A.P. the first term is 2 and the sum of the first 5 terms is one-fourth of the next 5 terms. Show that the 20^{th }term is -112.

**Solution:**

The first term(a)=2.

Let the common difference be d.

Sum of first 5 terms=10+10d.

Sum of next 5 terms=10+35d.

According to given condition ,

10+10d=¼(10+35d)

⇒40+40d=10+35d

⇒5d=-30

⇒d=-6

a

_{20}=2+(20-1)(-6)=2-(19)6=2-114=-112

Hence, proved

### Question 3. If the sum of first n terms of an A.P. is (pn + qn^{2}), where p and q are constants. Find the common difference.

**Solution:**

We know that,

S_{n }= n[2a + (n – 1)d]/2According to the given condition,

n[2a+(n-1)d]/2 = pn+qn

^{2}⇒n[2a+nd-d]/2 = pn+qn

^{2}⇒na+n

^{2}(d/2)-n(d/2) = pn+qn^{2}Comparing the coefficients of n

^{2}on both sides we get,d/2=q ⇒ d=2q

Therefore, the common difference is 2q.

### Question 4. How many terms of the A.P. -6, -11/7, -5,… are needed to give the sum -25?

**Solution:**

Let the number of terms to give the sum -25 be n.

We know

Swhere n is the number of terms, a is the first term and d is the common difference._{n}= n(2a+(n-1)d)/2Here, a=-6 and d=(-11/2)+6=(-11+12)/2=1/2.

Substituting these values in the formula of sum we get,

-25=n[2(-6)+(n-1)(1/2)]/2

⇒-50=n[-12+(n/2)-(1/2)]

⇒-50=n[(-25/2)+(n/2)]

⇒-100=n(-25+n)

⇒n

^{2}-25n+100=0⇒n

^{2}-5n-20n+100=0⇒n(n-5)-20(n-5)=0

⇒n=20 or 5

Therefore, the number of terms required is 5 or 20.

### Question 5. In an A.P. if the p^{th} term is 1/q and the q^{th }term is 1/p. Prove that the sum of the first pq terms is (pq+1)/2 where p ≠ q.

**Solution:**

It is known that the general term of an A.P. is

a._{n}=a+(n-1)dGiven,

p

^{th}term=a_{p}=a+(p-1)d=1/q —(1)q

^{th}term = a_{q}= a+(q-1)d = 1/p —(2)Subtracting (2) from (1):

(p-1)d-(q-1)d=(1/q)-(1/p)

⇒(p-1-q+1)d=(p-q)/pq

⇒(p-q)d=(p-q)/pq

⇒d=1/pq

Putting value of d in (1):

a+(p-1)/pq=1/q

⇒a=(1/q)-(1/q)+(1/pq)=1/pq

Therefore,

S

_{pq }= pq[2a+(pq-1)d]/2= pq[(2/pq)+((pq-1)/pq)]/2

=1+(pq-1)/2

=(pq+1)/2

The sum of the first pq terms of the A.P. is (pq+1)/2.

### Question 6. If the sum of certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term.

**Solution:**

Let the number of terms required to give the sum of 116 be n

We know, S

_{n}=n[2a+(n-1)d]/2Given, a=25 and d= 22-25=-3

Substituting the values we get,

116=n[2(25)+(n-1)(-3)]/2

⇒232=n[50-3n+3]

⇒232=n(53-3n)=53n-3n

^{2}⇒3n

^{2}-53n+232=0⇒3n

^{2}-24n-29n+232=0⇒3n(n-8)-29(n-8)=0

⇒(n-8)(3n-29)=0

⇒n=8 or 29/3

But, n cannot be 29/3.Therefore, n=8.

a

_{8}is the last term=a+(n-1)d=25+(8-1)(-3)=25+(7)(-3)=25-21=4

Therefore, the last term is 4.

### Question 7. Find the sum of n terms of A.P. if it’s k^{th} term is given as 5k+1.

**Solution:**

Given, k

^{th }term is 5k+1k

^{th}term = a_{k}= a+(k-1)da+(k-1)d=5k+1

⇒ a + kd – d=5k+1

Comparing the coefficients on both sides, we get d=5

a-d=1⇒a-5=1⇒a=6

S

_{n}=n[2a+(n-1)d]/2= n[2(6)-(n-1)5]/2

=n[12+5n-5]/2

=

n(5n+7)/2.

### Question 8. Find the sum of all natural numbers between 100 and 1000, which are multiples of 5.

**Solution:**

The natural numbers between 100 and 1000 which are multiples of 5 are, 105, 110, . . . , 995. It can be seen as an A.P. starting with a= 105 and common difference d=5.

a+(n-1)d=995

⇒105+(n-1)5=995

⇒(n-1)5=995-105=890

⇒n-1=178

⇒n=179

S

_{n}=179[2(105)+(179-1)5]/2=179[2(105)+(178)5]/2

=179[105+(89)5]

=179[105+445]

=179[550]

=98450

The required sum is 98450.

### Question 9. The sum of n terms of two A.P. are in the ratio 5n+4:9n+6. Find the ratio of their 18^{th} terms.

**Solution:**

Let a

_{1}, a_{2}and d_{1}, d_{2}be the first term and the common difference of the first and the second A.P. respectively.Given,

(Sum of n terms of first A.P)/(Sum of n terms of second A.P.)=(5n+4)/(9n+6)

⇒{n[2a

_{1}-(n-1)d_{1}]/2}/{n[2a_{2}+(n-1)d_{2}]/2}=(5n+4)/(9n+6)⇒[2a

_{1}+(n-1)d_{1}]/[2a_{2}+(n-1)d_{2}]=(5n+4)/(9n+6) —(1)Substituting n=35 in (1),

(2a

_{1}+34d_{1})/(2a_{2}+34d_{2})=[5(35)+4]/[9(35)+6]⇒[a

_{1}+17d_{1}]/[a_{2}+17d_{2}]=179/321 —(2)(18th term of the first A.P)/(18th term of the second A.P.)=(a

_{1}+17d_{1})/(a_{2}+17d_{2}) —(3)R.H.S. of (3) =L.H.S. of (2). So, we get,

(18th term of the first A.P)/(18th term of the second A.P.)=179/321

Therefore, the ratio of the 18th term of both the A.P. is 179:321.

### Question 10. If the sum of first p terms of an A.P. is equal to the sum of the first q terms then find the sum of the first (p + q) terms.

**Solution:**

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

Given,

p[2a+(p-1)d]/2=q[2a+(q-1)d]/2

⇒p[2a+(p-1)d]=q[2a+(q-1)d]

⇒2ap+pd(p-1)=2aq+qd(q-1)

⇒2a(p-q)+d[p(p-1)-q(q-1)]=0

⇒2a(p-q)+d[p

^{2}-p-q^{2}+q]=0⇒2a(p-q)+d[(p-q)(p+q)-(p-q)]=0

⇒2a(p-q)+d[(p-q)(p+q-1)]=0

⇒2a+d(p+q-1)=0

⇒d=-2a/(p+q-1) —(1)

So, S

_{p+q}=(p+q)[2a+(p+q-1)d]/2Putting value of d ,

⇒S

_{p+q}=(p+q)[2a-2a]/2=0

Therefore, the sum of first (p+q) terms of the A.P. is 0.

### Question 11. Sum of first p, q, and r terms of an A.P. are a, b and c respectively. Prove that {a(q-r)/p}+{b(r-p)/q}+{c(p-q)/r=0.

**Solution:**

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

Given,

S

_{p}=p[2a1+(p-1)d]/2=a⇒2a1=(p-1)d=2a/p —(1)

Sq=q[2a1+(q-1)d]/2=b

⇒2a1+(q-1)d=2b/q —(2)

Sr=r[2a1+(r-1)d]/2=c

⇒2a1+(r-1)d=2c/r —(3)

Subtracting (2) from (1), we get

(p-1)d-(q-1)d=(2a/p)-(2b/q)

⇒d(p-1-q+1)=(2aq-2bq)/pq

⇒d=2(aq-bp)/pq(p-q) —(4)

Subtracting (3) from (2),we get

(q-1)d-(r-1)d=(2b/q)-(2c/r)

⇒d(q-1-r+1)=(2b/q)-(2c/r)

⇒d(q-r)=(2br-2qc)/qr

⇒d={2(br-qc)}/{qr(q-r)} —(5)

Equating both values of d in (4) and (5),we obtain

(aq-bp)/(pq(p-q))=(br-qc)/(qr(q-r))

⇒qr(q-r)(aq-bq)=pq(p-q)(br-qc)

⇒r(aq-bp)(q-r)=p(br-qc)(p-q)

⇒(aqr-bpr)(q-r)=(bpr-pqc)(p-q)

Dividing both sides by pqr,

{(a/p)-(b/q)}(q-r)={(b/q)-(c/r)}(p-q)

⇒(a/p)(q-r)-(b/q)(q-r+p-q)+(c/r)(p-q)=0

⇒(a(q-r)/p)+(b(r-p)/q)+(c(p-q)/r)=0.

Hence, proved

### Question 12. The ratio of sum of m and n terms of A.P. is m^{2}:n^{2}. Show that the ratio of m^{th} and n^{th} term is (2m-1):(2n-1).

**Solution:**

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

Given,

(Sum of m terms)/(Sum of n terms)=m

^{2}/n^{2}⇒[m(2a+(m-1)d)/2]/[n(2a+(n-1)d)/2]=m

^{2}/n^{2}⇒[2a+(m-1)d]/[2a+(n-1)d]=m/n —(1)

Putting m=2m-1 and n=2n-1 in (1), we get

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

⇒[a+(m-1)d]/[a+(n-1)d]=[2m-1]/[2n-1] —(2)

[m

^{th}term of A.P.]/[n^{th}term of A.P.]=[a+(m-1)d]/[a+(n-1)d] —(3)R.H.S. of (3) = L.H.S. of (2). So, we get,

[m

^{th}term of A.P.]/[n^{th}term of A.P.]=(2m-1)/(2n-1)

Hence, proved

### Question 13. If the sum of n terms of an A.P. is 3n^{2}+5n and it’s m^{th} term is 164, find the value of m.

**Solution:**

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

a

_{m}=a+(m-1)d=164 —(1)S

_{n}=n[2a+(n-1)d]/2Given,

n[2a+nd-d]/2=3n

^{2}+5n⇒na+(n

^{2}d/2)=3n^{2}+5nComparing coefficients on both sides,

d/2=3⇒d=6 and a-(d/2)=5⇒a-3=5 ⇒ a=8

From (1),

8+(m-1)6=164

⇒ (m-1)6=164-8=156

⇒ m-1=26

⇒ m=27

Therefore, the value of m is 27.

### Question 14. Insert 5 numbers between 8 and 26 such that the resulting sequence is in A.P.

**Solution:**

Let A

_{1}, A_{2}, A_{3}, A_{4}and A_{5}be 5 numbers between 8 and 26 such that it forms an A.P.a=8, b=26, n=7

We can see, 26=8+(7-1)d

⇒6d=26-8=18

⇒d=3

A

_{1}=a+d=8+3=11A

_{2}=a+2d=8+2(3)=8+6=14A

_{3}=a+3d=8+3(3)=8+9=17A

_{4}=a+4d=8+4(3)=8+12=20A

_{5}=a+5d=8+5(3)=8+15=23

Therefore, the required 5 numbers between 8 and 26 to form an A.P. are 11, 14, 17, 20 and 23.

### Question 15. If (a^{n }+ b^{n })/(a^{n-1 }+ b^{n-1}) is the A.M. between a and b. Find the value of n.

**Solution:**

A.M. of a and b=(a+b)/2

According to given,

(a+b)/2=(a

^{n}+b^{n})/(a^{n-1}+b^{n-1})⇒(a+b)(a

^{n-1}+b^{n-1})=2((a^{n}+b^{n})⇒a

^{n}+ab^{n-1}+ba^{n-1}+b^{n}=2a^{n}+2b^{n}⇒ab

^{n-1}+a^{n-1}b=a^{n}+b^{n}⇒ab

^{n-1}-b^{n}=a^{n}-a^{n-1}b⇒b

^{n-1}(a-b)=a^{n-1}(a-b)⇒b

^{n-1}=a^{n-1}⇒(a/b)

^{n-1}=1=(a/b)^{0}⇒n-1=0

⇒

n=1

### Question 16. Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7^{th }and (m-1)^{th }term is 5:9. Find the value of m.

**Solution:**

Let A

_{1}, A_{2}, . . ., A_{m}be m numbers such that 1, A_{1}, A_{2}, . . ., A_{m}, 31 is an A.P.a=1, b=31, n=m+2

31=1+(m+2-1)d

⇒30=(m+1)d

⇒d=30/(m+1) —(1)

A

_{1}=a+dA

_{2}=a+2dA

_{3}=a+3dA

_{4}=a+4d . . .A

_{7}=a+7dA

_{m}=a+(m-1)dAccording to given,

(a-7d)/[a+(m-1)d]=5/9

⇒[1+7(30/(m+1))]/[1+(m-1)(30/m+1)]=5/9 [From (1)]

⇒[m+1+7(30)]/[m+1+30(m-1)]=5/9

⇒[m+1+210]/[m+1+30m-30]=5/9

⇒[m+211]/[31m-29]=5/9

⇒9m+1899=155m-145

⇒155m-9m=1899+145

⇒146m=2044

⇒m=14

Therefore, the value of m is 14.

### Question 17. A man starts repaying a loan as first installment of Rs. 100. If he increases the installment by Rs. 5 every month, what amount he will pay in the 30^{th }installment?

**Solution:**

Given, he first installment of the loan is Rs 100. The second installment of the loan in Rs 105 and so on. We can see that the amount that the man pays every month forms an A.P.

First term, a=100. Common difference, d=5

a

_{30}=a+(30-1)d=100+(29)5

=100+145

=245

Therefore, the installment to be paid in the 30^{th}installment is Rs 245.

### Question 18. The angles of a polygon will form an A.P. with common difference d as 5° and first term a as 120°. Find the number of sides of the polygon.

**Solution:**

It is known that the sum of all the angles of a polygon with n sides is 180°(n-2).

That is, S

_{n}=180°(n-2)⇒n[2a+(n-1)d]/2=180°(n-2)

⇒n[240+(n-1)5]=360(n-2)

⇒240n+5n

^{2}-5n=360n-720⇒5n

^{2}+235n-360n+720=0⇒5n

^{2}-125n+144=0⇒n

^{2}-16n-9n+144=0⇒n(n-16)-9(n-16)=0

⇒(n-9)(n-16)=0

⇒

n = 9 or 16