Your cart is empty.
Practice each question, then open the answer or worked solution.
Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that
Let a1 and d be the first term and the common difference of the A.P., respectively.
Then according to the question, we have
Now, subtracting (2) from (1), we get
The ratio of the sums of m and n terms of an A.P. is m2: n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).
Let’s consider a and b to be the first term and the common difference of the A.P., respectively.
Then from the question, we have
Hence, the given result is proved.
If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.
Let’s consider a and b to be the first term and the common difference of the A.P., respectively.
am = a + (m – 1)d = 164 … (1)
The sum of the terms is given by,
Sn = n/2 [2a + (n-1)d]
Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.
Let’s assume A1, A2, A3, A4, and A5 to be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, and 26 are in an A.P.
Here we have,
a = 8, b = 26, n = 7
So,
26 = 8 + (7 – 1) d
6d = 26 – 8 = 18
d = 3
Now,
A1 = a + d = 8 + 3 = 11
A2 = a + 2d = 8 + 2 × 3 = 8 + 6 = 14
A3 = a + 3d = 8 + 3 × 3 = 8 + 9 = 17
A4 = a + 4d = 8 + 4 × 3 = 8 + 12 = 20
A5 = a + 5d = 8 + 5 × 3 = 8 + 15 = 23
Therefore, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.
If 
is the A.M. between a and b, then find the value of n.
The A.M. between a and b is given by, (a + b)/2
Then according to the question,
Thus, the value of n is 1.
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.
Let’s consider a1, a2, … am be m numbers such that 1, a1, a2, … am, 31 is an A.P.
And here,
a = 1, b = 31, n = m + 2
So, 31 = 1 + (m + 2 – 1) (d)
30 = (m + 1) d
d = 30/ (m + 1) ……. (1)
Now,
a1 = a + d
a2 = a + 2d
a3 = a + 3d …
Hence, a7 = a + 7d
am–1 = a + (m – 1) d
According to the question, we have
Therefore, the value of m is 14.
A man starts repaying a loan as the first instalment of Rs. 100. If he increases the instalment by Rs. 5 every month, what amount will he pay in the 30th instalment?
Given,
The first instalment of the loan is Rs. 100.
The second instalment of the loan is Rs. 105, and so on, as the instalment increases by Rs. 5 every month.
Thus, the amount that the man repays every month forms an A.P.
And the A.P. is 100, 105, 110, …
Where, the first term, a = 100
The common difference, d = 5
So, the 30th term in this A.P. will be
A30 = a + (30 – 1)d
= 100 + (29) (5)
= 100 + 145
= 245
Therefore, the amount to be paid in the 30th instalment will be Rs. 245.
The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°, find the number of the sides of the polygon.
It’s understood from the question that the angles of the polygon will form an A.P. with a common difference d = 5° and first term a = 120°.
And, we know that the sum of all angles of a polygon with n sides is 180° (n – 2).
Thus, we can say
Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.