CBSE Class 10 Answered

Numbers between 101 and 304, which are divisible by 3 are 102,105,……303.

Clearly, it is an AP with first term = a = 102 and common difference = d = 3

Let there be n terms in this AP

Then, n^th term = 303

a + (n-1)d=303

102+ (n-1)3=303

n-1=67

n=68 … (1)

Numbers between 101 and 304, which are divisible by 5 are 105,110,……300.

Clearly, it is an AP with first term = a = 105 and common difference = d = 5

Let there be n terms in this AP

Then, n^th term = 300

a + (n-1)d=300

105+ (n-1)5=300

n-1=39

n = 40 … (2)

Numbers between 101 and 304, which are divisible by 15 are 105,120,……300.

Clearly, it is an AP with first term = a = 105 and common difference = d = 15

Let there be n terms in this AP

Then, n^th term = 300

a + (n-1)d=300

105+ (n-1)15=300

n-1=13

n = 14 … (3)

From (1), (2) and (3), we find that, there are (68+40-14) = 94 terms between 101 and 304 which are divisible by 3 or 5.

Their sum = S₆₈+S₄₀-S₁₄

= 34[102x2+67x3] +20[105x2+39x5] - 7[105x2+13x15]

=34[204+201]+20[210+195]-7[210+195]

= 13770+8100-2835

= 19035

OR

Here, AP (in rupees) is 6500, 6390, 6280, ……

The show will cease to be profitable on the night when the receipts are just Rs 1000. Let it happen on the nth night. It means a_n1000.

Here, a=6500, d=-110

Now, a_n=a+(n-1)d =6500-110(n-1)

Equating it with 1000, we get,

6500-110(n-1)=1000

110(n-1)=5500

n-1=50

n=51

Hence, on 51st night the show will cease to be profitable.