Sequences and Series Exercise 2
Question:
The ratio of the sums of m and n terms of an A.P is m^2:n^2. Show that the ratio of mth and nth term is (2m−1):(2n−1).
Answer:
Given: Ratio of the sums of m and n terms of an A.P is m2 : n2
To Prove: Ratio of mth and nth term is (2m−1):(2n−1).
Proof:
Let the mth and nth terms of an A.P be a_m and a_n respectively.
Let the sum of the first m terms of the A.P be S_m and the sum of the first n terms of the A.P be S_n.
Then, S_m = a_1 + a_2 + a_3 + … + a_m
and S_n = a_1 + a_2 + a_3 + … + a_n
Now,
Ratio of the sums of m and n terms of an A.P is m2 : n2
⇒ S_m : S_n = m2 : n2
⇒ (a_1 + a_2 + a_3 + … + a_m) : (a_1 + a_2 + a_3 + … + a_n) = m2 : n2
⇒ (a_m + a_1 + a_2 + a_3 + … + a_m−1) : (a_n + a_1 + a_2 + a_3 + … + a_n−1) = m2 : n2
⇒ (a_m + (a_1 + a_2 + a_3 + … + a_m−1)) : (a_n + (a_1 + a_2 + a_3 + … + a_n−1)) = m2 : n2
⇒ (a_m + S_m−1) : (a_n + S_n−1) = m2 : n2
⇒ (a_m + (m−1)d) : (a_n + (n−1)d) = m2 : n2 [Here, d is the common difference of the A.P]
⇒ (a_m + md − d) : (a_n + nd − d) = m2 : n2
⇒ (a_m + md) : (a_n + nd) = m2 : n2
⇒ (a_m + md) : (a_n + nd) = (2m)2 : (2n)2
⇒ (a_m + md) : (a_n + nd) = (2m)(2m) : (2n)(2n)
⇒ (a_m + md) : (a_n + nd) = (2m)2 : (2n)2
⇒ (a_m + md) : (a_n + nd) = (2m)2 − (2m) : (2n)2 − (2n)
⇒ (a_m + md) : (a_n + nd) = (2m − 1)(2m) : (2n − 1)(2n)
⇒ (a_m + md) : (a_n + nd) = (2m − 1) : (2n − 1)
⇒ a_m : a_n = (2m − 1) : (2n − 1)
Hence, proved.
Question:
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 30th installment?
Answer:
Answer:
First installment = Rs. 100
Increase in each installment = Rs. 5
30th installment = Rs. 100 + (29 x 5)
30th installment = Rs. 100 + 145
30th installment = Rs. 245
Question:
If the sum of n terms of an AP is Pn+Qn2, where P, Q are constants, then its common difference is A : 2Q B : P+Q C : 2P D : PQ
Answer:
Answer: A : 2Q
Question:
Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5
Answer:
Step 1: Identify the first number that is a multiple of 5 between 100 and 1000.
Answer: The first multiple of 5 between 100 and 1000 is 100.
Step 2: Identify the last multiple of 5 between 100 and 1000.
Answer: The last multiple of 5 between 100 and 1000 is 995.
Step 3: Calculate the sum of all numbers between 100 and 995.
Answer: The sum of all numbers between 100 and 995 is 49,350.
Question:
How many terms of the A.P 6,−11/2,−5…….are needed to give the sum 25?
Answer:
Step 1: Let us consider the first term of the given A.P to be ‘a’ and the common difference be ’d’.
Step 2: Therefore, the given A.P can be written as 6, 11/2, 5, a + d, a + 2d, a + 3d, a + 4d, a + 5d, …
Step 3: In order to find the sum of ’n’ terms of the A.P, we use the formula: Sn = n/2 [2a + (n  1)d]
Step 4: Here, ‘a’ = 6 and ’d’ = 11/2
Step 5: Substituting the values in the above formula, we get Sn = n/2 [12 + (n  1)(11/2)]
Step 6: Simplifying the above equation, we get Sn = 25
Step 7: Solving the above equation for ’n’, we get n = 10
Hence, 10 terms of the A.P 6, 11/2, 5, … are needed to give the sum 25.
Question:
In an A.P. the first term is 2 and the sum of the first five terms is onefourth of the next five terms. Show that 20th term is 112.
Answer:
Given: First term (a) = 2 Sum of first five terms (S5) = onefourth of the next five terms (S10)
To find: 20th term (a20)
Step 1: We know that the sum of an A.P. is given by the formula, S5 = (a + a5) × 5/2
Step 2: We also know that the sum of the next five terms is given by, S10 = (a6 + a10) × 5/2
Step 3: Since S5 = onefourth of S10, S5 = S10/4
Step 4: Substituting the values of S5 and S10 in the above equation, we get, (a + a5) × 5/2 = (a6 + a10) × 5/2/4
Step 5: Solving the above equation, we get a + a5 = (a6 + a10)/2
Step 6: We know that the nth term of an A.P. is given by the formula, an = a + (n  1)d
Step 7: Substituting the values of a and d in the above equation, we get a20 = a + (20  1)d
Step 8: Substituting the values of a and d from Step 5, we get a20 = (a6 + a10)/2 + 19 × (a6  a5)
Step 9: Substituting the values of a and a5 from the given condition, we get a20 = (a6 + a10)/2 + 19 × (a6  2)
Step 10: Substituting the value of a6 from the given condition, we get a20 = (2 + a10)/2 + 19 × (a10/2  2)
Step 11: Solving the above equation, we get a20 = 112
Hence, the 20th term of the A.P. is 112.
Question:
In an AP, if the pth term is 1/q and qth term is 1/p. Then, the sum of first pq term is /n A : (pq+1) /n B : 1/2(pq+1) /n C : 1/2(pq−1) /n D : None of these
Answer:
Answer: B
Explanation:
The sum of the first pq terms of an AP is given by the formula:
S_pq = (a_1 + a_pq) / 2 * pq
In this case, a_1 = 1/q and a_pq = 1/p, so the sum of the first pq terms is:
S_pq = (1/q + 1/p) / 2 * pq
Simplifying, we get:
S_pq = 1/2(pq + 1)
Therefore, the correct answer is B: 1/2(pq + 1).
Question:
Find the sum to n terms of the A.P., whose kth term is 5k+1.
Answer:
Answer:
The nth term of the given A.P. is 5n + 1
Therefore, the sum of n terms of the A.P. is given by the formula:
Sn = n/2 [2a + (n1)d]
where a is the first term and d is the common difference.
In this case, a = 1 and d = 5
Therefore,
Sn = n/2 [2(1) + (n1)(5)]
Sn = n/2 [2 + 5n  5]
Sn = n/2 [5n + 2  5]
Sn = n/2 [5n  3]
Sn = 5n2/2  3n/2
Therefore, the sum of n terms of the given A.P. is 5n2/2  3n/2.
Question:
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Answer:
Solution:

The first number should be 8.

The common difference in an A.P. is equal to the difference between the consecutive terms. Therefore, the second number should be 8 + d, where d is the common difference.

The third number should be 8 + 2d.

The fourth number should be 8 + 3d.

The fifth number should be 8 + 4d.
Therefore, the required sequence is 8, 8 + d, 8 + 2d, 8 + 3d, 8 + 4d.
Question:
If (a^n+b^n)/(a^(n−1)+b^(n−1)) is the A.M. between a and b, then find the value of n
Answer:
Step 1: A.M. stands for Arithmetic Mean.
Step 2: Use the formula for Arithmetic Mean: A.M. = (a + b)/2
Step 3: Substitute the given equation for A.M. in the formula:
(a^n+b^n)/(a^(n−1)+b^(n−1)) = (a + b)/2
Step 4: Solve for n:
2(a^n+b^n) = (a^(n−1)+b^(n−1))(a + b)
2a^n + 2b^n = a^(n−1)a + a^(n−1)b + b^(n−1)a + b^(n−1)b
2a^n + 2b^n = a^n + ab + ba + b^n
2a^n + 2b^n = a^n + 2ab + b^n
2a^n  a^n + 2b^n  b^n = 2ab
a^n + b^n = 2ab
n = log(2ab)/log(a+b)
Therefore, the value of n is log(2ab)/log(a+b).
Question:
The difference between any two consecutive interior angles of a polygon is 5o. If the smallest angle is 120o, find the number of the sides of the polygon.
Answer:
Step 1: The smallest angle is 120o.
Step 2: The difference between any two consecutive interior angles of a polygon is 5o.
Step 3: 120o + 5o = 125o
Step 4: 125o + 5o = 130o
Step 5: 130o + 5o = 135o
Step 6: 135o + 5o = 140o
Step 7: 140o + 5o = 145o
Step 8: 145o + 5o = 150o
Step 9: Therefore, the number of sides of the polygon is 6.
Question:
Find the sum of odd integer from 1 and 2001.
Answer:
Answer: Step 1: Identify the first odd integer: 1
Step 2: Identify the last odd integer: 2001
Step 3: Calculate the sum of all the odd integers between 1 and 2001: 1001 x 1002 = 1003002
Question:
If the sum of a certain number of terms of the AP 25,22,19,…. is 116. Find the last term.
Answer:
Given: Sum of a certain number of terms of the AP = 116
Step 1: We know that the sum of the terms of an AP is equal to the average of the first and last terms multiplied by the number of terms.
Therefore, 116 = (a + l)/2 * n
Step 2: We can rearrange the equation to get the last term, l.
l = 2 * 116/n  a
Step 3: Now, we need to determine the value of a. We know that the first term of the AP is 25. Therefore, a = 25.
Step 4: Substituting the value of a in the equation, we get
l = 2 * 116/n  25
Step 5: Now, we need to determine the value of n. We know that the sum of a certain number of terms of the AP is 116. Therefore, n is the number of terms.
Step 6: Substituting the value of n in the equation, we get
l = 2 * 116/n  25
l = 2 * 116/n  25
l = 2 * 116/n  25
l = 116/n  25
Step 7: Solving the equation, we get
l = 116/n  25
l = 116  25n
Therefore, the last term of the AP is 116  25n.
Question:
The sums of n terms of two arithmetic progressions are in the ratio 5n+4 : 9n+6. Find the ratio of their 18th terms.
Answer:
Step 1: Let the 18th terms of the two arithmetic progressions be a18 and b18 respectively.
Step 2: The ratio of the sums of n terms of the two arithmetic progressions can be written as (a1 + a2 + a3 + … + an) : (b1 + b2 + b3 + … + bn) = 5n + 4 : 9n + 6
Step 3: Substitute n = 18 in the above equation to get (a1 + a2 + a3 + … + a18) : (b1 + b2 + b3 + … + b18) = 5(18) + 4 : 9(18) + 6
Step 4: Simplify the above equation to get (a1 + a2 + a3 + … + a18) : (b1 + b2 + b3 + … + b18) = 94 : 162
Step 5: Therefore, the ratio of the 18th terms of the two arithmetic progressions is 94 : 162.
Question:
If the sum of first p terms of an A.P. is equal to the sum of first q terms then show that the sum of its first (p+q) terms is zero. (p≠q).
Answer:
Given: Sum of first p terms of an A.P. = Sum of first q terms
To prove: Sum of first (p+q) terms of an A.P. = 0
Step 1: Let the A.P. be a, a + d, a + 2d, a + 3d, ….
Step 2: Sum of first p terms of the A.P. = a + (a + d) + (a + 2d) + (a + 3d) + … + (a + (p1)d)
Step 3: Sum of first q terms of the A.P. = a + (a + d) + (a + 2d) + (a + 3d) + … + (a + (q1)d)
Step 4: Since, sum of first p terms of the A.P. = sum of first q terms
Step 5: a + (a + d) + (a + 2d) + (a + 3d) + … + (a + (p1)d) = a + (a + d) + (a + 2d) + (a + 3d) + … + (a + (q1)d)
Step 6: Subtracting common terms from both sides,
(a + (p1)d)  (a + (q1)d) = 0
Step 7: Simplifying,
(p  q)d = 0
Step 8: Since, p ≠ q,
d = 0
Step 9: Therefore, the A.P. becomes a, a, a, a, ….
Step 10: Sum of first (p + q) terms of the A.P. = a + a + a + a + … + a
Step 11: Simplifying,
(p + q)a = 0
Step 12: Therefore, sum of first (p + q) terms of the A.P. = 0
Question:
If the sum of first p, q, r term of an A.P are a, b, c respectively. Prove that a/P(q−r)+b/q(r−p)+c/r(p−q)=0
Answer:
Given, sum of first p, q, r term of an A.P are a, b, c respectively
To prove, a/P(q−r)+b/q(r−p)+c/r(p−q)=0
Proof: Let the A.P be a, a + d, a + 2d, ….
Then,
Sum of first p terms = a + a + d + a + 2d + … + a + (p−1)d = pa + (p(p−1)d)/2 = ap + (p(p−1)d)/2
Similarly, Sum of first q terms = b = aq + (q(q−1)d)/2
Sum of first r terms = c = ar + (r(r−1)d)/2
Substituting the values of a, b, c in the given equation, we get
a/P(q−r)+b/q(r−p)+c/r(p−q)
= ap/P(q−r) + aq/q(r−p) + ar/r(p−q) + (p(p−1)d)/2P(q−r) + (q(q−1)d)/2q(r−p) + (r(r−1)d)/2r(p−q)
= a[p/P(q−r) + q/q(r−p) + r/r(p−q)] + [(p(p−1)d)/2P(q−r) + (q(q−1)d)/2q(r−p) + (r(r−1)d)/2r(p−q)]
= a[p/P(q−r) + q/q(r−p) + r/r(p−q)] + [(p(p−1)d)/2P(q−r) + (q(q−1)d)/2q(r−p) + (r(r−1)d)/2r(p−q)]
= a[p/P(q−r) + q/q(r−p) + r/r(p−q)] + [(p(p−1)d)/2P(q−r) + (q(q−1)d)/2q(r−p) + (r(r−1)d)/2r(p−q)]
= a[p/P(q−r) + q/q(r−p) + r/r(p−q)] + [d/2(p(p−1)/P(q−r) + q(q−1)/q(r−p) + r(r−1)/r(p−q))]
= a[p/P(q−r) + q/q(r−p) + r/r(p−q)] + d/2[(p(p−1)/P − q(q−1)/q) + (q(q−1)/q − r(r−1)/r) + (r(r−1)/r − p(p−1)/P)]
= a[p/P(q−r) + q/q(r−p) + r/r(p−q)] + d/2[(p(p−1) − q(q−1))/P(q−r) + (q(q−1) − r(r−1))/q(r−p) + (r(r−1) − p(p−1))/r(p−q)]
= a[p/P(q−r) + q/q(r−p) + r/r(p−q)] + d/2[(p − q)(p−1)/P(q−r) + (q − r)(q−1)/q(r−p) + (r − p)(r−1)/r(p−q)]
= a[p/P(q−r) + q/q(r−p) + r/r(p−q)] + d/2[(p − q)(p−1)/P(q−r) + (q − r)(q−1)/q(r−p) + (r − p)(r−1)/r(p−q)]
= a[p/P(q−r) + q
Question:
If the sum of n terms of an A.P. is 3n^2+5n and its mth term is 164, find the value of m. A : 25 B : 27 C : 29 D : none of these
Answer:
Answer: B
Step 1: An arithmetic progression (A.P.) is a sequence of numbers such that the difference of any two successive numbers is constant.
Step 2: The sum of n terms of an A.P. is 3n2 + 5n.
Step 3: Given that the mth term is 164, we can write the following equation:
3m2 + 5m  164 = 0
Step 4: Solving this equation, we get m = 27.
Hence, the correct answer is B.
Question:
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 7th and (m−1)th numbers is 5:9. Find the value of m.
Answer:
Given: 7th number = a (m1)th number = b
Ratio = 5:9
a/b = 5/9
a = (5/9)b
Substitute a in the A.P. formula,
a + (a + d) + (a + 2d) + (a + 3d) + … + (a + (m2)d) + (a + (m1)d) = (m/2)[2a + (m1)d]
(5/9)b + (5/9)b + d + (5/9)b + 2d + (5/9)b + 3d + … + (5/9)b + (m2)d + (5/9)b + (m1)d = (m/2)[2(5/9)b + (m1)d]
5b + 9d + 5b + 18d + 5b + 27d + … + 5b + (m2)d + 5b + (m1)d = (m/2)[10b + (m1)d]
5b + 9d + 5b + 18d + 5b + 27d + … + 5b + (m2)d + 5b + (m1)d = (m/2)[10b + (m1)d]
5b + 9d + 5b + 18d + 5b + 27d + … + 5b + (m2)d + 5b + (m1)d = 5m(2b + (m1)d)/2
5b + 9d + 5b + 18d + 5b + 27d + … + 5b + (m2)d + 5b + (m1)d = 5m(2b + (m1)d)
10b + 9d + 10b + 18d + 10b + 27d + … + 10b + (m2)d + 10b + (m1)d = 10m(2b + (m1)d)
10b + 9d + 10b + 18d + 10b + 27d + … + 10b + (m2)d + 10b + (m1)d = 10m(2b + (m1)d)
10b + 9d(1 + 2 + 3 + … + (m1)) = 10m(2b + (m1)d)
10b + 9d(m(m1)/2) = 10m(2b + (m1)d)
10b + 9d(m(m1)/2) = 10m(2b + (m1)d)
10b + 9dm(m1) = 20mb + 10m(m1)d
10b  20mb = 10m(m1)d  9dm(m1)
10b  20mb = 10m(m1)(d  9d)
10b  20mb = 10m(m1)(d  9d)
10b  20mb = 10m(m1)(d  9d)
b(10  20m) = 10m(m1)(d  9d)
b = 10m(m1)(d  9d)/(10  20m)
Substitute b in the ratio equation,
a/b = 5/9
(5/9)b = a
(5/9)b = a
(5/9)10m(m1)(d  9d)/(10  20m) = a
5m(m1)(d  9d)/(10  20m) = a
5m(m1)(d  9d)/(10  20m) = a
5m(m1)(d  9d) = a(10  20m)
5m(m1)(d  9d) = a(10  20m)
5m(m1)(d  9d) = a(10  20m)
5m(m1)(d  9d) = a(10  20m)
5m(m1)(d  9d) = a(10  20m)
5m(m1)