# Show that a1, a2, ........., an form an AP where an is defined as below :(i) an = 3 + 4n (ii) an = 9 - 5nAlso find the sum of the first 15 terms in each case.

(i)Let us take different values for a, i.e. 1, 2, 3 and so on

a = 3 + 4 = 7

a2 = 3 + 4 x 2 = 11

a3 = 3 + 4 x 3 = 15

a4 = 3 + 4 x 4 = 19

Here; each subsequent member of the series is increasing by 4 and hence it is an AP.

The sum of "n" terms is given as: where, n = no. of terms

a= first term

d = common difference

∴ The sum of first 15 terms is given as:

= (15/2) [2 × 7 + (15-1)×4]

= (15/2) [14 + (14)×4]

= (15/2) 

= 525

(ii) Let us take values for a, i.e. 1, 2, 3 and so on

a = 9 – 5 = 4

a2 = 9 – 5 x 2 = - 1

a3 = 9 – 5 x 3 = - 6

a4 = 9 – 5 x 4 = - 11

Here; each subsequent member of the series is decreasing by 5 and hence it is an AP.

∴ The sum of first 15 terms is given as:

= (15/2) [2 × 4 + (15-1)×-5]

= (15/2) [8 + (14)×-5]

= (15/2) [8-70]

= 15(-31)
= -465

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