Question

Question 10 Show that a1, a2 … , an , … form an AP where an is defined as below (i) an = 3 + 4n (ii) an = 9 − 5n Also find the sum of the first 15 terms in each case.

Class 10 - Math - Arithmetic Progressions Page 113

Answer 1

(i) an = 3 + 4n
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
It can be observed that
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4
i.e., ak + 1 − ak is same every time. Therefore, this is an AP with common difference as 4 and first term as 7.
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(7) + (15 - 1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35
= 525

(ii) an = 9 − 5n
a1 = 9 − 5 × 1 = 9 − 5 = 4
a2 = 9 − 5 × 2 = 9 − 10 = −1
a3 = 9 − 5 × 3 = 9 − 15 = −6
a4 = 9 − 5 × 4 = 9 − 20 = −11
It can be observed that
a2 − a1 = − 1 − 4 = −5
a3 − a2 = − 6 − (−1) = −5
a4 − a3 = − 11 − (−6) = −5
i.e., ak + 1 − ak is same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.
Sn = n/2 [2a + (n - 1)d]
S15 = 15/2 [2(4) + (15 - 1) (-5)]
= 15/2 [8 + 14(-5)]
= 15/2 (8 - 70)
= 15/2 (-62)
= 15(-31)
= -465

Answer 2

Sum of n terms of an AP

The sum of first n terms of an AP with first term 'a' and common difference 'd' is given by

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

Sn=n /2 [ a + l ]       (l = last term)
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Solution:

(i)  an = 3 + 4n

on putting  n=1,2,3, 4.......
a1 = 3 + 4(1) = 7
a2 = 3 + 4(2) = 3 + 8 = 11
a3 = 3 + 4(3) = 3 + 12 = 15
a4 = 3 + 4(4) = 3 + 16 = 19
 
we get the list of numbers 7, 11, 15, 19........

find d common difference of these numbers
a2 − a1 = 11 − 7 = 4
a3 − a2 = 15 − 11 = 4
a4 − a3 = 19 − 15 = 4

we observe that d is same every time. Therefore, this is an AP with common difference as 4 and first term as 7. (d= 4 , a= 7)

S = n/2 [2a + (n - 1)d]
here n= 15 (given)

S15 = 15/2 [2(7) + (15 - 1) × 4]
= 15/2 [(14) + 56]
= 15/2 (70)
= 15 × 35

S15= 525



(ii)  an = 9 − 5n

on putting  n=1,2,3, 4.......
a1 = 9 − 5 × 1 = 9 − 5 = 4
a2 = 9 − 5 × 2 = 9 − 10 = −1
a3 = 9 − 5 × 3 = 9 − 15 = −6
a4 = 9 − 5 × 4 = 9 − 20 = −11

we get the list of numbers 4, -1, -6, -11........

find d common difference of these numbers

a2 − a1 = − 1 − 4 = −5
a3 − a2 = − 6 − (−1) = −5
a4 − a3 = − 11 − (−6) = −5

we observe that d is same every time.. Therefore, this is an A.P. with common difference as −5 and first term as 4. (d= -5, a= 4) 

Sn = n/2 [2a + (n - 1)d]

Here n=15    (given)

S15 = 15/2 [2(4) + (15 - 1) (-5)]
= 15/2 [8 + 14(-5)]
= 15/2 (8 - 70)
= 15/2 (-62)
= 15(-31)

S15= -465 
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Hope this will help you....