Arithmetic Progression NCERT Exercise 5.3 part five Class Ten Mathematics

# Arithmetic Progression

## Exercise 5.3 Part 5

Question 7: Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.

Solution: We have; n = 22, d = 7 and a22 = 149

We know;

a_n = a + (n – 1)d

Or, 149 = a + 21 xx 7

Or, a = 149 – 147 = 2

The sum can be calculated as follows:

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

=(22)/(2)(2+149)

=11xx151=1661

Question 8: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution: We have; a2 = 14, a3 = 18 and n = 51

Here;

a_3 – a_2 = 18 – 14 = 4

Hence, d = 4

Or, a_2 – a = 4

Or, 14 – a = 4

Or, a = 10

Now, sum can be calculated as follows:

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

=(51)/(2)(2xx10+50xx4)

=51(10+50xx2)

=51xx110=5610

Thus, sum = 5610

Question 9: If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Solution: Here;

S_7=7/2(2a+6d)=49

Or, 7(a+3d)=49

Or, a+3d=49÷7=7

This is the 4th term

Similarly;

S_(17)=(17)/(2)(2a+16d)=289

Or, 17(a+8d)=289

Or, a+8d=289÷17=17

This is the 9th term

Subtracting 4th term from 9th term, we get;

a + 8d – a – 3d = 17 – 7

Or, 5d = 10

Or, d = 2

Using the value of d in equation for fourth term, we can find a as follows:

a + 3d =7

Or, a + 6 = 7

Or, a = 1

Using the values of a and d; we can find the sum of first n terms as follows:

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

=n/2[2xx1+(n-1)2]

=n(1+n-1)=n^2

Thus, sum of n terms of this AP = n2