Arithmetic Progression NCERT Exercise 5.2 part six Class Ten Mathematics

Arithmetic Progression

Exercise 5.2 Part 6

Question 16: Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.

Solution: Solution: Given a3 = 16 and a7 – a5 = 12

`a_3 = a + 2d = 16`

`a_5 = a + 4d`

`a_7 = a + 6d`

As per question;

`a + 6d – a – 4d = 12`

Or, `2d = 12`

Or, `d = 6`


Substituting the value of d in third term, we get;

`a + 2 xx 12 = 16`

Or, `a + 24 = 16`

Or, `a = 16 – 24 = - 8`

Thus, the AP can be given as follows:

-8, 4, 16, 28, 40, ………..

Question 17: Find the 20th term from the last term of the AP; 3, 8, 13, …………….253.

Solution: a = 3, d = 5

Now, `253 = a + (n + 1) d`

`⇒ 253 = 3 + (n -1) xx 5`

`⇒ 253 = 3 + 5n – 5 = – 2`

`⇒ 5n = 253 + 2 = 255`

`⇒ n = (255)/(5) = 51`

Therefore, 20th term from the last term `= 51 – 19 = 32`

`a_(32) = a + 31d`

`= 3 + 31 xx 5`

`= 3 + 155 = 158`

Thus, required term is 158


Question 18: The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Solution: Given, a8 + a4 = 24 and a10 + a6 = 44

`a_8 = a + 7d`

`a_4 = a + 3d`

As per question;

`a + 7d + a + 3d = 24`

Or, `2a + 10d = 24`

Or, `a + 5d = 12` …………(1)

`a_(10) = a + 9d`

`a + 9d + a + 5d = 44`

Or, `2a + 14d = 44`

Or, `a + 7d = 22` ………….(2)

Subtracting equation (1) from equation (2);

`a + 7d – a – 5d = 22 – 12`

Or, `2d = 10`

Or, `d = 5`


Substituting the value of d in equation (1), we get;

`a + 5 xx 5 = 12`

Or, `a + 25 = 12`

Or, `a = - 13`

Hence, first three terms of AP: – 13, – 8, – 3

Question 19: Subha Rao started work in 1995 at an annual salary of Rs. 5000 and received an increment of Rs. 200 each year. In which year did his income reach Rs. 7000?

Solution: Here, a = 5000, d = 200 and an = 7000

We know, `a_n = a + (n – 1)d`

Or, `7000 = 5000 + (n – 1)200`

Or, `(n -1)200 = 7000 – 5000`

Or, `(n – 1)200 = 2000`

Or, `n – 1 = 10`

Or, `n = 11`

Thus, `1995 + 10 = 2005`

Hence, his salary reached at Rs. 7000 in 2005.

Question: 20 – Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs. 20.75, find n.

Solution: Here, a = 5, d = 1.75 and an = 20.75

We know, `a_n = a + (n -1)d`

Or, `20.75 = 5 + (n – 1)1.75`

Or, `(n – 1)1.75 = 20.75 – 5`

Or, `(n – 1)1.75 = 15.75`

Or, `n – 1 = 15.75/1.75 = 9`

Or, `n = 10`



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