CBSE Class 10 Mathematics Exam 2019: Important 4 Marks Questions with Solutions

Q. For any positive whole number n, demonstrate that n3 – n is distinct by 6. 

Arrangement. 

n3 – n = n(n2 – 1) = n(n+1)(n – 1) = (n – 1)n(n+1) = result of three successive positive whole numbers. 

Presently, we need to demonstrate that the result of three back to back positive whole numbers is separable by 6. 

We realize that any positive whole number n is of the structure 3q, 3q + 1 or 3q + 2 for some

positive number q. 

Presently three back to back positive whole numbers are n, n + 1, n + 2. 

Case I. In the event that n = 3q. 

n(n + 1) (n + 2) = 3q(3q + 1) (3q + 2) 

In any case, we realize that the result of two back to back numbers is a significantly whole number. 

∴ (3q + 1) (3q + 2) is a significantly whole number, say 2r. 

⟹ n(n + 1) (n + 2) = 3q × 2r = 6qr, which is distinct by 6. 

Case II. On the off chance that n = 3n + 1. 

∴ n(n + 1) (n + 2) = (3q + 1) (3q + 2) (3q + 3) 

= (significantly number say 2r) (3) (q + 1) 

= 6r (q + 1), 

which is distinct by 6. 

Case III. On the off chance that n = 3q + 2. 

∴ n(n + 1) (n + 2) = (3q + 2) (3q + 3) (3q + 4) 

= different of 6 for each q 

= 6r (state), 

which is distinct by 6. 

Thus, the result of three back to back numbers is detachable by 6. 

Q. The first and the last terms of an AP are 10 and 361 individually. On the off chance that its normal contrast is 9, at that point locate the quantity of terms and their all out aggregate? 

Sol. 

Given, first term, a = 10 

Last term, al = 361 

What's more, typical distinction, d = 9 

al = a + (n −1)d 

361 = 10 + (n − 1)9 

361 = 10 + 9n − 9 

361 = 9n + 1 

9n = 360 

n = 40 

Along these lines, all out number of terms in AP = 40 

Presently, aggregate of all out 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 

In this way, whole of each of the 40 terms of AP = 7420