- 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
Sn = n/2 [2a + (n – 1)d]
Sn = 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
Sn = n/2 [2a + (n – 1)d]
Sn = 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 20th 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
a20 = a + (20-1)d = 2+(19)(-6) = 2 -114 = -112
Thus the 20th 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
Sn = 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)
n2 – 25n + 100 = 0
n2 – 5n – 20n + 100 = 0
n(n-5) – 20(n – 5) = 0
n = 20 or 5
- In an A.P. if pth term is 1/q and qth 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 an = a + (n – 1)d
According to the given information
pth term = ap = a +(p – 1)d = 1/q ———————(1)
qth term = aq = 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
Spq = pq/2 [2a +(pq – 1)d]
= pq/2 [2/pq + (pq – 1)1/pq]
= 1 + 1/2(pq – 1)
= 1/2pq + 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
Sn = n/2 [2a + (n – 1)d]
Here a = 25 and d = 22 – 25 = -3
Sn = n/2 [2 x 25 + (n – 1)(-3)]
116 = n/2 [50 – 3n + 3]
232 = n(53 – 3n) = 53n – 3n2
3n2 – 53n + 232 = 0
3n2 – 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, a8 = 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 kth term is 5k + 1.
Solution:
It is given that the k th term of the A.P. is 5k + 1
k th term = ak=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
Sn = 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 + qn2), where p and q are constants find the common difference.
Solution:
It is known that Sn = n/2[2a + (n – 1)d]
According to the given condition,
n/2[2a + (n – 1)d] = pn + qn2
n/2[2a + nd – d] = pn + qn2
na + n2 .d/2 – n.d/2 = pn + qn2
Comparing the coeffients of n2 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 18th terms
Solution:
Let a1, a2 and d1, d2 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.
Sp = p/2[2a +(p-1)d]
Sq = 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[p2-p-q2+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
Sp+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
