In this article we are providing the CBSE Class 10 Mathematics: Important 4 Marks Questions to prepare for the board exam 2018. All these questions are completely solved.
In CBSE Class 10 Mathematics Board Exam 2018, Section - D will comprise 8 questions of 4 marks each.
All these Class 10 Mathematics: Important 4 Marks Questions have been prepared after carrying the thorough analysis of the trend followed in previous years’ board exams and the latest syllabus. Questions have been picked from the most important topics of class 10 Maths. Thus practicing these questions will help cover the important part of the curriculum thus making an effective preparation for the board exams 2018. Moreover, the solutions provided here will give an idea to present your answers in the best way to grab maximum scores.
Given below are some sample questions for CBSE Class 10 Mathematics: Important 4 Marks Questions:
Q. For any positive integer n, prove that n3 – n is divisible by 6.
= product of three consecutive positive integers.
Now, we have to show that the product of three consecutive positive integers is divisible by 6.
We know that any positive integer n is of the form 3q, 3q + 1 or 3q + 2 for some positive integer q.
Now three consecutive positive integers are n, n + 1, n + 2.
Case I. If n = 3q.
n(n + 1) (n + 2) = 3q(3q + 1) (3q + 2)
But we know that the product of two consecutive integers is an even integer.
∴ (3q + 1) (3q + 2) is an even integer, say 2r.
⟹ n(n + 1) (n + 2) = 3q × 2r = 6qr, which is divisible by 6.
Case II. If n = 3n + 1.
∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3)
= (even number say 2r) (3) (q + 1)
= 6r (q + 1),
which is divisible by 6.
Case III. If n = 3q + 2.
∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4)
= multiple of 6 for every q
= 6r (say),
which is divisible by 6.
Hence, the product of three consecutive integers is divisible by 6.
Q. The first and the last terms of an AP are 10 and 361 respectively. If its common difference is 9 then find the number of terms and their total sum?
Given, first term, a = 10
Last term, al = 361
And, common difference, d = 9
al =a + (n −1)
361 = 10 + (n − 1)9
361 = 10 + 9n − 9
361 = 9n + 1
9n = 360
n = 40
Therefore, total number of terms in AP = 40
Now, sum of total number of terms of an AP is given as:
Sn = n/2 [2a + (n − 1)d]
S40 = 40/2 [2 x 10 + (40 − 1)9]
= 20|20 + 39 x 9]
=20[20 + 351]
=20 x 371 = 7420
Thus, sum of all 40 terms of AP = 7420
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