**Find the sum of odd integers from 1 to 2001**

Solution

The odd integers from 1 to 2001 are 1, 3, 5 … 1999,2001

This sequence from an A.P.

Here first term is a = 1

Common difference d = 2

Then, a + (n – 1)d = 2001

1 + (n – 1)2 = 2001

n = 1001

S_{n} = ^{n}/_{2} [2a + (n – 1)d]

S_{n} = ^{1001}/_{2} [2×1 + (1001 – 1)x2]

= ^{1001}/_{2} x [2 + 100×2]

= 1001 x 1001

= 1002001

Thus the sum of odd numbers from 1 to 2001 is 1002001

**Find the sum of all natural numbers lying between 100 to 1000 which are multiples of 5.**

Solution:

The natural numbers lying between 100 and 1000 which are multiples of 5 are 105, 110 … 995

Here a = 105 and d = 5

a + (n – 1)d = 2001

105 + (n-1)5 = 995

(n-1)5 = 995 – 105 = 890

n – 1 = 178

n = 179

S_{n} = ^{n}/_{2} [2a + (n – 1)d]

S_{n} = ^{179}/_{2} [2×105 + (179 – 1)x5]

= 179[105 + 89×5]

= 179 x 550

= 98450

Thus the sum of all natural numbers lying between 100 and 10000 which are multiples of 5 is 98450

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

Solution:

First term = 2

Let d be the common difference of the A.P.

Therefore the A.P is 2, 2 + d, 2 + 2d, 2 + 3d…

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

According to the given condition

10 + 10d = ^{1}/_{4}(10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a_{20} = a + (20-1)d = 2+(19)(-6) = 2 -114 = -112

Thus the 20^{th} term of the is -112

**How many terms o the A.P -6, –**^{11}/_{2}, -5… are needed to give the sum -25?

Solution:

Let the sum of n terms of the given A.P. be -25

It is known that

S_{n} = ^{n}/_{2} [2a + (n – 1)d]

Here a = -6

d = –^{11}/_{2} + 6 = ^{-11+12}/_{2} = ^{1}/_{2}

Therefore, we obtain

-25 = ^{n}/_{2} [ 2x(-6)+(n-1)(^{1}/_{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

**In an A.P. if p**^{th}term is^{1}/_{q}and q^{th}term is^{1}/_{p}prove that the sum of first pq terms is^{1}/_{2}(pq + 1), where p≠q

Solution:

It is known that the general term of an A.P. is a_{n} = a + (n – 1)d

According to the given information

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 (1) from (1), we obtain

(p – 1)d – (q-1)d = ^{1}/_{q} – ^{1}/_{p}

(p-1-q+1)d = ^{p-q}/_{pq}

d = ^{1}/_{pq}

Putting the value of d in (1), we obtain

a+(p-1)^{1}/_{pq} = ^{1}/_{q}

a = ^{1}/_{q } – ^{1}/_{q} + ^{1}/_{pq} = ^{1}/_{pq}

S_{pq} = ^{pq}/_{2} [2a +(pq – 1)d]

= ^{pq}/_{2} [^{2}/_{pq} + (pq – 1)^{1}/_{pq}]

= 1 + ^{1}/_{2}(pq – 1)

= ^{1}/_{2}pq + 1 – ^{1}/_{2} = ^{1}/_{2 }pq + ^{1}/_{2}

= ^{1}/_{2}(pq + 1)

Thus the sum of first pq terms of the A.P. is ^{1}/_{2}(pq + 1)

**If the sum of a certain number of terms of the A.P 25, 22, 19,… is 116, find the last term.**

Solution:

Let the sum of n terms of the guven A.P. be the 116

S_{n} = ^{n}/_{2} [2a + (n – 1)d]

Here a = 25 and d = 22 – 25 = -3

S_{n} = ^{n}/_{2} [2 x 25 + (n – 1)(-3)]

116 = ^{n}/_{2} [50 – 3n + 3]

232 = n(53 – 3n) = 53n – 3n^{2}

3n^{2} – 53n + 232 = 0

3n^{2} – 24n – 29n + 232 = 0

(n – 8)(3n – 29) = 0

n = 8 or n = ^{29}/_{3}

However n cannot be equal to ^{29}/_{3}. Therefore n = 8

Thus, a_{8 }= last term = a + (n – 1)d = 25 + (8 – 1)(-3)

= 25 + 7(-3) = 25 – 21

= 4

Thus, the last term of the A.P. is 4

**Find the sum to n terms of the A.P., whose k**^{th}term is 5k + 1.

Solution:

It is given that the k ^{th} term of the A.P. is 5k + 1

k th term = a_{k}=a+(k – 1)d

a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

Comparing the coefficient of k, we obtain d = 5, a – d = 1

a – 5 = 1

a = 6

S_{n} = ^{n}/_{2}[2a + (n – 1)d]

= ^{n}/_{2}[2(6) + (n – 1)5]

= ^{n}/_{2}[12 + 5n – 5]

= ^{n}/_{2}(5n + 7)

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

Solution:

It is known that S_{n} = ^{n}/_{2}[2a + (n – 1)d]

According to the given condition,

^{n}/_{2}[2a + (n – 1)d] = pn + qn^{2}

^{n}/_{2}[2a + nd – d] = pn + qn^{2}

na + n^{2} .^{d}/_{2} – n.^{d}/_{2} = pn + qn^{2}

Comparing the coeffients of n^{2} on both sides, we obtain

^{d}/_{2} = q

d = 2q

Thus the common difference of A.P. is 2q

**The sum of n terms of two arithmetic progressions 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 terms and the common difference of the first and second arithmetic progression respectively.

According to the given condition,

^{Sum of n terms of first A.P.}/ _{Sum of n terms of second A.P }= ^{5n+4}/_{9n+6}

Subtracting n = 35 in (1), we obtain

**If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then ind the sum of the first (p + q) terms.**

Solution:

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

S_{p} = ^{p}/_{2}[2a +(p-1)d]

S_{q} = ^{q}/_{2}[2a +(q-1)d]

According to the given condition

^{p}/_{2}[2a +(p-1)d] = ^{q}/_{2}[2a +(q-1)d]

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

2ap + pd(p – 1) = 2aq + qd(q – 1)

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}

S_{p+q} = ^{p+q}/_{2}[2a+d(p+q-1)]

= ^{p+q}/_{2}[2a+(^{-2a}/_{p+q-1} )(p+q-1)]

= ^{p+q}/_{2}[2a – 2a]

=0

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