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.

Solution.

n³ – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = 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?

Sol.

Given, first term, a = 10

Last term, a_l = 361

And, common difference, d = 9

a_l = a + (n −1)d

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:

S_n = 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