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



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