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Arithmetic

Progressions

  • Sums based on Sn formula

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10) Show that a1, a2,…an,…form an AP where an = 3 + 4n. Also find the sum of the first 15 terms.

Sol:

an = 3 + 4n

a1 =

3 + 4

(1)

a1 =

3 + 4

a1 =

7

a2 =

3 + 4

(2)

a2 =

3 + 8

a2 =

11

a3 =

3 + 4

(3)

a3 =

3 + 12

a3 =

15

d = a2 – a1

= 11 – 7

= 4

d = a3 – a2

= 15 – 11

= 4

  • As ‘d’ is constant, the given list of numbers is an AP.

Sn =

∴ S15 =

[2(7)

4]

+ (15 – 1)

=

[14

+ 14 × 4]

=

[14

+ 56]

=

× 70

=

15 × 35

∴ S15 =

525

Lets find the difference between the consecutive terms

To find S15

For S15 substitute

n = 15, a = 7 & d = 4

∴Sum of first 15 terms is 525

Lets find a1, a2 & a3

To find a1 put n = 1

To find a2 put n = 2

To find a3 put n = 3

35

Exercise 5.3 10(i)