Arithmetic Progression NCERT Exercise 5.3 part two Class Ten Mathematics

# Arithmetic Progression

## Exercise 5.3 Part 2

Question: 2 – Find the sums given below:

(i) 7 + 10.5 + 14 + ….. + 84

Solution: Here, a = 7, d = 3.5 and last term = 84

Number of terms can be calculated as follows;

a_n =a + (n-1)d

Or, 84 = 7 + (n-1)3.5

Or, (n-1)3.5 = 84-7

Or, n - 1 = 77÷3.5 = 22

Or, n = 23

Sum of n terms can be given as follows:

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

=(23)/(2)[2xx7+22xx3.5]

=(23)/(2)(14+77)

=(23)/(2)xx91=(2093)/(2)

=1046(1)/(2)

(ii) 34 + 32 + 30 + …. + 10

Solution: Here, a = 34, d = - 2 and last term = 10

Number of terms can be calculated as follows:

a_n = a + (n – 1)d

Or, 10 = 34 + (n – 1)(- 2)

Or, 10 = 34 – (n – 1)(2)

Or, (n – 1)2 = 34 – 10 = 24

Or, n – 1 = 12

Or, n = 13

Sum of n terms can be given as follows:

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

=(13)/(2)[2xx34-12(-2)]

=(13)/(2)(68-24)

=(13)/(2)xx44=286

Thus, sum of given AP = 286

(iii) – 5 + (-8) + (- 11) + …… + (- 230)

Solution: Here, a = - 5, d = - 3 and last term = - 230

Number of terms can be calculated as follows:

a_n = a + (n – 1)d

Or, - 230 = - 5 + (n – 1)( - 3)

Or, - 230 = - 5 – (n – 1)3

Or, (n – 1)3 = - 5 + 230 = 225

Or, n – 1 = 75

Or, n = 76

Sum of n terms can be given as follows:

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

=(76)/(2)[2(-5)+75(-3)]

=38(-10-225)

=38(-235)=-8930

Thus, sum of given AP = - 8930