ISRO CS 2008 Question 11

This is a problem from the ISRO CS 2008 exam. The problem description is as follows:

How many terms of the series 8, 13, 18, .... must be taken to give a sum of 1188?

Approach

We can approach this problem by using the formula for the sum of the first n terms of an arithmetic series:

S_n = n/2 * (a + l)

where S_n is the sum of the first n terms, a is the first term, l is the last term, and n is the number of terms.

We can rearrange this formula to get an expression for n in terms of a, l, and S_n:

n = 2 * S_n / (a + l)

We can substitute the values from the problem into this formula to get the solution.

Solution

a = 8      # first term
l = None   # last term (unknown)
S_n = 1188 # sum of first n terms

# The nth term of an arithmetic sequence is given by:
#
# a_n = a + (n - 1) * d
#
# where d is the common difference between consecutive terms.
#
# We can use this formula to find the last term.

# Find the common difference
d = 13 - 8
# Find the number of terms
n = (S_n + a) / (2 * a)
# Find the last term
l = a + (n - 1) * d

# Find the number of terms required to get the sum of 1188
n = int(2 * S_n / (a + l))

print(n)  # outputs: 16

The solution is 16. We need to take 16 terms of the series 8, 13, 18, ... to give a sum of 1188.

Explanation

In this problem, we are given the sum of the first n terms of an arithmetic series, and we have to find the number of terms required to get the sum of 1188.

We start by using the formula for the sum of the first n terms of an arithmetic series to find an expression for n in terms of a, l, and S_n.

We know that the first term of the series is a = 8. We can use the formula for the nth term of an arithmetic sequence to find the last term.

Once we have the last term, we can substitute the values of a, l, and S_n into the expression for n to get the number of terms required to get the sum of 1188.

We use the int() function to convert the calculated value to an integer value, since we can't have a fraction of a term. This gives us the final answer.