NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 10(ii)

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 10(ii)

Question:

Show that a1, a2, . . ., an, . . . form an AP where an is defined as below : an = 9 – 5n. Also find the sum of the first 15 terms in each case.

Given:

an = 9 – 5n

To Find:

Whether the sequence forms an AP and the sum of the first 15 terms.

Formula:

The common difference of an AP is given by d = an+1 - an. 

The sum of first n terms of an AP is given by Sn = n2(2a + (n-1)d)

Solution:

an = 9 – 5n

an+1 = 9 – 5(n+1) 

⇒ an+1 = 9 – 5n – 5 

⇒ an+1 = 4 – 5n

d = an+1 – an 

⇒ d = (4 – 5n) – (9 – 5n) 

⇒ d = 4 – 5n – 9 + 5n 

⇒ d = -5

Since the common difference is constant (-5), the sequence forms an AP.

a1 = 9 – 5(1) = 4

n = 15

d = -5

⇒ S15 = 152 (2(4) + (15-1)(-5)) 

⇒ S15 = 152 (8 + 14(-5)) 

⇒ S15 = 152 (8 -70) 

⇒ S15 = 152 (-62) 

⇒ S15 = 15 (-31) 

⇒ S15 = -465

Result:

The sequence forms an arithmetic progression with a common difference of -5. 

The sum of the first 15 terms is -465.

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