# Arithmetic Progression

## Exercise 5.3 Part 6

Question 10: Show that a1, a2, …., an, …. Form an AP where an is defined as below:

(i) an = 3 + 4n

Solution: Let us take different values for a, i.e. 1, 2, 3 and so on

a = 3 + 4 = 7

a_2 = 3 + 4 xx 2 = 11

a_3 = 3 + 4 xx 3 = 15

a_4 = 3 + 4 xx 4 = 19

Here; each subsequent member of the series is increasing by 4 and hence it is an AP.

(ii) an = 9 – 5n

Solution: Let us take different values for a, i.e. 1, 2, 3 and so on

a = 9 – 5 = 4

a_2 = 9 – 5 xx 2 = - 1

a_3 = 9 – 5 xx 3 = - 6

a_4 = 9 – 5 xx 4 = - 11

Here; each subsequent member of the series is decreasing by 5 and hence it is an AP.

Question 11: If the sum of the first n terms of an AP is 4n – n2, what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.

Solution: We can find the first term as follows:

S_1 = 4 xx 1 - 1^2 = 3

Now; sum of first two terms can be calculated as follows:

S_2 = 4 xx 2 - 2^2

= 8 - 4 = 4

Hence; second term = 4 – 3 = 1

And d = 1 – 3 = - 2

So, 3rd term can be calculated as follows:

a_3 = a + (n – 1)d

= 3 + 2( - 2) = 3 – 4 = - 1

Similarly, 10th term can be calculated as follows:

a_10 = a + 9d

= 3 + 9(- 2) = 3 – 18 = - 15

Question 12: Find the sum of the first 40 positive integers divisible by 6.

Solution: The smallest positive integer which is divisible by 6 is 6 itself and its 40th multiple will be 6 x 40 = 240

So, we have a = 6, d = 6, n = 40 and 40th term = 240.

Sum of first 40 positive integers divisible by 6 can be calculated as follows:

S=n/2[2a+(n-1)d]

=(40)/(2)(2xx6+39xx6)

=20(12+234)

=20xx246=4920